Math 11 Advanced Student Workbook: Pre-Calculus

Table of contents :
MATHEMATICS 2200 STUDENT WORKBOOK unit 1
MATHEMATICS 2200 STUDENT WORKBOOK unit 2
MATHEMATICS 2200 STUDENT WORKBOOK unit 3
MATHEMATICS 2200 STUDENT WORKBOOK unit 4
MATHEMATICS 2200 STUDENT WORKBOOK unit 5
MATHEMATICS 2200 STUDENT WORKBOOK unit 6
MATHEMATICS 2200 STUDENT WORKBOOK unit 7
MATHEMATICS 2200 STUDENT WORKBOOK unit 8
MATHEMATICS 2200 STUDENT WORKBOOK unit 9
MATHEMATICS 2200 STUDENT WORKBOOK unit 99 solutions

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Teacher: ___________________

Student: ___________________

Math 11 Advanced / Pre-Calculus

Student Workbook

Sharpe Mathematics  2017

FOREWORD This Student Workbook for Mathematics 2200 was written as a student-centered resource for the course. In doing so, this workbook can be used by students to develop and practice their skills in an easily followed manner. This workbook can be used as a portfolio of student work, containing many practice questions and review material, and will enable students to keep their work organized as opposed to having countless worksheets to supplement the notes and extra practice used to ensure mastery of skills. It will hopefully allow students to gain a better grasp of the material by seeing complete examples worked out in full, as well as being able to practice the concepts with additional exercises. Where suitable, there are also hints, suggestions, and thought questions to keep students on the right track and to hopefully avoid common pitfalls along the way. The use of graphing calculators enables students to realize the importance and applicability of technology in mathematics. Where necessary, the appropriate keystrokes have been included to aid with this objective. In addition, students considering taking Advanced Mathematics 3200 are encouraged to try the more challenging problems for further enrichment and extension of the topic. I would like to extend my sincere gratitude to my wife, Debbie, and my children, Joshua and Eric, for their never-ending support and constant encouragement to complete this massive undertaking. I would also like to thank my students for their feedback and their suggestions for improvement. Thank you! I hope you find this to be a useful resource for both students and teachers alike. I welcome your feedback. Todd Sharpe (B.Sc., B.Ed., M.Ed.) e-mail: [email protected] webpage: https://www.facebook.com/sharpemathematics

This book is copyright. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior written permission of the author. Sharpe Mathematics  2017

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Mathematics 2200

TABLE OF CONTENTS CHAPTER 1: SEQUENCES & SERIES 1.0 1.1 1.2 1.3 1.4 1.5

SEQUENCES ..................................................................................... 2 ARITHMETIC SEQUENCES ................................................................. 3 ARITHMETIC SERIES ........................................................................ 13 GEOMETRIC SEQUENCES .................................................................. 22 GEOMETRIC SERIES ......................................................................... 36 INFINITE GEOMETRIC SERIES ........................................................... 46

CHAPTER 2: TRIGONOMETRY 2.1 2.2

2.3 2.4 2.5

A. ANGLES IN STANDARD POSITION ................................................. B. SPECIAL ANGLES ........................................................................ TRIGONOMETRIC RATIOS OF ANY ANGLE ........................................... A. DETERMINING THE ANGLE θ FROM THE POINT (C0Sθ , SINθ) ....... B. EVALUATING TRIGONOMETRIC EXPRESSIONS ............................... C. THE PRIMARY TRIGONOMETRIC RATIOS (REVIEW) ........................ NON RIGHT-ANGLED TRIANGLE TRIGONOMETRY ................................ A. THE SINE LAW ........................................................................... B. THE AMBIGUOUS CASE OF THE SINE LAW ................................... THE COSINE LAW ............................................................................ APPLICATIONS OF THE SINE AND COSINE LAW ..................................

54 58 63 63 69 74 77 77 79 82 86

CHAPTER 3: QUADRATIC FUNCTIONS 3.0 3.1

3.2 3.3

CHARACTERISTICS OF A QUADRATIC FUNCTION ................................ 89 VERTEX FORM OF A QUADRATIC FUNCTION ....................................... 91 A. HORIZONTAL TRANSLATION y = (x – p)2 ..................................... 95 B. VERTICAL TRANSLATION y = x2+q ............................................... 98 C. VERTICAL STRETCH y = ax2 ........................................................ 101 D. VERTICAL REFLECTION y= –x2 .................................................... 105 E. OVERVIEW ................................................................................... 109 F. VERTEX FORM OF A QUAD FUNCTION FROM ITS GRAPH ................. 114 STANDARD FORM OF A QUADRATIC FUNCTION .................................. 119 A. DETERMINING THE NUMBER OF X-INTERCEPTS ............................ 125 COMPLETING THE SQUARE ................................................................ 126 A. USING COMPLETING THE SQUARE TO CHANGE FORMS ................. 128 B. MAXIMUM/MINIMUM WORD PROBLEMS ...................................... 132

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CHAPTER 4: QUADRATIC EQUATIONS 4.1 4.2 4.3 4.4

SOLVING QUADRATIC EQUATIONS BY GRAPHING .......................................... 140 SOLVING QUADRATIC EQUATIONS BY FACTORING ......................................... 148 A. FACTORING REVIEW .............................................................................. 148 B. SOLVING QUADRATIC EQUATIONS BY FACTORING ................................... 154 SOLVING QUAD EQUATIONS BY COMPLETING THE SQUARE ........................... 156 A. THE QUADRATIC FORMULA ................................................................... 159 B. THE NATURE OF ROOTS OF A QUADRATIC EQUATION ............................. 164 C. WORD PROBLEMS USING QUADRATIC EQUATIONS .................................. 169 D. THE SUM AND PRODUCT OF ROOTS *ENRICHMENT* ............................. 174

CHAPTER 5: RADICAL EXPRESSIONS & EQUATIONS 5.1

5.2 5.3

WORKING WITH RADICALS .......................................................................... 177 A. ORDERING RADICALS ............................................................................ 182 B. RESTRICTIONS ON VARIABLES ............................................................... 183 C. ADDING AND SUBTRACTING RADICALS .................................................... 184 A. MULTIPLYING RADICAL EXPRESSIONS .................................................... 186 B. DIVIDING RADICAL EXPRESSIONS ........................................................... 189 C. DIVIDING RADICAL EXPRESSIONS USING CONJUGATES ........................... 192 SOLVING RADICAL EQUATIONS .................................................................... 195

CHAPTER 6: RATIONAL EXPRESSIONS & EQUATIONS 6.1 6.2 6.3 6.4 6.5

RATIONAL EXPRESSIONS ........................................................................... 206 MULTIPLYING RATIONAL EXPRESSIONS ...................................................... 210 DIVIDING RATIONAL EXPRESSIONS .............................................................. 212 A. ADDING & SUBTRACTING RATIONAL EXPRESSIONS.................................. 214 B. COMPLEX FRACTIONS ........................................................................... 217 A. SOLVING RATIONAL EQUATIONS ............................................................ 220 B. APPLICATIONS OF RATIONAL EQUATIONS .............................................. 223

CHAPTER 7: ABSOLUTE VALUE & RECIP FUNCTIONS 7.1 7.2 7.3 7.4

ABSOLUTE VALUE .....................................................................................228 ABSOLUTE VALUE FUNCTIONS ................................................................... 230 ABSOLUTE VALUE EQUATIONS .................................................................. 244 RECIPROCAL FUNCTIONS ........................................................................... 253 A. RECIPROCALS OF LINEAR FUNCTIONS ................................................... 254 B. RECIPROCALS OF QUADRATIC FUNCTIONS .............................................258

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CHAPTER 8: SYSTEMS OF EQUATIONS 8.0

8.1 8.2

REVIEW OF LINEAR-LINEAR SYSTEMS OF EQUATIONS ....................... 275 A. SOLVING A LINEAR-LINEAR SYSTEM GRAPHICALLY ....................... 276 B. USING SUBSTITUTION TO SOLVE A LINEAR-LINEAR SYSTEM ........ 279 C. USING ELIMINATION TO SOLVE A LINEAR-LINEAR SYSTEM ........... 282 SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY ............................. 285 A. USING TECHNOLOGY TO SOLVE A SYSTEM GRAPHICALLY ............ 292 SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY .......................... 298

CHAPTER 9: LINEAR & QUADRATIC INEQUALITIES 9.1 9.2

9.3

LINEAR INEQUALITIES IN TWO VARIABLES ........................................ 314 A. USING A GDC TO GRAPH A LINEAR INEQUALITY ........................... 318 QUADRATIC INEQUALITIES IN ONE VARIABLE .................................... 324 A. SOLVING GRAPHICALLY .............................................................. 324 B. SOLVING ALGEBRAICALLY USING A SIGN DIAGRAM ...................... 327 QUADRATIC INEQUALITIES IN TWO VARIABLES .................................. 335

SOLUTIONS TO PRACTICE EXERCISES ................... 343

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Chapter 1: Sequences & Series

CHAPTER

1

1

SEQUENCES & SERIES Contents..............................................................

1.0

SEQUENCES ..................................................................................... 2

1.1

ARITHMETIC SEQUENCES .................................................................

1.2

ARITHMETIC SERIES ........................................................................ 13

1.3

GEOMETRIC SEQUENCES .................................................................. 22

3

1.4 GEOMETRIC SERIES ......................................................................... 36 1.5

INFINITE GEOMETRIC SERIES ........................................................... 46

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Chapter 1: Sequences & Series

1.0 SEQUENCES Much of this chapter will involve the recognition of patterns through the use of sequences. A SEQUENCE is an ordered arrangement of numbers, pictures or symbols whereby each item or term is determined through a formula or rule. A sequence is written as a list of numbers in the form { t 1 ,t 2 ,t 3 ,t 4 ,...} , where t 1 is the first term in the sequence, t 2 is the second term, and so on. Note that the terms begin at “1” and not at “0”. We will examine two types of sequences throughout the unit: finite and infinite. A FINITE sequence is one which eventually terminates. The number of terms can be counted and is of the form { t 1 ,t 2 ,t 3 ,...,t n } , where t n is the last or nth term in the sequence. For example:

(a)

{ 1,

3, 5, 7, 9}

(b)

{ 2,

t1 t 2 t 3 t 4 t 5

4, 8, 16, 32, 64}

t1 t 2 t 3 t 4

t5

t6

An INFINITE sequence is one which does not terminate. The number of terms cannot be counted and is of the form { t 1 ,t 2 ,t 3 ,t 4 ,...} , where the ellipsis (“...”) indicates an infinitely continuing sequence. For example:

(c)

{ 5,

10, 15, 20, ...}

t1 t 2

t3

(d)

{ 1,

t4

2, 5, 14, 41, 122, ...}

t1 t 2 t 3 t 4

t5

t6

Can you identify/describe the rule or formula for each sequence (use words if you wish)? a. ______________________________

b. ______________________________

c. ______________________________

d. ______________________________

Just as with linear and quadratic relations, we can identify the domain and range of a sequence. In any sequence, the DOMAIN refers to all the possible values of n (the independent variable), whereas the RANGE refers to the possible values for t n (the dependent variable). For example, in the sequence

{ 1,

4, 9, 16, 25, ...} , think of it instead as a table of values: n

1

2

3

4

5

tn

1

4

9

16

25

The DOMAIN (n) is the values 1, 2, 3, 4, 5, ... or the positive integers. ∴ D= {n | n ∈ N } The RANGE ( t n ) is the values 1, 4, 9, 16, 25, ... or the values of n 2 . ∴ R= {t n= |t n n 2 , n ∈ N }

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1.1 ARITHMETIC SEQUENCES A sequence can be represented in various forms, such as pictorial, algebraic, graphical or as a table of values. For example, consider the sequence { 3, 5, 7, 9, 11, ...} , where = t 1 3= , t 2 5= , t 3 7= , t 4 9= , t 5 11 and so on. i. pictorial

ii. algebraic

t= 2n + 1 n

iii. table of values

iv. graphical tn

n

tn

1

3

2

5

3

7

4

9

5

11

10 8 6 4 2

2

4

6

n

In each of the forms outlines above, the same sequence is represented in different ways. Sometimes one form may be preferable over another, depending on the problem at hand. In many instances, however, the algebraic form is most beneficial in that we can identify the value for any term. Let us revisit the first sequence { 3, 5, 7, 9, 11, ...} . What do you notice about the change in value of each term? Evidently, each consecutive term is increasing by two. This change in value is referred to as a common difference ( or “d”), since the difference between a term and the preceding term is the same in each case. In general, the common difference for any two terms may be found by calculating d= t n − t n − 1 or= d t n + 1 − t n . This is more easily seen as follows:

t1 t2 t3 t4 t5

{ 3,

2

5, 7, 9, 11, ...} 2

2

2

common difference = 2

This sequence is an example of what is referred to as an ARITHMETIC SEQUENCE. In an arithmetic sequence, the terms increase/decrease by a constant value, d, and the data appears linear in shape when graphed. Sharpe Mathematics  2017

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Chapter 1: Sequences & Series

TRY THESE! 1.

a. How would you describe the shape of the graph? b. By examining the graph, what is the value of t 1 ? c. By examining the graph, what is the common difference, d? d. What does the common difference remind you of if the graph were that of a function with the dots connected? e. What is the formula for the sequence shown? f. How can you use your answer in (e) to determine the value of t 20 (ie. when n = 20)?

2.

a. How would you describe the shape of the graph? b. By examining the graph, what is the value of t 1 ? c. By examining the graph, what is the common difference, d? d. What does the common difference remind you of if the graph were that of a function with the dots connected? e. What is the formula for the sequence shown? f. How can you use your answer in (e) to determine the value of n if t n = −22 ?

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Chapter 1: Sequences & Series

5

You should have noticed that the common difference , d, of an arithmetic sequence is similar to the slope of a linear function. An arithmetic sequence with n terms can be defined by the formula

t n =t 1 + (n − 1) d where t1 is the first term in the sequence and d is the common difference.

Using Example 1 from the previous page we had: To determine the formula of the sequence we have:

EXAMPLE 1 A sequence is defined by the equation t= 2n + 1 . n a. Identify the value of t 6 and t 100 .

t1 t2 t3 t4 t5

{ 3,

5, 7, 9, 11, ...}

t n =t 1 + (n − 1) d =3 + (n − 1) 2 =3 + 2n − 2 = 2n + 1

b. Which term has a value of 129?

Solution a. To find the value of t 6 we let n=6:

t= n = t6 t= 6 t6 =

b. Let t n = 129 and solve for n:

To find the value of t 100 we let n=100:

t= 2n + 1 n 129 = 2n + 1 128 = 2n n = 64 ∴ t 64 = 129

2n + 1 t= n

2n + 1

= t 100 2(100) + 1

2(6) + 1 12 + 1

= t 100 200 + 1

13

t 100 = 201

EXAMPLE 2 Determine the formula for each arithmetic sequence. a.

{ 1,

5, 9, 13, 17, ...}

b.

{ 5,

− 2, − 9, − 16, − 23, ...}

c.

{ 3.4,

3.6, 3.8, 4.0, 4.2, ...}

Solution a. t n =t 1 + (n − 1) d =1 + (n − 1) 4 =+ 1 4n − 4 = 4n − 3 Sharpe Mathematics  2017

b. t n =t 1 + (n − 1) d =5 + (n − 1)( −7) =5 − 7n + 7 = −7n + 12 DO NOT COPY!

c. t n =t 1 + (n − 1) d = 3.4 + (n − 1)(0.2) =3.4 + 0.2n − 0.2 = 0.2n + 3.2 Mathematics 2200

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Chapter 1: Sequences & Series

PRACTICE 1.

Using t 1 = 3 , write the first five terms of an arithmetic sequence which has a common difference of: a. 5



b. 2



c. –3  d. –6  2.

Using d = 4 , write the first five terms of an arithmetic sequence where the first term is: a.

2 

b. –4  c.

0 

d. –17  3.

4.

Use each function given to generate the first five terms of each sequence: a. t n = 2n



b. t n = −3n + 2



c. = t n 0.2n + 5



d. t n = −4n − 0.2



A fence is to be constructed in sections using iron rods in the shape of regular octagons as illustrated. a. Create a sequence for t 1 to t 6 representing the number of iron rods required to construct the first six sections of the fence. b. Determine the formula or rule to find the nth term of the sequence representing the number of iron rods required to construct n sections of the fence.

c. How many iron rods are required to build a fence which has 80 sections? d. How many sections can be constructed if the contractor only ordered 435 iron rods?

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Chapter 1: Sequences & Series

5.

Use the sequence

{ − 3,

7

− 8, − 13, − 18, − 23, ...} to answer the following questions.

a. Verify that the sequence is arithmetic.

b. Determine a formula to find the nth term of the sequence.

c. Determine the value of the 10th term of the sequence (i.e. find t 10 ).

d. Which term has the value –213 (i.e. if t n = −213 find the value of n) ?

e. State the domain and range of the sequence.

6. Use the number of dots to write the first five terms of each arrangement below as a sequence. If the sequence is arithmetic, explain why and identify the rule to find the nth term of the sequence. a.

b.

c.

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Chapter 1: Sequences & Series

7. Determine whether each sequence is arithmetic. If the sequence is arithmetic, identify the rule to find the nth term of the sequence. a.

{ 1,

5, 9, 13, 17, ...}

b.

{ 1,

4, 9, 16, 25, ...}

c.

{ 8,

d.

{ 2,

e.

{ 5,

6, 8, 11, 15, ...}

f.

{ 0.5,

g.

{ 4.1,

3.9, 3.5, 2.9, ...}

h.

{ 1,

2, 4, 8, 16, ...}

i.

{ 61 ,

j.

{ 5,

10, 15, 20, 25, ...}

k.

{ 41 , 0,

l.

{ 1,

4.5, 7, 9.5, 12, ...}

m. { x − 2, x + 1, x + 4, ...}

−1 4

}

, −21 , −43 , ...

n. { x + 1, 2x − 3, 3x − 7, ...}

5, 2, − 1, − 4 ...} 1.5, 2.5, 3.5, 4.5, ...}

1 5 2, 6,

}

1 61 , 1 21 , ...

8, 27, 64, 125, ...}

o. { 2x + 1, 4 x + 1, 6x + 1, ...}

8. Identify the common difference between successive terms for the sequence generated by: a. t= 2n + 5 n

b. t n = −4n + 1

9.

c. = t n 0.3n + 2

− 43 n + 3 d. t n =

a. List the first four terms of the sequence plotted. 8

b. Identify the common difference of the sequence. 6

c. How can you conclude that the sequence is arithmetic?

4 2

d. Identify the formula for the sequence. 2

-2

4

6

e. If the points were joined, would the data plotted be representative of a sequence? Explain. f. State the domain and range of the sequence.

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Chapter 1: Sequences & Series

10.

9

Match each graph to the correct formula which generates each arithmetic sequence. b. t= 2n + 1 n

a. t= 2n − 1 n A.

tn

c. t n = −2n + 3

B.

tn

9

-1

8

1

2

3

4

5

6

n

9 8

1

-7

3

-7

2

-8

2

-8

-9

1

1

11.

2

3

6

7

n

b. t 20

b. t 20

{ 13,

-9 -10

2

3

4

5

6

7 n -11

d. t 50

{ − 3,

e. t 100

f. t 200

1, 5, 9, 13, 17...} , find each value. d. t 50

c. t 27

{ − 7,

e. t 100

f. t 200

− 2, 3, 8, 13, 18...} ,

a. find the value of the 54th term.

b. find the value of t 65 .

c. which term has a value of 213?

d. which term has a value of 578?

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10, 7, 4, 1, − 2...} , find each value.

c. t 27

Given the arithmetic sequence

5

-6

1

-11

For the arithmetic sequence a. t 10

13.

5

For the arithmetic sequence a. t 10

12.

4

4

-5

3

-10

3

-4

4

1

2

-2

-6

4

1

-3

5

-5

D.

-1

6

-4

5

7

7

-3

6

tn

10

-2

7

C.

tn

1

10

d. t n = −2n + 2

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7

n

10

Chapter 1: Sequences & Series

EXAMPLE 3 Given an arithmetic sequence with t 3 = 7 and t 15 = 55 . Determine the general term, t n , to define the sequence. Solution STEP 1: We must first determine the common difference, d. Since t 3 =7 we get:

Since t 15 =55 we get:

t n =t 1 + (n − 1) d t 3 = t 1 + (3 − 1) d = 7 7 t 1 + 2d =

t n =t 1 + (n − 1) d t 15 =t 1 + (15 − 1) d =55 55 t 1 + 14d =

t1 = −2d + 7 

t1 = −14d + 55 

Solving equations  =  we get: − 2d + 7 =−14d + 55 14d − 2d = 55 − 7 12d = 48 d = 4*

STEP 2: We can now determine the value of t 1 by choosing either  or  :

Using  we get: t1 = −2d + 7

t1 = −2(4) + 7 t 1 =−8 + 7 t 1 = −1 STEP 3: We can now determine the general term, t n :

t n = t 1 + (n − 1)d t n = −1 + (n − 1)(4) t n = −1 + 4n − 4 = t n 4n − 5 *NOTE: We could have determined the value of d by interpreting the terms as points and thus interpreting d as the slope! = d

t 15 − t 3

= 15 − 3

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55 − 7 48 = = 4 12 12

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Chapter 1: Sequences & Series

11

TRY THESE! 1.

Given an arithmetic sequence with t 5 = 18 and t 21 = 66 . Determine the general term, t n , to define the sequence.

2.

Given an arithmetic sequence with t 7 = 12 and t 9 = 17 . Determine the general term, t n , to define the sequence.

In an arithmetic sequence, a missing term located between two known terms, a and b, is known as the arithmetic mean and is equal to

3.

Determine how many terms are in the arithmetic sequence

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{ 5,

a +b . 2

8, 11, ... , 188} .

Mathematics 2200

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Chapter 1: Sequences & Series

PRACTICE 14.

The 7th term of an arithmetic sequence is 37. If the common difference is -4, find the first three terms of the sequence.

15.

Given an arithmetic sequence with t 12 = 95 and t 21 = 167 . Determine the general term, t n , to define the sequence.

16.

Given an arithmetic sequence with t 9 = −13 and t 35 = −65 . Determine the general term, t n , to define the sequence.

17.

The 12th term of an arithmetic sequence is 49. If the first term is 5, determine the value of the 13th, 14th, and 15th terms.

18.

Determine how many terms are in the arithmetic sequence

{ − 5,

− 1, 3, ... , 191} .

19.

Determine how many terms are in the arithmetic sequence

{ − 5,

− 2, 1, 4, ... , 217} .

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Chapter 1: Sequences & Series

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1.2 ARITHMETIC SERIES A SERIES is the sum of the terms of a sequence. An ARITHMETIC SERIES is the sum of the terms of an arithmetic sequence. For example, if given the arithmetic sequence { 3, 5, 7, 9, 11 , ...} , the resulting arithmetic series would be 3 + 5 + 7 + 9 + 11 + … . The term Sn is used to represent the sum of the first n terms. Using the previous arithmetic sequence we have: S1 = t 1 S 4 = t1 + t 2 + t 3 + t 4 S 3 = t1 + t 2 + t 3 S2= t1 + t2 S1 = 3 S2= 3 + 5 S3 = 3 + 5 + 7 S4 = 3 + 5 + 7 + 9 S2 = 8

S3 = 15

S4 = 24

These are referred to as PARTIAL SUMS. For instance, the 3rd partial sum would be 15. We can use partial sums to find the sum of a sequence containing n terms. We can do this by writing the series on one line, writing it again on another line in reverse order, and then adding the two lines vertically. Consider the first n terms of the arithmetic sequence, t n = t1 + d ( n − 1), where n ∈ N . Thus, t1 = t1 + d (1 − 1) = t1 t2 = t1 + d (2 − 1) = t1 + d t3 = t1 + d (3 − 1) = t1 + 2d  t n − 2= t n − 2d t n −= tn − d 1 tn = tn Since

Sn = t1 + t2 + t3 + ... + t n − 1 + t n we can now write Sn as:

Sn =t1 + ( t1 + d ) + ( t1 + 2d ) + ( t1 + 3d ) + ... + ( t n − 2d ) + ( t n − d ) + t n . Now if we write the second series in reverse and add it to the first we get:

( t1 + d ) + ( t1 + 2d ) t n + ( t n − d ) + ( t n − 2d ) + Sn= 2 Sn =( t1 + t n ) + ( t1 + t n ) + ( t1 + t n ) Sn=

t1

+

S n n ( t1 + t n ) 2= Sn =

+ + +

( t1 + 3d ) ( tn − 3d ) ( t1 + t n )

( tn − 2d ) + .... + ( t1 + 2d ) + .... + ( t1 + t n )

+ .... +

+ + +

( tn − d ) ( t1 + d ) ( t1 + t n )

+

tn

+

t1

+

( t1 + t n )

n times

n (t + t ) 2 1 n

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Chapter 1: Sequences & Series

When the last term of the series is not known, however, another formula for the sum of n terms of an arithmetic series can be formed by replacing t n = t1 + d ( n − 1) . Hence, n (t + t ) 2 1 n n Sn= t +  t + ( n − 1) d  2 1 1 n S= ( 2t1 + ( n − 1) d ) n 2

Sn =

(

)

THE SUM OF n TERMS OF AN ARITHMETIC SERIES Given an arithmetic series with common difference d, first term t1 , and nth term t n . The sum of the first n terms, Sn can be found using the formula: = Sn

n (t + t ) 2 1 n

or

S= n

n ( 2t + ( n − 1) d ) 2 1

EXAMPLE 4 Determine the sum of the first 10 terms of the arithmetic sequence { − 3, 1, 5, 9, 13, ...} . Solution We know t1 = −3 , d = 4 , and n = 10. Since we do not know the value of t10 we will use the second version of the formula. n S= ( 2t + ( n − 1) d ) n 2 1 10 S= ( 2(−3) + ( 10 − 1) 4 ) 10 2 = 5 ( −6 + ( 9 ) 4 )

= 5 ( −6 + 36 ) = 5 ( 30 ) = 150 The sum of the first 10 terms is 150.

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Chapter 1: Sequences & Series

15

EXAMPLE 5 An arithmetic series has t1 = 6.5 and d = −2 . Determine the value of S25 . Solution Since we are trying to find S25 , we know n = 25. Since we do not know the value of t25 we will use the second version of the formula. n S= ( 2t + ( n − 1) d ) n 2 1 25 S25 = ( 2(6.5) + (25 − 1)( −2) ) 2 25 = ( 13 + ( −48) ) 2 = 12.5 ( −35 )

= −437.5

EXAMPLE 6 An arithmetic series has S20 = 430 , t20 = 50 , d = 3 . Write the first three terms of the corresponding arithmetic sequence. Solution Since we are trying to find S20 , we know n = 20. Since we do know the value of t20 we can use the first version of the formula to find t1 . n = Sn (t + t ) 2 1 n 20 = S20 (t + t ) 2 1 20 430 10 ( t1 + 50 ) = 43= t1 + 50 = t1 50 − 43 t 1 = −7

Thus, the first term is –7 and the common difference is 3. So, the first three terms of the arithmetic sequence are { − 7, − 4 , − 1, ...} .

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Chapter 1: Sequences & Series

TRY THESE! 1. Determine the sum of the first 12 terms of the arithmetic sequence { 18, 15, 12, 9, 6, ...} .

2. An arithmetic series has d = −2 and S16 = 144 . Determine the value of t1 .

3. Write the first three terms of the arithmetic sequence with S17 = 106.25 and t17 = 8.25 .

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17

PRACTICE 20.

21.

Determine the sum of the first 20 terms of each arithmetic series. a) 8 + 13 + 18 + 23 + ...

b)

17 + 12 + 7 + 2 + ...

c) 10 + 8 + 6 + 4 + …

d) –21 – 15.5 – 10 – 4.5 – …

e) 2.6 + 3.8 + 5 + 6.2 + …

f) –18.7 – 18.3 – 17.9 – 17.5 – …

Determine the sum of each arithmetic series. a) –2 + 3 + 8 + ... + 163

b)

c) 32 + 36 + 40 + … + 108

d) –14 – 11 – 8 – … + 61

e) 6.2 + 7.4 + 8.6 + … + 21.8

f) –8 – 11.5 – 15 – … – 39.5

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12 + 9 + 6 + ... + (–42)

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Chapter 1: Sequences & Series

PRACTICE 22.

23.

For each arithmetic series, determine the indicated value. a) (–12) + (–7) + (–2) + 3 + ... determine S18

b)

9 + 7 + 5 + 3 + ... determine S20

c) 3.5 + 5 + 6.5 + 8 + … determine S14

d) –8.2 – 8.4 – 8.6 – … determine S25

For each arithmetic series, determine the indicated value. a) If S20 = b) If S15 322 −850 and t20 = −90 , = = .5 and t1 4 , determine t1 . determine d .

c) If t1 1= = .5 and t20 58.5 , determine S15 .

24.

d)

Sn = −126, t1 = −1 and t n = −20 , determine n.

Determine a rule for the sum of the first n natural numbers:

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1 + 2 + 3 + … + n.

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19

PRACTICE 25.

Christian’s parents bought him a motor scooter but he had to repay them the money in monthly instalments. He agreed to repay $15 at the end of the first month, $20 at the end of the second month, $25 at the end of the third month, and so on. If Christian was able to repay the loan in 10 months, how much did the scooter originally cost?

26.

The local movie theatre has 30 seats in the first row, 36 seats in the second row, 42 seats in the third row, and so on. If the theatre has 20 rows of seats, how many seats are in the theatre?

27.

Mr. Smith wishes to reward his daugther, Sarah, for doing her weekly chores. He agrees to give his daughter an allowance of $5.00 in the first week of January with an increase of 25 cents each week until the last week in December. a) What was the amount of Sarah’s allowance in the last week of the year?

b) What was the total amount of money she earned in allowances for the entire year?

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Chapter 1: Sequences & Series

EXAMPLE 7 Find the sum of the multiples of 3 from 20 to 100. Solution We need to determine the sum of the arithmetic sequence { 21, 24 , 27, ... , 99 } whereby t1 = 21 , t n = 99 and d = 3 . We need to first determine the number of terms in the sequence.

t n = t 1 + (n − 1)d 99 = 21 + (n − 1)(3) 99 − 21 = (n − 1)(3) (n − 1)(3) = 78 (n − 1) = 26 n = 27 Thus we need to determine the value of S27 . n (t + t ) 2 1 n 27 = S27 (t + t ) 2 1 27 27 = S27 ( 21 + 99 ) 2 27 S27 = ( 120 ) 2 S27 = 1620 = Sn

Thus, the multiples of 3 from 20 to 100 have a sum of 1620.

PRACTICE 28.

Find the sum of the multiples of 4 from 101 to 250.

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PRACTICE 29.

Find the sum of the multiples of 3 from 200 to 300.

30.

In an arithmetic= series, t1 10 = and S9 144 . Find the sum of the first 20 terms by first determining the common difference between the terms in the sequence.

31.

The sum of the first two terms of an arithmetic series is 13 and the sum of the first four terms is 46. Determine the sum of the first ten terms.

32.

The sum of the first five terms of an arithmetic series is 170 and the sum of the first six terms is 225. If the common difference is 7, what are the first five terms of the series?

33.

Determine the arithmetic series that has S 4 = 26 , S5 = 40 and S6 = 57 .

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Chapter 1: Sequences & Series

1.3 GEOMETRIC SEQUENCE As we saw in the previous section, arithmetic sequences have a common difference and have a linear pattern when graphed. There are other types of sequences that have other patterns as well. Let us examine another sequence and investigate the results. EXAMPLE: Jack and Jill are planning a 15-day vacation and are looking for a baby-sitter. They are having trouble getting anyone to babysit their little darling, Chucky, and decide to post the rate of pay based on a daily salary. They stipulate the rate of pay as 1¢ for the first day of service, 2¢ for the second, 4¢ for the third day, and so on. Alternatively, they state that the babysitter can instead choose a single payment of $100.00 upon their return. Which rate of pay would you choose? a.

Complete the table of values for the given situation and calculate the difference between each pair of terms d= t n − t n − 1 . Day (n) 1 2 3 Wages ( t n ) 0.01 0.02 0.04

4

5

6

7

8

9

11

10

12

13

14

15

d

b.

What do you notice about the difference, d?

tn 110 100

c.

Is the sequence arithmetic? How can you tell?

90 80 70 60

d.

50

Plot the graph of n vs t n .

40 30

e.

Would you connect the dots? Why or why not?

20 10 2

f.

4

6

8

10

12

14

What do you think the graph of a geometric sequence would look like? (How would you describe the shape of the graph?).

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Chapter 1: Sequences & Series

23

In the previous example, you should have noticed that the sequence of common differences kept repeating the terms of the original sequence. Unlike those of arithmetic sequences, this new type of sequence does not have the same pattern as before. A pattern between successive terms can, however, be discovered by dividing rather than subtracting them. Day (n) 1 2 3 Wages ( t n ) 0.01 0.02 0.04

t n+1 tn

t2 t1

t3 t2

4

5

6

t5 t4

t4 t3

t6 t5

7

9

8

t8 t7

t7 t6

t9 t8

11

10

t10 t9

12

t12 t11

t11 t10

13

t13 t12

14

t14 t13

15

t15 t14

The common value obtained above is referred to as a COMMON RATIO, and identifies what is referred to as a GEOMETRIC SEQUENCE. A GEOMETRIC SEQUENCE is a sequence obtained by multiplying successive terms by the same t number. This constant is referred to as the common ratio, r, and is obtained by calculating n +1 . tn EXAMPLE 8 Determine if the sequence { 8, 4 , 2, 1, 21 , 41 , ...} is geometric. Solution Determine if there is a common ratio: r=

(

tn+ 1 tn

{

8 , 4 , 1 2

1 2

1 2

2 , 1 , 1 2

1 2

, 41 , ...

}

1 2

)

Since there is a common ratio r = 21 , the given sequence is geometric. EXAMPLE 9 32 Determine the common difference for { − 31 , 32 , − 43 , 83 , − 16 3 , 3 , ...} .

Solution Calculate the ratio r =

tn+ 1 : tn

32 { − 31 , 32 , − 43 , 83 , − 16 3 , 3 , ...}

r=

tn+ 1 tn

–2

–2

–2

–2

–2

Thus the common ratio is r = −2 . Sharpe Mathematics  2017

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Chapter 1: Sequences & Series

You can determine any term of a geometric sequence if you know its first term and the common ratio. Subsequent terms can be generated by multiplying the previous term by the common ratio. EXAMPLE 10 Generate the next 5 terms of the geometric sequence Solution

{ − 2, 3, .... } .

tn+ 1 t 3 3 = −  r = n+ 1 = tn 2 −2 tn

Calculate the common ratio r =

0 1 2 3 4    3  3  3  3  3 ⇒  − 2 ×  −  , − 2 ×  −  , − 2 ×  −  , − 2 ×  −  , − 2 ×  −  , ....   2  2  2  2  2      3 9  27   81  =  − 2 × ( 1) , − 2 ×  −  , − 2 ×   , − 2 ×  − , − 2 ×   , ....    2 4  8   16   

6 18 54 162   = ,− , ....   − 2, , − , 2 4 8 16   81 9 27   , − , ....  =  − 2, 3, − , 8 2 4  

The GENERAL TERM, t n , of a geometric sequence is given by: tn = t1 ⋅ r n − 1 , n ∈ N , t1 , r ∈ R where a is the first term, n is the term number and r is the common ratio. EXAMPLE 11 Given the geometric sequence { 3, 6, 12, 24 , 48, ...} :

a) determine the value of the 14th term b) determine which term is 3072.

Solution a) Use n = 14 and determine t14 : t n= t1 ⋅ r n − 1

b) we want to find n if t n = 3072 : t n= t1 ⋅ r n − 1 3072= 3 ⋅ 2n − 1

t14= 3 ⋅ 2

14 − 1

1024 = 2n − 1

t14= 3 ⋅ 2

13

210 = 2n − 1

t14= 3 ⋅ 8192

2 10 = 2 n − 1 10= n − 1 n = 11

t14 = 24 576 Thus, the 14th term is 24 576. (ie. t14 = 24 576 ) Sharpe Mathematics  2017

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EXAMPLE 12 In a geometric sequence the third term is 20 and the eighth term is –4860. Identify the first five terms of the sequence. Solution The third term is 20, so we can write t3 = 20 , where n = 3. Thus: t n= t1 ⋅ r n − 1

The eighth term is –4860, so we can write t8 = −4860 , where n = 8. Thus: t n= t1 ⋅ r n − 1

t3 = t1 ⋅ r 3− 1 = 20 t1 ⋅ r 2 = 20

t8 = t1 ⋅ r 8 − 1 = −4860



t1 ⋅ r 7 = −4860

We now have a system of equations that can be solved by dividing  by  : t1 ⋅ r 7 −4860 = 20 t1 ⋅ r 2

Substituting r = – 3 back into equation  (or ) we get: 20 t1 ⋅ r 2 = t1 ⋅ ( −3 ) =20 2

t1 ⋅ r 7

−4860 = 2 20 t1 ⋅ r

20 t1 ⋅ ( 9 ) =

r 5 = −243 r=

( −243 )



1

t1 = 5

20 9

r = −3 20 and the common ratio is – 3. 9 20  20  So the first five terms of the sequence are  ,− , 20 , − 60 , 180 , ...  . 3  9 

Thus the first term is

It is interesting to note that a geometric sequence can be expressed as a multiple of powers. That is, its terms can be expressed as a product of the first term, t1 , multiplied by powers of the common ratio, r. Thus, a geometric sequence can be written as

{t , 1

t1 r , t1 r 2 , t1 r 3 , t1 r 4 , . . . } .

EXAMPLE 13 Express the geometric sequence { 2, 6, 18, 54 , 162, 486, ...} as a multiple of a power. Solution First determine the common ratio. Verify for yourself that the common ratio is r = 3 . ∴ the sequence can be expressed as multiples of powers of 3 as: { 2 × 1 , 2 × 3 , 2 × 9 , 2 × 27, 2 × 81, 2 × 243, ...} = {2 × 30 , 2 × 31 , 2 × 32 , 2 × 33 , 2 × 34 , 2 × 35 , ...} Sharpe Mathematics  2017

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Chapter 1: Sequences & Series

TRY THESE! 1.

Determine the common ratio, r, for each geometric sequence. a)

2.

{ 2, 6, 18, 54 , 162, ... }

b)

{ 6, − 2, 32 , − 92 , 272 , ... }

Find the general term, tn , for each of the following: a)

{ 3, 6, 12, 24 , 48, ... }

b)

{ 0.5, 0.05, 0.005, 0.0005, ... }

( )

3.

Write the first 5 terms of the geometric sequence defined by t n = 4 81

4.

Find the missing terms in the geometric sequences given: a)

5.

6, 12, 24, ____, 96

b) 6, ____, ____, ____, 96

Given a geometric sequence where t3 = 18 and t7 = 1458 , find the value of t1 .

n−1

.

c) ____, 5, ____, 125

A missing term between two known terms, a and b , in a geometric sequence is known as the

Geometric Mean and equals

6.

ab .

If x − 3, x + 1, and 4 x − 2 are consecutive terms in a geometric sequence, find x .

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PRACTICE 34. Determine the common ratio, r, for each geometric sequence. a.

{ 1, 2, 4 , 8, 16, 32, ... }

b.

{ 5, 10, 20, 40, 80, ... }

c.

{ 8, 4 , 2, 1, 21 , 41 , ... }

d.

{ 2, 0.2, 0.02, 0.002, ... }

e.

{ 36, 9, 2.25, 0.5625, ... }

f.

{ 6, 2, 32 , 92 , 272 , ... }

g.

{ 21 , − 2, 8, − 32, 128, ... }

h.

{ 1.2, 6, 30, 150, 750, ... }

i.

{

2 , 2, 2 2 , 4 , 4 2 , ...

}

35. Determine the missing values for each geometric sequence. a.

{ 2, 4 , 8, ____, ____, ... }

b.

{ 8, 4 , ____, ____, ____, ... }

c.

{ − 9, 3, − 1, ____, ____, ... }

d.

{ 1.2, 0.6, ___, ___, ___, ... }

e.

{ 10, 1, ____, ____, ____, ... }

f.

{ 4 , ____, 41 ____, ____, ... }

g.

{ 3, ____, 12 ____, ____, ... }

h.

{ 0.5, 0.25, ___, ___, ___, ... }

i.

{4

2 , 4 , ___, ___, ___, ...

36. Express the terms of each sequence as a multiple of a power. a.

{ 1, 3, 9, 27, 81, ... }

b.

{ 2, 6, 18, 54 , 162, ... }

c.

{ 1, 21 , 41 , 81 , 161 , ... }

d.

{ 5, 10, 20, 40, 80, ... }

e.

{ 12, 3, 43 , 163 , 643 , ... }

f.

{ 8, 12, 18, 27, 40.5, ... }

g.

{ 200, 100, 50, 25, 12.5, ... }

h.

{ 1, 0.2, 0.04 , 0.008, ... }

i.

{ 43 , 32 , 3, 6, 12, ... }

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}

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Chapter 1: Sequences & Series

37. Identify the common ratio using the graph of each geometric sequence below. a.

b.

13 12

c.13

13 12

12

11

11

11

10

10

10

9

9

9

8

8

8

7

7

7

6

6

6

5

5

5

4

4

4

3

3

2

2

1

1

1

2

3

4

5

1

6

2

3

4

5

(4,8/3) (5,16/9)

3

(4,1.5) (5,0.75)

2 1

6

1

2

3

4

5

6

38. Determine the number of terms in each geometric sequence. a.

{ 3, 6, 12, ..., 1536}

b.

4 { 12, 4 , 43 , 94 , ..., 243 }

c.

1 { 21 , 41 , 81 , 161 , ..., 2048 }

d.

{ 5, − 10, 20, − 40, ..., 5120}

e.

{ 2, 6, 18, 54 ,

f.

{ 8, 12, 18, 27, ..., 2187 16 }

g.

{ 256, 192, 144 , ..., 729 16 }

i.

3 , ..., − 3 { 3, − 103 , 100 1000000 }

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h.

..., 1458}

{ 10, 2, 0.4 , ..., 0.00000512}

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Chapter 1: Sequences & Series

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39. Peter has found a pumpkin pattern in a magazine that he would like to enlarge to use on his pumpkin for Halloween. The space on the pumpkin cannot exceed 20 cm by 24 cm. If the original pattern measures 10 cm by 12 cm, and the photocopier enlarges the image by 10% each time, how many times can the pattern be enlarged before exceeding the maximum space? Length (cm) Width (cm)

10 12

40. If x − 4 , x + 2 , and 2 x + 20 are consecutive terms of a geometric sequence, find the value(s) of x.

41. If x − 1 , 3 x , and 10 x + 8 are consecutive terms of a geometric sequence, find the value(s) of x.

42. If t1 + t2 = 2 and t3 + t 4 =, 18 identify the first four terms of the geometric sequence(s). t + t 4 18 (Hint: Let ). = t1 a= , t2 ar= , t3 ar 2= , t 4 ar 3 and evaluate 3 = t1 + t 2 2

43. If t3 + t 4 = 108 , identify the first five terms of the geometric sequence(s). 36 and t 4 + t5 =

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Chapter 1: Sequences & Series

Geometric sequences have many applications in the real world. Several common applications involve the study of bacteria population growth, financial investments and the distance travelled by a bouncing ball. Let us now examine a few of these applications. EXAMPLE 14 The spore population of a certain type of bacteria was found to double every day. If the initial population was 120 spores, answer the following questions. a) Write the first six terms of the geometric sequence for the situation described. b) Identify the first term, t1 , and the common ratio, r, of the sequence. c) Write the general term for the sequence generated. d) How many spores would be present on the 10th day? e) On what day will the spore population first exceed 30 million? Solution a) The first six terms of the sequence would be

{ 120, 240, 480, 960, 1920, 3840, ... }

b) t1 = 120 and r = 2 c)

t n= t1 ⋅ r n − 1

 = t n 120 ⋅ (2)n − 1

d) On the 10th day n=10. Thus, we wish to determine the value of t10 . t n 120 ⋅ (2)n − 1 = = t10 120 ⋅ (2)10 − 1 t10 120 ⋅ (2)9 = = t10 120 ⋅ (512) t10 61440 =

∴ There are 61440 spores present on the 10 th day.

e) To determine when the population first exceeds 30 million, we will determine the value of n when t n = 30 , 000 , 000 . = t n 120 ⋅ (2)n − 1 30000000 = 120 ⋅ (2)n− 1 30000000 120 = ⋅ ( 2) n − 1 120 120 120 ⋅ ( 2) n − 1 250000 = 120 250000 = (2)n − 1 Using trial and error we get n ≈ 18.93 \ The population will first exceed 30 million on the 19th day. Sharpe Mathematics  2017

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THE BOUNCING BALL PROBLEM! EXAMPLE 15 A ball is dropped from an initial height of 40 m. After each bounce, it rises to 60% of its previous height. a) Write the general term for the sequence. b) What height does the ball reach after the eighth bounce? c) After how many bounces will the ball reach an approximate height of 15 cm? Solution a) The first few terms of the sequence would be : 40 , 40 × 0.60 , 40 × (0.60 )2 , 40 × (0.60 )3 , 40 × (0.60 )4 , ...

{

}

= { 40 , 24 , 14.4 , 8.64 , 5.184 , ... }

Thus, t1 = 40 and r = 0.60 Hence, t n= t1 ⋅ r n − 1



t n 40 ⋅ (0.60 )n − 1 =

b) After the 8th bounce n=8. Thus, we wish to determine the value of t8 . = t n 40 × (0.60 )n − 1 = t8 40 × (0.60 )8 − 1 = t8 40 × (0.60 )7 t8 ≈ 1.12 ∴ The ball which bounce to an approximate height of 1.12 m after the 8th bounce.

c) To determine when the ball will reach an approximate height of 15 cm, we will determine the value of n when t n = 0.15 m.

= t n 40 × (0.60 )n − 1 = 40 × (0.60 )n − 1 0.15 0.15 40 = × (0.60 )n − 1 40 40 40 × (0.60 )n − 1 0.00375 = 40 0.00375 = (0.60 )n − 1 Using trial and error we get n ≈ 11.94 \ The ball will reach an approximate height of 15 cm after the 12th bounce.

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Chapter 1: Sequences & Series

TRY THESE! 1.

Since the province of NL started using a very successful tourism, the population of Fogo Island has been increasing by 20% every year. If the initial population was 840, answer the following questions. a) Write the first five terms of the geometric sequence for the situation described.

b) Identify the first term, t1 , and the common ratio, r, of the sequence.

c) Write the general term for the sequence generated.

d) What would be the expected population in 8 years?

e) How long will it take for the population to reach 10 000?

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Chapter 1: Sequences & Series 2.

33

A ball is dropped from an initial height of 24 m. After each bounce, it rises to 2 of its 3 previous height. a) Write the general term for the sequence.

b) What height does the ball reach after the sixth bounce?

c) After how many bounces will the ball reach an approximate height of 16 cm?

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Chapter 1: Sequences & Series

PRACTICE 44.

The population of a certain bacteria was found to triple every 20 minutes. If there were initially 50 bacteria, answer the following questions. a) Write the first six terms of the geometric sequence for the situation described.

b) Identify the first term, t1 , and the common ratio, r, of the sequence.

c) Write the general term for the sequence generated.

d) How many bacteria would be present after 5 hours?

45.

The community of Gambo has been hosting a weekly “Chase the Ace” fundraiser to raise money for its local church. The initial jackpot started at $1200 and each week that the ace is not selected from a deck of cards, the jackpot has been found to increase at a rate modelling a geometric sequence. After 34 weeks the jackpot reaches $250 000. a) What was the weekly growth rate in the jackpot? (Round to two decimal places).

b) Write the general term for the sequence generated. c) Assuming the jackpot continues to increase at the same rate each week, what will be the jackpot’s expected value if the ace is not selected until the very last week?

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Chapter 1: Sequences & Series 46.

35

A ball is dropped from an initial height of 64 m. After each bounce, it rises to 40% of its previous height. a) Write the general term for the sequence.

b) What height does the ball reach after the sixth bounce?

47.

The value of a rare stamp increases by 12% each year. If the stamp was originally purchased for $120, what would be its value in 8 years?

48.

A truck was purchased for $25000 but depreciates in value by 15% each year. How much will the truck be worth after 5 years?

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Chapter 1: Sequences & Series

1.4 GEOMETRIC SERIES As with an arithmetic series, the sum of the terms in a geometric series can be represented by Sn = t1 + t2 + t3 + ... + t n − 1 + t n . For example, if given the geometric sequence

{ 5, 10, 20, 40, 80, ... } , the resulting geometric series would be 5 + 10 + 20 + 40 + 80 + … .

As before, Sn is used to represent the sum of the first n terms in a series, as well as the variables r, n and t n . To develop a formula that can be used to find the sum of a geometric series, let us first consider adding the terms of a geometric sequence, t n = t1 ( r )

n −1

, n∈ N ,

Hence, S n =t1 + t1 r + t1 r 2 + .... + t1 r n − 2 + t1 r n − 1



Next we multiply both sides of the equation by r : rS n = t1 r + t1 r 2 + .... + t1 r n − 2 + t1 r n − 1 + t1 r n  

Subtracting the equations  –  we get: rSn =

−

(S

n

t1 r + t1 r 2 + .... + t1 r n − 2 + t1 r n − 1 + t1 r n

= t1 + t1 r + t1 r 2 + .... + t 1 r n − 2 + t1 r n − 1

)

rSn − Sn = − t1 + t1 r n

Thus,

rSn − Sn = t1 r n − t1

Sn ( r − 1= ) t1 ( r n − 1) Sn ( r − 1)

( r − 1)

=

t1 ( r n − 1)

( r − 1)

t1 ( r n − 1) t1 ( 1 − r n ) = = Sn Sn , or r −1 1− r

r≠1

THE SUM OF n TERMS OF A GEOMETRIC SERIES Given a geometric series with common ratio, r, first term t1 , and nth term t n . The sum of the first n terms, Sn can be found using the formula: t1 ( r n − 1) t1 ( 1 − r n ) or , = Sn = Sn 1− r r −1

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r≠1

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Chapter 1: Sequences & Series

37

EXAMPLE 16 Determine the sum of the first 10 terms of the geometric sequence { 91 , 31 , 1, 3, 9, ...}. Solution We know t1 = 91 , r = 3 , and n = 10. Since we want the sum of the first 10 terms, we want to determine the value of S10 . t1 ( r 10 − 1) S10 = = r −1

1 9

=

(3

1 9

10

− 1)

3−1 ( 59049 − 1) 2

1 (29524 ) 9 4 = 3280 9 =

Thus the sum of the first 10 terms is 3280

4 . 9

EXAMPLE 17 Determine the sum of the geometric series 3 − 12 + 48 −  + 196 608 . Solution We must first find the number of terms, n, in the series using t1 = 3 , r = −4 , t n = 196 608 . Since we want the sum of the first 10 terms, we want to determine the value of S10 . t n= t1 ⋅ ( r )n − 1 t n = 3 ⋅ ( −4 )n − 1 = 196 608 3 ⋅ ( −4 )n− 1 196 608 = 3 3 ( −4 ) n − 1 = 65536

Now we can use the general formula for the sum of a geometric series. t1 ( r n − 1) Sn = r −1 3 ( ( −4 )9 − 1) S9 = ( −4 ) − 1

=

( −4 ) n − 1 = ( −4 )8 ( −4 )

n− 1

= ( −4 )

=

8

n−1= 8 n=9 ∴ There are 9 terms in the series. Sharpe Mathematics  2017

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3 ( −262 144 − 1) −5 3 ( −262 145 )

−5 = 3(52 429 ) = 157 287 ∴ The sum of the series is 157 287. Mathematics 2200

38

Chapter 1: Sequences & Series

For a finite geometric series where the first term t1 , the last term t n and the common ratio r are known, there is an alternate formula that can also be used to determine the sum. ALTERNATE FORMULA: SUM OF A GEOMETRIC SERIES WHEN # OF TERMS IS UNKNOWN Given a geometric series with common ratio, r, first term t1 , and nth term t n . The sum of the first n terms, Sn can be found using the formula: = Sn

r t n − t1 t1 − r t n or , = Sn 1− r r −1

r≠1

The major benefit of this version of the formula is that we do not need to first determine the number of terms in the series. While this formula can make solving the problem easier, it is NOT provided on any assessments and, as a result, must be memorized if you wish to use it on any evaluations. To see how this formula would simplify this type of problem, let us re-visit the previous example. EXAMPLE 18 Determine the sum of the geometric series 3 − 12 + 48 −  + 196 608 . Solution Using the new alternate formula above we have t1 = 3 , r = −4 , and t n = 196 608 . r t n − t1 r −1 ( −4 )(196 608) − (3) = −4 − 1 −786 432 − (3) = −5 −786 435 = −5 = 157 287

Sn =

Thus the sum of the series is 157 287. (Notice that with this formula we did not need to determine the number of terms!)

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Mathematics 2200

Chapter 1: Sequences & Series

39

TRY THESE! 1. Determine the sum of the first 12 terms of the geometric sequence

{ 12, 3, 43 , 163 , 643 , ... } .

1 − ... − 1 . 2. Determine the sum of the geometric series 41 − 81 + 16 2048

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Mathematics 2200

40

Chapter 1: Sequences & Series

EXAMPLE 19 The sum of 4 + 20 + 100 +  + t n is 15 624. (a) How many terms are in the series? (b) Determine the value of the final term t n . Solution From the given series we have t1 = 4 , r = 5 , and Sn = 15 624 . (a)

t1 ( r n − 1)

Sn =

r −1 4 ( 5n − 1) Sn = = 15 624 5−1 4 ( 5n − 1) = 15 624 4

(b)

5n − 1 = 15 624

rt n − t1 r −1 5t n − 4 Sn = = 15 624 5−1 5t n − 4 = 15 624 4 5t n − 4 = 62 496 Sn =

5t n = 62 500

n

5 = 15 625

5 tn

5n = 56 n=6

=

62 500 5

5 t n = 12 500

\ there are 6 terms in the series.

\ the final term is 12 500.

EXAMPLE 20 In a geometric series t 4 = 30 and t9 = 960 . Determine the value of S9 . Solution Dividing  ÷  we get:

Using t 4 = 30 we get: t n= t1 ⋅ r n − 1

t1 ⋅ r 8 960  ⇒ = 30  t1 ⋅ r 3

t4 = t1 ⋅ r 4 − 1 = 30 t1 ⋅ r 3 = 30



t1 ⋅ r 8 t1 ⋅ r 3

Using t9 = 960 we get:

= 32

r 5 = 32 1

t n= t1 ⋅ r n − 1

r = ( 32 ) 5

t9 = t1 ⋅ r 9 − 1 = 960

r=2

t1 ⋅ r 8 = 960

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Thus

t1 ⋅ r 3 = 30 t1 ⋅ (2)3 = 30 30 8 15 t1 = 4 t1 =

t1 ( r 9 − 1) S ∴ 9 = r −1 15 29 − 1 ( ) = 4 2−1 15 511 ) 7665 4 ( = = 1 4 Mathematics 2200

Chapter 1: Sequences & Series

41

TRY THESE! 1.

The sum of 3 − 6 + 12 − 24 +  + t n is –1023. (a) How many terms are in the series?

(b) Determine the value of the final term t n .

2.

In a geometric series t 4 = 54 and t6 = 24 . Determine the value of S12 .

3.

In a geometric series S2 = 12 and S3 = 30 . Identify the first five terms of the series.

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42

Chapter 1: Sequences & Series

PRACTICE 49.

Determine the sum of the first ten terms of each geometric sequence. a)

{ 2, 10, 50, 250,  }

3 3 3  c)  , , , 3,   8 4 2 

50.

b)

{ 5, − 10, 20, − 40,  }

d)

{ 80, 8, 0.8, 0.08,  }

Determine the sum of the first ten terms of each geometric sequence. a)= t1 6= , r 2

b) t1 120 = = , r

c) t1 = −5, r = −1.2

d)= t1

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1 2

2 3 = , r 3 4

Mathematics 2200

Chapter 1: Sequences & Series 51.

52.

43

Determine the sum of the terms of each geometric series. a)

4 − 12 + 36 − 108 +  + 2916

b)

1 1 1 + + +  + 4096 64 16 4

c)

−2 + 4 − 8 + 16 −  − 2048

d)

12 + 8 +

16 256 ++ 3 243

Determine the value of the final term t n for each geometric series. a) The sum of 3 − 6 + 12 − 24 +  + t n is –4095.

b) The sum of 2 + 6 + 18 + 54 +  + t n is 6560 .

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44

53.

Chapter 1: Sequences & Series

Determine the number of terms, n, in each geometric series. a) 2 + 8 + 32 +  where Sn = 174 762

b) −6 − 12 − 24 −  where Sn = −378 .

54.

In a geometric series t3 = 8 and t6 = 27 . Determine the value of S10 .

55.

A geometric sequence was found to have a common ratio of 4 and t6 = 4096 . Determine the value of S8 .

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Chapter 1: Sequences & Series 56.

45

A ping pong ball is dropped from a height of 90 m. With each bounce the ball rebounds to 2/3 of its previous vertical height. a) Determine the total vertical distance travelled by the ball when it makes contact with the floor at the sixth bounce. Round your answer to the nearest tenth.

b) How many times does the ball need to bounce in order to travel a total vertical distance of 290 m?

57.

A basketball is dropped on to a paved driveway from a roof 10 feet above the ground. With each rebound the ball loses 20% of its previous vertical height. Determine the total vertical distance travelled by the ball when it makes contact with the ground at the fifth bounce. Round your answer to the nearest tenth.

58.

A ball is dropped from an initial height of 4 m. After each bounce, it rises to 75% of its previous height. What is the total vertical height travelled by the ball after the tenth bounce? Round your answer to the nearest tenth.

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46

Chapter 1: Sequences & Series

1.5 INFINITE GEOMETRIC SERIES At the start of this chapter we identified two types of sequences: finite and infinite. A FINITE sequence is one which eventually terminates. The number of terms can be counted and is of the form { t 1 ,t 2 ,t 3 ,...,t n } , where t n is the last or nth term in the sequence. For example:

(a)

{ 1,

3, 5, 7, 9}

(b)

{ 2,

4, 8, 16, 32, 64}

An INFINITE sequence is one which does not terminate. The number of terms cannot be counted and is of the form { t 1 ,t 2 ,t 3 ,t 4 ,...} , where the ellipsis (“...”) indicates an infinitely continuing sequence. For example:

(a)

{ 5,

10, 15, 20, ...}

(b)

{ 1,

2, 5, 14, 41, 122, ...}

TRY THIS! 1. A Convergent Infinite Geometric Series Joshua buys a cookie. Instead of eating it all at once he comes up with a plan to make it last longer. On the first day he decides to eat half of the cookie, the next day another half of the remaining piece, then the next day another half of the remaining piece, and so on. a) Complete the table. n tn Sn

1

2

3

1 2 1 2

1 4 3 4

1 8 7 8

4

5

6

7

8

9

10

b) Find a formula for Sn .

c) Plot the values of Sn on the grid. d) What value does Sn appear to approach? Sharpe Mathematics  2017

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Chapter 1: Sequences & Series

47

This type of series is called CONVERGENT as the sums converge closer and closer to a finite value. In the previous example, the terms of the infinite series get closer and closer to 1 but will never actually reach 1. We say that the sum of the infinite geometric series is 1, and use the symbol S∞ to represent its sum.

TRY THIS! 2. A Divergent Infinite Geometric Series Joshua loves cookies! Instead of eating just one he decides to keep eating double the amount of cookies with each passing day. On the first day he decides to eat 2 cookies, the next day he eats 4 cookies, then the next day 8 cookies, and so on. a) Complete the table. n

1

2

3

tn

2

4

8

Sn

2

6

14

4

5

6

7

8

9

10

b) Find a formula for Sn .

c) Plot the values of Sn on the grid. d) What value does Sn appear to approach?

You should have noticed that as n gets larger, the value of Sn gets larger and larger and does NOT converge to a finite value. This kind of series is called DIVERGENT and does not have a finite sum as n approaches infinity. Sharpe Mathematics  2017

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Mathematics 2200

48

Chapter 1: Sequences & Series

t1 ( 1 − r n ) t1 − t1 r n Using the formula for a geometric series we have , Sn = = 1− r 1− r If −1 < r < 1 then the value of r n approaches 0 as n gets larger and larger. t1 Therefore, = S∞ , where − 1 < r < 1 . 1− r

An infinite geometric series is CONVERGENT when the sequence of partial sums approaches a fixed value. This occurs when –1 < r < 1. The sum of a convergent infinite geometric series can be determined using the formula: S∞ =

t1 1− r

where t n is the first term of the series r is the common ratio S∞ represents the sum of an infinite number of terms. An infinite geometric series is DIVERGENT when the sequence of partial sums does not approach a fixed value. The sum of this series cannot be calculated. This occurs when r > 1 or r < –1.

EXAMPLE 21 A ball is dropped from an initial height of 25 m. After each bounce, it rises to 40% of its previous height. In theory, what is the total vertical distance travelled by the ball? Solution The first few terms of the sequence would be :

{ 25, 25 × 0.40, 25 × (0.40 ) , 25 × (0.40 ) , 25 × (0.40 ) 2

3

4

, ...

}

= { 25, 10 , 4 , 1.6, 0.64 , ... }

Thus, t1 = 25 and r = 0.40 . So the series is converegent since −1 < r < 1 . Hence = S∞

t1 25 = 1 − r 1 − 0.40 25 = 0.60 2 = 41 3

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∴ the ball's total vertical distance travelled was DO NOT COPY!

41

2 m. 3

Mathematics 2200

Chapter 1: Sequences & Series

49

TRY THESE! 1.

Determine whether each of the following infinite geometric series are convergent or divergent. If convergent, find the sum. a)

6+2+

2 2 + + 3 9

b)

2 8 32 + + + 3 9 27

c) 4 − 2 + 1 −

1 1 +  2 4

3 , determine the value of x. 7

2.

If the geometric series 3 + 6 x + 12 x 2 +  converges to

3.

Find the first three terms of an infinite geometric series given that t1 = 40 and S∞ = 100 .

4.

A ball is dropped from an initial height of 30 m. After each bounce it rebounds to 60% of its previous height. What is the total vertical distance travelled by the ball?

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50

Chapter 1: Sequences & Series

PRACTICE 59.

60.

Identify the common ratio in each infinite geometric series, and use it to determine whether the series is convergent or divergent. a)

12 + 6 + 3 + 

b)

4 − 1+

d)

1+ 2+ 4 + 8 +

e)

g)

16 + 12 + 9 + 

j)

7 7 7 + + + 10 1 000 100 000

1 + 4

c)

4 + 6 + 9 +

10 − 8 + 6.4 − 

f)

1− 1+ 1−

h)

2 1 1 − + − 3 3 6

i) 3 5 + 3 4 + 3 3 + 

k)

2+ 2 + 1+

1 2

+

l)

0.001 + 0.01 + 0.1 + 

c)

20 − 12 + 7.2 − 

f)

24 + 23 + 22 + 

i)

−25 + 10 − 4 + 

Find the infinite sum of each infinite series. a) 32 + 16 + 8 + 4 + 

b) 3 + 1 +

1 + 3

3 3 3 + + + 10 200 4000

d)

0.9 + 0.27 + 0.81 + 

e)

g)

3 1 1 2 − + − + 4 2 3 9

h) 3 3 + 3 + 3 + 

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Mathematics 2200

Chapter 1: Sequences & Series

61.

62.

51

Given the first term of an infinite geometric series and its infinite sum, find the value of r. a) = t1 6= , S∞ 12

b) = t1 8= , S∞ 20

d) = t1 4= , S∞ 10

e) t1 2= = .8 , S∞ 8.4

c) t1 12 = = , S∞ 28

f) = t1

2 4 = , S∞ 3 3

Given the common ratio of an infinite geometric series and its infinite sum, find t1 . a) r =

1 = , S∞ 14 2

b)

3 r= − , S∞ = 20 4

c) r =

2 = , S∞ 48 3

d) r =

5 = , S∞ 12 6

e)

2 21 r= − , S∞ = 3

f) r =

3 = , S∞ 40 8

63.

A ball is dropped from a height of 24 m and rebounds to 75% of its previous height with each bounce. Determine the total vertical distance traveled by the ball.

64.

A ball is dropped from the top of a tower and rebounds to 2/3 of its previous height with each bounce. If the ball travels a total vertical distance of 81 m, determine the initial height from which the ball was dropped.

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Mathematics 2200

52 65.

66.

Chapter 1: Sequences & Series Find the sum of each infinite series and write as an exact fraction. (Hint: 0.06 =0.06 + 0.006 + 0.0006 +  ) a) 0.06

b) 0.13

c) 0.27

d) 2.4

e) 0.26

f) 5.01

An infinite geometric series is given by 1 − 2 x + 4 x 2 − 8 x 3 +  . If its infinite sum is determine the value of the common ratio.

67.

2 , 3

Each shaded square represents ¼ of the area of the larger square surrounding it. The shading pattern continues indefinitely. (a) Write the first five terms of the infinite series that represents this situation.

(b) How much of the largest square’s area is shaded?

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Chapter 2: Trigonometry

CHAPTER

2

TRIGONOMETRY Contents..............................................................

2.1

A. ANGLES IN STANDARD POSITION .................................................. 54 B. SPECIAL ANGLES ........................................................................ 58

2.2

TRIGONOMETRIC RATIOS OF ANY ANGLE ........................................... 63 A. DETERMINING THE ANGLE θ FROM THE POINT (C0Sθ , SINθ) ...... 63 B. EVALUATING TRIGONOMETRIC EXPRESSIONS ............................... 69 C. THE PRIMARY TRIGONOMETRIC RATIOS (REVIEW) ........................ 74

2.3

NON RIGHT-ANGLED TRIANGLE TRIGONOMETRY ................................ 77 A. THE SINE LAW ............................................................................ 77 B. THE AMBIGUOUS CASE OF THE SINE LAW ................................... 79

2.4 THE COSINE LAW ............................................................................. 82 2.5

APPLICATIONS OF THE SINE AND COSINE LAW .................................. 86

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53

54

Chapter 2: Trigonometry

2.1 ANGLES IN STANDARD POSITION In Math 1201 you were introduced to trigonometry as a way of solving unknown measures in a right-angled triangle. In this chapter our focus is again trigonometry, but our attention will be on examing its relationship within a circle. We use the point (1,0) as the starting point and rotate it through a rotation of θ degrees. Counter-clockwise rotations are interpreted as positive rotations (+θ ) whereas clockwise rotations are interpreted as negative rotations ( – θ ). y y

+ (ccw) θ (1,0)

(1,0)

x

θ

x

– (cw) The coordinate grid can be divided into four main regions called quadrants. y

Depending in which quadrant a point or coordinate lies, we can determine whether its value will be positive or negative. y

1

SECOND QUADRANT

1

FIRST QUADRANT

-1

1

THIRD QUADRANT

x

-1

1

-1

y

Using trigonometry, this is also known as the “CAST” Rule, which indicates where cos (C), sin (S), tan (T), and all (A) the ratios have a positive value within each quadrant, where: sinθ = y ,

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x

(+,–)

(–,–)

FOURTH QUADRANT

-1

cosθ = x ,

(+,+)

(–,+)

S

A x

T

tanθ = yx

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C Mathematics 2200

Chapter 2: Trigonometry

55

PRACTICE 1.

y

Using the relationship (x, y)  ( cosθ , sinθ ), match each rotation θ with a point on the unit circle. A.

150°

B.

90°

C.

225°

D.

–30°

E.

180°

F.

135°

G. –90°

H.

–180°

J.

–405°

I.

420°

A x

y 2.

i.

C cos 45° = ___

iii. sin 120° = ___ v.

sin 210° = ___

vii. cos

180°

= ___

ix. cos 450° = ___

3.

sin (–90°) = ___

iv. cos (–30°) = ___ vi. cos 315° viii. sin x.

(–270°)

(C, D)

(I, J)

= ___ = ___

(A, B) x

(K, L)

(U, V)

(M, N)

sin (–390°) = ___

(S, T)

(O, P)

(Q, R)

In which quadrant(s) do the following conditions occur? a. b. c. d. e. f.

4.

ii.

(E, F)

(G, H)

Identify the letter that could represent the value of each given expression.

sines of θ are positive cosines of θ are positive sines of θ are negative and cosines of θ are negative sines of θ are negative and cosines of θ are positive sines of θ are positive and cosines of θ are negative sines of θ are positive and cosines of θ are positive

Complete the following chart based on the CAST Rule.

__________ __________ __________ __________ __________ __________

     

QUADRANT Value

I

II

III

IV

cosθ sinθ

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Mathematics 2200

56

Chapter 2: Trigonometry

An angle of rotation is said to be in standard position if the initial arm is on the positive x-axis and the vertex is at the origin. COTERMINAL ANGLES

Coterminal angles are angles that have the same terminal arm. In other words, they are at the

same location on the unit circle, and can be found by adding or subtracting multiples of 360o. y

Example:

terminal point

terminal arm

y 40o and -320o are coterminal angles. - 320o

θ

θ

x

40o

x

initial point

initial arm

REFERENCE ANGLES The acute angle formed between the terminal arm and the x-axis is referred to as the reference angle. Coterminal angles have the same reference angle, as they share the same terminal arm. EXAMPLE 1 Identify the reference angle. a.

y

b.

y

c.

120o

y

50o

θ

θ

x

x

x

θ 320o

Solution a. reference angle = 50o Sharpe Mathematics  2017

b.

reference angle = 60o DO NOT COPY!

c.

reference angle = 40o Mathematics 2200

Chapter 2: Trigonometry

57

PRACTICE 5.

Name one positive and one negative coterminal angle for each rotation below. Sketch each angle on the given diagram, indicating the location of the terminal point/arm as well as the direction of rotation from the initial arm. a.

60o y

b. 60o

-30o y

c.

x

x

45o

f.

-150o

g.

i.

j.

83o

k.

-420o & -300o 6.

-56o

x

x

-420o & -300o l.

y

x

-400o y

-420o & -300o

y

y

h.

x

-420o & -300o

137o

270o

x

-420o & -300o

x

y

x

135o y

-420o & -300o

-420o & -300o

y

y

d.

x

-420o & -300o

420o & -300o e.

120o y

708o y

x

-420o & -300o

-420o & -300o

x

-420o & -300o

Identify the reference angle in Practice Exercise #5 above. a. ref. ∠ = _____

b. ref. ∠ = _____

c. ref. ∠ = _____

d. ref. ∠ = _____

e. ref. ∠ = _____

f. ref. ∠ = _____

g. ref. ∠ = _____

h. ref. ∠ = _____

i. ref. ∠ = _____

j. ref. ∠ = _____

k. ref. ∠ = _____

l. ref. ∠ = _____

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58

Chapter 2: Trigonometry

B. SPECIAL ANGLES As you have seen, when using trigonometry the values for various degrees of rotation, θ, are often approximations. We will now investigate these decimal values, and attempt to determine their exact values, as opposed to their approximate decimal values. Let’s focus our attention on several key angles of rotation in the first quadrant: 30 o, 45 o, 60 o. y 120o

90o 1

135o 150o

30o

0.5

1

0o

180o -1

-0.5

225o

240o

-1 o 270

90o 60o 45o 30o

x

1 360o

0.5 -0.5

210o

y

y

60o 45o

330o 300o

315o

0

30 o rotation

60 o rotation

y 1

1

1

0o

x

45 o rotation

y 1

C

E

A ?

1 2

30o

O ?

B 1

ΔAOB is a right triangle ∴We can use the Pythagorean Theorem

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x

? 60o

O

1 2

D

1

ΔCOD is a right triangle ∴We can use the Pythagorean Theorem

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x

45o

O

?

F

1

ΔEOF is an isosceles right Δ ∴We can use the Pythagorean Theorem

Mathematics 2200

x

Chapter 2: Trigonometry

59

SOLUTION: From these diagrams we can determine the exact value of each missing side using the Pythagorean Theorem: C E A 1

1 2

30°

O

1

x

B

O

(2)

2

D

1 2

( ) 1 2

x +1= 1 4

2

x2 + x2 = 12

3 4

y =

x2 = 1 2

x =

2

2

2

= x

4 y = 3

60°

1 2

3

2

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1× 2 2 2

x = 2 4 x = 2 2

We now have these measurements for each special triangle: 30° - 60° - 90° triangle

1 2

x = 1

3 4

y = 3

4 x = 3

Why?

2x 2 = 1

4 2 4 −1 y= 4 4 y2 = 3 4

x = 3

30°

x2 + y 2 = 12

+ y2 = 12

y2 = 1 − 1

4 2 4 −1 x= 4 4 x2 = 3 4

1

OF2 + EF2 = OE2

2

x2 = 1 − 1

F

x

1+y = 1 4

2

x =

O

OD2 + CD2 = OC2

= 12

y

45°

60°

OB2 + A B2 = OA 2 x2 + 1

1

y

Complete the following table to indicate the exact measure of the side opposite each angle:

45° - 45° - 90° triangle

45°

1

θ 2

2

30°

45°

opposite side

45° 2

2

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Mathematics 2200

60°

60

Chapter 2: Trigonometry

By thinking of the circle as a bicycle wheel, we can imagine a nail embedded in the tire as it is rotated in a counter-clockwise direction. By using a coordinate grid we can also determine the nail’s location. 90o

1m

This circle is called a “unit circle” since its radius is one unit.

1m 180o

0m

0o 360o nail

-1 m

270o y

y 90o

1

120o

-0.5

30o

150o

0.5

-1

60o

0.5

1

0o 360o

180o

x

-0.5

330o

210o 240o

-1

x

270o

300o

Using the grid above, determine the nail’s approximate position (x, y) for each rotation θ indicated in the table below. Round your answers to one decimal place.

θ

0o

x

1

y

0

30o

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60o

90o

120o

150o

180o

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210o

240o 270o 300o 330o 360o

Mathematics 2200

Chapter 2: Trigonometry

61

From the previous exercise, you may have noticed the repetition of the same values as the angle of rotation, θ , changed. The values identified in the ordered pairs can be expressed using trigonometry. Given a rotation of θ degrees, the following relationship exists: y Using the SOHCAHTOA rule we have the following ratios: (x, y)

= cosθ

1

θ

y

adjacent =  cosθ x

hypotenuse 1 opposite = sin θ =  sin θ 1y hypotenuse

x

x

y

(cosθ, sinθ) Thus we have the equality:

1

x = cosθ y = sin θ

θ

sinθ

cosθ

x

In other words, we can evaluate the location of the nail in the tire from the angle of rotation, θ, where the x-coordinate is equivalent to cosθ and the y-coordinate is equivalent to sinθ. As a result looking back at the previous exercise, the table of values you created can now be refined using cosine and sine. Complete the table below using your calculator, rounding your answers to 3 decimal places where appropriate.

θ

0o

cosθ

1

sinθ

0

30o

60o

90o

120o

150o

180o

210o

240o 270o 300o 330o 360o

Comparing your new table to the one you previously completed, how close were your answers?

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Chapter 2: Trigonometry

TRY THIS! Complete the following table to indicate the exact measure for each given rotation:

1.

θ

0o

x

1

y

0

30o

45o

60o

90o

120o

135o

150o

180o

210o

225o

240o

270o

300o

315o

330o

360o

2.

How do your values compare with those in the table you completed earlier in this section?

3.

Use the exact values you obtained in the table to identify the coordinates of each point indicated.

Careful with the “–” signs!

( , ) ( , ) 120 ( , ) 135

o

90o

1

( , ) 60o

o

( , ) ( , )

150o

30o

1 360o

-1

330o

210o

( , ) 240 ( , )

o

(1,0)

( , )

315o

225o

-1

270o

( , ) ( , )

300o

( , ) Complete the following reference angle chart based on the values obtained above. Do you see a pattern in your answers?

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( , ) 0o

180o

( , )

4.

( , )

45o

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Exact value

Exact value

sin 30°

cos 30°

sin 45°

cos 45°

sin 60°

cos 60°

Mathematics 2200

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63

2.2 TRIGONOMETRIC RATIOS OF ANY ANGLE A. DETERMINING THE ANGLE θ FROM THE POINT (cosθ, sinθ) Much of your previous work in this unit will help you to solve trigonometric equations algebraically, including coterminal angles, the CAST rule, and reference angles. When determining the value of θ , for simplicity we restrict the domain to be 0° ≤ θ ≤ 360° . For example, let us explore the possible values of θ in the following trigonometric equation: sinθ = 3 , where 0° ≤ θ ≤ 360° . 2 We need to refer to the unit circle, and identify the angles of rotation where the coordinates have a y-value of 3 . Using the values that we created in an earlier exercise: 2

(− (

2, 2 2 2

− 3 ,1 2 2

)

(−

)

120o

)

1

90o

(

1, 3 2 2

60o

135o

150o

( −1, 0 )

180o

(− (−

)

3 ,− 1 2 2

1, 3 2 2

( 0 ,1)

)

1

225o

(−

315o 240o

1 ,− 3 2 2

)

-1

270o

( 0 , −1)

(

300o

1 ,− 3 2 2

)

(

(

)

3 ,1 2 2

0o 360o

330o

210o

)

2, 2 2 2

30o

-1

2 ,− 2 2 2

(

45o

(

)

(1, 0 ) 3 ,− 1 2 2

2 ,− 2 2 2

)

)

As is evident from the chart above, the y-coordinate is 3 at the angles 60° and 120°. 2

Thus, for the exercise:

“Solve the trigonometric equation sinθ = 3 , for θ, where 0° ≤ θ ≤ 360° ”, 2 the solution would be:

θ = 60° and 120°

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Chapter 2: Trigonometry

This method obviously requires that you memorize the unit circle chart and its values. To do so, you could memorize the values in Quadrant I, and then simply take reflections of the values for the remaining quadrants. Alternatively, you could use a more intuitive approach to solve trigonometric equations... EXAMPLE 2 Determine the values of θ for the trigonometric equation sinθ = 3 , where 0° ≤ θ ≤ 360° . 2 Solution: y

STEP 1: Identify the principal angle.

60o

sin θ = 23

( )

θ = sin −1 23 θ= 60°

x

STEP 2: Use the CAST Rule to determine the quadrants in which the solutions occur. y

sin θ = 23 > 0 ∴

60o

Using the CAST Rule, the solutions must occur in Quadrants I & II (reject Quadrants III and IV).

STEP 3: Identify the reference angle and reflect the known angle over in the available quadrant. Mark in the reference angle, and identify the “new” angle.

x

y 120o

60o

60o

60o

∴θθ==60° 60° and and 120° 120° ∴ CHECK: How can you verify your results?

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Chapter 2: Trigonometry

65

Up to now, all of the values in the examples are “nice” values which can be verified on the unit circle. What if the values and angles are not contained in the unit circle chart? Where you could use the unit circle before, instead you are required to use the calculator to determine the solution... EXAMPLE 3 Find the values of θ in standard position for cos θ = 0.6128 , where 0° ≤ θ ≤ 360° . Solution:

y

STEP 1: Use the calculator to identify the principal angle.

52o

Using the calculator: x

cos θ = 0.6128

θ = cos −1 ( 0.6128 ) Press 2nd cos 0.6128 ENTER θ ≈ 52° STEP 2: Use the CAST Rule to determine the quadrants in which the solutions occur. y cos θ = 0.6128 > 0

52o

∴ Using the CAST Rule,

the solutions must occur in Quadrants I & IV (reject Quadrants II and III).

x

y STEP 3: Identify the reference angle and reflect the known angle over in the available quadrant. Mark in the reference angle, and identify the “new” angle. ∴ θ ≈ 52° and 308°

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52o

52o 52o

x

308o

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66

Chapter 2: Trigonometry

PRACTICE 7.

Determine all possible values of θ in each trigonometric equation, where 0° ≤ θ ≤ 360° .

x

x

e. sin θ = − y

1 2

f. cos θ = y

2 2

8.

x

y

y

5 j. sin θ = − 8 y

x

m. sin θ = 0.9272 y

n. cos θ = 0.8387 y

x

x

3 2

x 4 k. cos θ = − 5 y

x

x

h. cos θ =

g. sin θ = 0

x

x 3 i. cos θ = 4 y

1 d. cos θ = − 2 y

c. sin θ = −1 y

b. cos θ = 0 y

a. sin θ = 1 y

x

l. sin θ = −0.4 y x

x

o. sin θ = −0.2756 p. cos θ = −0.8829 y y x

x

Determine all possible values of θ in each trigonometric equation, where 0° ≤ θ ≤ 360° . a. sin θ = 0.6691

b. cosθ = 0.2756

c. sin θ = 0.9877

d. cosθ = −0.9336

e. sin θ = −0.7431

f. sin θ = −0.4226

g. cosθ = 0.9781

h. sin θ = 0.9945

i. cosθ = −0.7660

j. sin θ = −0.5736

k. cosθ = −0.1564

l. cosθ = 0.3420

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Chapter 2: Trigonometry

67

As we have seen, the values identified in the ordered pairs of a unit circle can be expressed using trigonometry. Given a rotation of θ degrees, the following relationship exists for the unit circle: = cos θ = sin θ

x   x cos θ = 1 y =   y sin θ 1

We can go even further to expand this relationship to circles of any radius, and not just the unit circle. Instead of using ‘1’ for the hypotneuse (radius) we can use the variable ‘r’ instead. y Thus we have the relationships:

(cosθ, sinθ) r

x =   r cos θ r x y   r sin θ = = sin θ r sin θ  tan θ = = tan θ cos θ = cos θ

sinθ

θ

cosθ

x y y x

EXAMPLE 4 If the terminal arm of an angle in standard position passes through the given point P(θ), determine the value θ to the the nearest tenth of a degree. a. P(3,4) Solution

b.

P(–6,8)

c.

P(12,–5)

y

y

a.

b.

(3,4) 4

θ 3

(–6,8)

c. 10

5

8

x

y

θ

α –6

θ 12

x

α

x –5

13 (12,–5)

y tan θ = x 4 tan θ = 3 θ ≈ 53.1°

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y x 8 tan α = −6 α ≈ 53.1° ∴ θ ≈ 180° − 53.1° θ ≈ 126.9° tan α =

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y x −5 tan α = 12 α ≈ 22.6° ∴ θ ≈ 360° − 22.6° θ ≈ 337.4°

tan α =

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Chapter 2: Trigonometry

PRACTICE 9.

A circle has its center at the origin and a radius of 17. If the terminal arm of an angle in standard position passes through the point (–15, 8), determine the measure of θ to the nearest degree.

10.

A circle has its center at the origin and a radius of 41. If the terminal arm of an angle in standard position passes through the point (9, –40), determine the measure of θ to the nearest degree.

11.

A circle has its center at the origin and a radius of 25. If the terminal arm of an angle in standard position passes through the point (–24, –7), determine the measure of θ to the nearest degree.

12.

A point P(θ) on the terminal arm of an angle on the unit circle has coordinates (–0.4067, 0.9135). Determine the measure of θ to the nearest degree.

13.

A point P(θ) on the terminal arm of an angle on the unit circle has coordinates (–0.7880, –0.6157). Determine the measure of θ to the nearest degree.

14.

A point P(θ) on the terminal arm of an angle on the unit circle has coordinates (0.9397, –0.3420). Determine the measure of θ to the nearest degree.

15.

If a point P(3,0) on a circle with centre at the origin is rotated 150o from standard position on a circle of radius 3, what are the new exact coordinates of P?

16.

If a point P(4,0) on a circle with centre (0,0) is rotated –240o about the origin, what are the new exact coordinates of P?

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Chapter 2: Trigonometry

69

B. EVALUATING TRIGONOMETRIC EXPRESSIONS Using the unit circle diagram that you just created, we are able to determine the exact location of certain rotations of the point (1,0). The coordinates of these rotations can be written using a mapping rule as: Rθ (1, 0) → (cos θ , sin θ ) This means that when the initial point (1,0) is rotated about the origin θ coordinates of the resulting image point can be determined by (cosθ , sinθ ).

degrees, the

EXAMPLE 5 Use the unit circle to determine the exact value of each mapping rule. a.

R30o (1, 0 )

b.

R135o (1, 0 )

Solution a. R30o (1, 0 ) → (cos 30°, sin 30°) →

( 23 , 21 )

b. R135o (1, 0 ) → (cos 135°, sin 135°) →

( 22 , 22 ) −

We can also evaluate trigonometric expressions directly without using the mapping rule to determine values of cosθ and sinθ for any value of θ. Using the unit circle on the previous page (or the mapping rule above), remember that the cosθ and sinθ are the x- and ycoordinates, respectively, for the image point. cos θ → x sin θ → y EXAMPLE 3 Determine the exact value of each trigonometric expression using the unit circle. a.

cos 45°

b.

sin 240°

c.

cos 300°

d.

sin 270°

b.

− 23

c.

1 2

d.

−1

Solution a.

2 2

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70

Chapter 2: Trigonometry

USING TECHNOLOGY By recognizing patterns that appear on the calculator, we can determine the exact values directly from the resulting decimal approximations. For example, using the calculator we get 23 ≈ 0.8660 and 22 ≈ 0.7071. Therefore, we can identify the exact values by recognizing these two key decimal approximations. EXAMPLE 4 Use the calculator to determine the exact value of each trigonometric expression. a.

cos 30o

b.

sin 315o

c.

cos 120o

Solution a.

b.

cos 120° = −0.5

∴ cos 120° = − 21

sin 315° ≈ −0.7071

∴ sin 315° = −

2 2

c.

cos 120° = −0.5

∴ cos 120° = − 21

PRACTICE 17.

18.

19.

Determine the exact coordinates of the image point (1,0) under the given rotation. a.

R60o (1, 0)

b.

R180o (1, 0)

c.

R45o (1, 0)

d.

R330o (1, 0)

e.

R210o (1, 0)

f.

R90o (1, 0)

g.

R225o (1, 0)

h.

R300o (1, 0)

i.

R−30o (1, 0)

j.

R420o (1, 0)

k.

R−120o (1, 0)

l.

R510o (1, 0)

Determine the approximate value of each trigonometric expression to four decimal places. a.

cos 23°

b.

sin 38°

c.

cos 125°

d.

sin 297°

e.

cos 18°

f.

cos 152°

g.

sin (-308°)

h.

sin 277°

i.

sin (-100°)

j.

cos 341°

k.

sin 165°

l.

cos (-30°)

Determine the exact value of each trigonometric expression. a.

sin 30°

b. cos 90°

c.

sin 150°

d.

cos 210°

e.

cos 45°

f.

sin 330°

g.

cos 270°

h.

sin 225°

i.

sin 300°

j.

cos (-120°)

k.

cos 180°

l.

sin 360°

o.

cos 480°

p.

sin (-315°)

m. cos (-240°)

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Chapter 2: Trigonometry

71

Multiple trigonometric expressions can be combined in one expression. Using your skills with simplifying radicals, they can then be reduced to simplest exact form. EXAMPLE 5 Determine the exact value of each trigonometric expression. sin 2 120° + cos 135°

a.

b.

sin 30° − cos 210° sin 225°

b.

sin 30° − cos 210° sin 225°

Solution a.

sin 2 120° + cos 135°

=

(sin 120°)2 + cos 135°

( ) +( ) ( 23 )( 23 ) + ( −22 ) ( 43 ) + ( −22 ) 3 2

= = =

3 4

= =

2

+

− 2 2

−2 2 4

3−2 2 4

Note that sin = 120° (sin 120°)2 2

=

−

( −22 )

( −23 )

1+ 3    2   =

 Evaluate each expression

 Simplify numerator

( −22 )

1+ 3  − 2  =  2  ÷  2   Rewrite      1 + 3   −2   Multiply by reciprocal =  2  ×  2    1+ 3  −2   Simplify = ×  2   2      = = =

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( 21 )

−1 − 3

 Simplify

2 −1 − 3 2

×

2 2

− 2− 6 2

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 Rationalize denominator  Simplify

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Chapter 2: Trigonometry

PRACTICE 20.

21.

22.

Determine the exact value of each trigonometric expression. a.

sin 30° + cos 60°

b.

sin 240° + cos 225°

c.

cos 90° + sin 270°

d.

cos 360° − cos 180°

e.

sin 120° + cos 135°

f.

4 sin 330°

g.

sin 2 30°

h.

−2 cos 135°

i.

cos 2 210°

j.

cos 150° sin 300°

k.

sin 270° cos 60°

l.

cos 210° cos 135°

m.

sin 315° sin 45°

n.

4 sin 2 150° cos( −30°)

o.

2 cos( −90°) sin 2 (135°)

p.

sin 2 30° + cos 2 30°

q.

sin 2 45° + cos 2 45°

r.

1 − sin 2 120°

s.

sin 60° cos 135° + cos 330° sin 210°

t.

2 cos 2 240° − cos( −180°) sin 30°

Determine the exact value of each trigonometric expression. a.

sin 30° cos 240°

b.

sin 120° cos 330°

c.

sin 135° cos 210°

d.

sin 30° sin 2 225°

e.

sin 60° cos 45° + cos 30° sin 150°

f.

cos 120° + sin 30° sin 150°

g.

cos 30° + sin 270° sin 330°

h.

3 sin 90° − cos 180° sin 210°

i.

2 cos 180° + cos 135° sin 135°

j.

sin 240° − cos 2 45° cos 330°

k.

sin 2 60° + cos 210° sin 225° sin 210° + cos 0°

Determine the exact value of each trigonometric expression. a.

4 sin 90° 1 + cos 150°

b.

sin 300° 2 cos 30° − 1

c.

sin 2 135° 4 sin 240° + 2

d.

sin( −60°) cos 60 + sin 225°

e.

3 cos 150° cos 120° − sin 60° sin 225°

f.

cos 180° sin 30° + 4 cos 315° 2 cos 210°

g.

cos 150° + sin 240° cos 30° sin 330°

h.

sin 2 60° − 2 cos 30° sin 300°

i.

5 1 + 2 sin 135°

j.

sin 2 240° + sin 270° 4 cos 210° − 2 sin 135°

k.

cos 30° sin 60° − cos 150° sin 45° sin 240° + cos 315°

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Chapter 2: Trigonometry

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EXAMPLE 6 The terminal arm of an angle θ in standard position lies on the line with equation 2 y = x , for x ≥ 0 . Determine the value of θ. 3 Solution 2 x can be used to 3 determine the angle θ. Since x ≥ 0 the angle

y=

The slope of the line y =

we seek is in Quadrant I since cos x ≥ 0 .

θ

2 x 3

rise =2

run =3

y x 2 tan θ = 3 θ ≈ 34° tan θ =

PRACTICE 23.

The terminal arm of an angle θ in standard position lies on the line with equation 3 y = x , for x ≥ 0 . Determine the value of θ. 4

24.

The terminal arm of an angle θ in standard position lies on the line with equation 1 y = − x , for x ≤ 0 . Determine the value of θ. 3

25.

The terminal arm of an angle θ in standard position lies on the line with equation 4 y = x , for x ≥ 0 . Determine the value of θ. 5

26.

The terminal arm of an angle θ in standard position lies on the line with equation 2x − 5 y = 0 , for x ≤ 0 . Determine the value of θ.

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Chapter 2: Trigonometry

C. THE PRIMARY TRIGONOMETRIC RATIOS (REVIEW) We will now review the trigonometric ratios. The three primary trigonometric ratios refer to the sine, cosine, and tangent of an angle. The trig ratios are opp remembered with sin θ = SOHCAHTOA. hyp hypotenuse

cos θ =

tan θ =

opposite

reference angle

adj hyp

θ

opp adj

adjacent

These three formulae are used to determine sides and angles of right-angled triangles. EXAMPLE 7 Determine the missing measure for each triangle. a. 12

x

b.

13

θ

25o

8

Solution Label the sides as “a” (adj), “o” (opp) or “h” (hyp) and decide which trig ratio to use. a. 12 (o)

x (h) 25o opp

sin θ = hyp 12 sin 25° = x x=

12 sin 25°

x ≈ 28.4

b.

13 (h)

θ 8 (a) adj

cos θ = hyp 8 cos θ = 13  8   13  θ ≈ 52.0°

θ = cos −1 

To find the value of

θ on the calculator, press 2nd cos 8÷13

NOTE: Make sure to check that your calculator MODE is set on DEGREES. Sharpe Mathematics  2017

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Chapter 2: Trigonometry

75

PRACTICE 27.

Use your calculator to determine the missing measure. a. e. i.

28.

x 5 x tan 45  15 1 sin θ = 2 cos 32° =

b. sin 54° =

8 x

12 20 3 g. cos θ = 5 x k. sin 22° = 18

c. tan θ =

f. sin θ = 0.4695 j. tan θ = 3

d. cos θ = 0.8192 20 25 2 l. cos θ = 2

h. sin θ =

Determine the missing value for each of the following. x a. b. c.

35o

θ

x

14

d. 42

f. x

12

g.

73

6 θ

29. a. When the angle of elevation of the sun is 65°, a tree casts a shadow 12 m long. How tall is the tree?

θ

42

50

b. The angle of elevation from a person on the ground to the airplane is 32°. If the airplane is 2500 m from the person, what is the plane's altitude?

32o

12 m c. A kite is flying at the end of a string 48 m long, making an angle of 75° with the ground. How high is the kite above the ground?

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38

24

h.

θ

o

16

22o

37

25 e.

18

x

28o

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d. A ladder 6.5 m long is placed against a wall that is 2.5 m high. What angle does the ladder make with the wall?

Mathematics 2200

76

30.

Chapter 2: Trigonometry

Determine the missing sides and angles indicated in each figure. a.

b.

θ

13

18

x

x

θ

42° 5

8

143

y

31.

In ΔABC, ∠A=90°, ∠B=64° and BC=18. Find the measures of ∠C, AB and AC.

32.

Determine the area of an equilateral triangle with side length 12.8 cm.

33.

A flagpole casts a shadow of 8.2 m in length when the sun makes an angle of 46° with the ground. Determine the height of the flagpole.

34.

From the top of a cliff, the angle of depression to a sailboat in the water is 58°. If the cliff is 143 m high, how far is the sailboat from the base of the cliff?

35.

While standing on the ground, Jack notices that the angle of elevation to a nearby tree is 72°. If he is 12 m away from the base of the tree, determine the tree’s height.

 36.

From a hot dog stand located 30 m from the base of an apartment building, you look up at an angle of elevation of 58° and see a window cleaner. Looking even higher up at an angle of elevation of 64° you notice a pigeon on a balcony waiting to ruin his work. How far apart are the cleaner and the pigeon?

 37.

Determine the value of x in the given diagram. 130 m 57° x

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61°

Mathematics 2200

Chapter 2: Trigonometry

77

2.3 NON RIGHT-ANGLED TRIANGLE TRIGONOMETRY So far, all of your work with triangles has involved those with right angles. As a result, we can apply some common properties, namely the Pythagorean Theorem and right-angled trigonometry (SOHCAHTOA). Such instances are restrictive in that many real-life situations do not involve right-angled triangles. For non right-angled triangles, we have two properties that can be used: the sine law and the cosine law.

A. THE SINE LAW The Sine Law is a set of equations that enable us to find missing sides and angles for triangles with angles of any measure. To derive the Sine Law, we turn to the area formulae that we developed earlier in Math 1201, namely = Area 21= bc sin A 21= ac sin B 21 ab sin C . A sin B sin C . Dividing each expression by 21 abc gives sin = = a c b

SINE LAW

A

sin A sin B sin C = = a b c or a b c = = sin A sin B sin C

b

C

c

a

The Sine Law is used to find the missing side or angle in a triangle if you have: • any two angles and one side (AAS or ASA), or • two sides and an opposite angle (SSA) EXAMPLE 8

Solution

Find the missing side x.

Using the Sine Law :

A 80o B

32o 14

sin A = sin B a b sin80° = sin32° x 14

x C

x sin 80° =

14 sin 32°

° x = 14sin32 sin80°

x ≈ 7.5

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B

78

Chapter 2: Trigonometry EXAMPLE 9

Solution

Find the missing side x.

First identify the measure of ∠C=80 o

A x

72o

Then, using the Sine Law: 21 28o

C

sin B = sin C c b sin28° = sin80° x 21

B

x sin = 80°

21sin 28°

° x = 21sin28 sin80°

x ≈ 10.0 PRACTICE 38.

Determine the measure of x. Round your answer to one decimal place. a.

x

b. 27

63o

x

c.

34

o

112

32o

o

56o

6.8

x 81o

14

39.

In ∆ABC determine the length of: a. BC if ∠A = 57°

∠B = 43° AC = 12 cm

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b.

AC if ∠A = 74° ∠B = 38° AB = 21 cm

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c.

AB if ∠B = 32° ∠C = 59° BC = 8.2 cm

Mathematics 2200

Chapter 2: Trigonometry

79

B. THE AMBIGUOUS CASE OF THE SINE LAW The previous two examples involved finding the missing side of a triangle. When trying to find a missing angle, however, consideration must be given to the results. To clarify what is happening, think back to when you learned about congruent triangles. There were certain conditions that had to be met in order to ensure that two triangles were congruent. These conditions, or postulates, identified the various ways in which congruent triangles could be constructed, namely: • ASA • SAS • SSS • AAS • HL There is one postulate which does not guarantee congruent triangles. Using the same information for each, the SSA Postulate can sometimes result in two different non-congruent triangles. As a demonstration, label both diagrams below with the following information: ∠A = 42°, AB = 24 cm, BC = 18 cm B

B

A

C

A

C

As you can see, it is possible to construct two different triangles from the given information (SSA) and, as a result, there are two possible values of ∠C (and AC). B

A

C

C′

From the diagram above, ΔCBC′ is an isosceles triangle, where ∠BCC′ and ∠BC′C are equal in measure. In addition, ∠BCA and ∠BCC′ are supplementary angles, meaning that their sum is 180o. Therefore, if we find one measure of ∠C, we can then find the other possible measure by finding its supplementary angle! In other words, the other possible value can be found by subtracting from 180o. Sharpe Mathematics  2017

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Chapter 2: Trigonometry

EXAMPLE 11

Solution

Find the possible value(s) of θ.

Using the Sine Law :

A 18

14

θ

35o

B

sin B = sin C c b sin35° = sin θ 14 18

C

18 sin 35 = °

14 sin θ

° sin θ = 18sin35 14

sin θ ≈ 0.7375 θ ≈ sin -1 (0.7375) θ ≈ 48° or 180° − 48° ∴ θ ≈ 48° or 132°

When considering the ambiguous case, sometimes we must reject the alternative solution since it is impossible (the alternative angle may result in the sum of the angles of a triangle exceeding 180o).

EXAMPLE 12

Solution

Find the possible value(s) of θ.

Using the Sine Law :

E

sinD = sinE e d sin52° = sinθ 25 20

θ

20sin52° = 25sinθ

25

° sinθ = 20sin52 25

sinθ ≈ 0.6304

52o D

20

F

θ ≈ sin -1 (0.6304) θ ≈ 39° or 180° − 39° ∴ θ ≈ 39° or 141° But we must reject θ =141° since 141°+52° > 180°

∴θ

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81

PRACTICE 40.

Determine the possible value(s) of θ. Round your answer to the nearest degree. a.

b. 36

28

θ

c.

35o 60

43o

51

θ θ

114o

32

36

41.

In ∆ABC determine the possible value(s) of: a. ∠A if ∠B = 41° AC = 52

b. ∠B if ∠C = 45° AB = 32

BC = 65

c. ∠C

AC = 44

if ∠A = 72° BC = 26 AB = 20

42.

Solve ∆DEF where ∠D = 110o , ∠E = 32o and DE = 24 (i.e. find all missing measures).

43.

Find x and y in the given diagram.

y 95o x x

25o 120o 21

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2.4

Chapter 2: Trigonometry

THE COSINE LAW

As with the Sine Law, the Cosine Law is a set of equations used to find missing sides and angles for triangles. The Sine Law, however, requires certain conditions, namely any two angles and one side (AAS or ASA) or the ambiguous case (SSA). In all other cases, you must use the Cosine Law. To derive the Cosine Law, we require the use of the Pythagorean Theorem and right-triangle trigonometry. In ∆ABD  c 2 = h2 + m 2 = h2 + ( a − n )2 = h + a − 2an + n 2

2

(since m =a − n ) 2

= h2 + n2 + a 2 − 2an In ∆ACD  b 2 = h2 + n 2

∴

and

A

(FOIL) (rearrange the equation)

c

n = b cos C  n =b cos C

c 2 = b 2 + a 2 − 2an

h

B

= b 2 + a 2 − 2 a ( b cos C )

b

m

= b 2 + a 2 − 2 a b cos C

n

D a

Similarly, it can be shown that a 2 = b 2 + c 2 − 2bc cos A and b 2 = a 2 + c 2 − 2ac cos B . A

COSINE LAW a 2 = b 2 + c 2 − 2bc cos A

b

c

b = a + c − 2ac cos B 2

2

2

c 2 = a 2 + b 2 − 2ab cos C

C

B

a

The Cosine Law is used to find the missing side or angle in a triangle if you have: • two sides and an included angle (SAS), or • all three sides (SSS) NOTE: If the given angle is 90o, the Cosine Law reduces down to the Pythagorean Theorem.

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Chapter 2: Trigonometry

83

EXAMPLE 13

Solution

Find the missing side x.

Using the Cosine Law :

B 20

a 2 = b 2 + c 2 − 2bc cos B a 2 = 14 2 + 20 2 − 2 (14)(20) cos 38°

x

a 2 ≈ 196 + 400 − 441.28

38o

A

C

14

a 2 ≈ 154.72 a ≈ 154.72 a ≈ 12.4

EXAMPLE 14

Solution

Find the missing angle θ.

Using the Cosine Law :

A b 2 = a 2 + c 2 − 2ac cos B

20 35

θ

B

24 C

352 = 24 2 + 20 2 − 2 (24)(20) cos θ 1225 =576 + 400 − 960 cos θ = 976 − 960 cos θ 1225 −960 cos θ 1225 − 976 = 249 = −960 cos θ cos θ = −249 960

(

249 θ = cos −1 −960

)

θ ≈ 105°

There is NO ambiguous case with the Cosine Law.

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Chapter 2: Trigonometry

When using the Cosine Law to find the missing angle (i.e. given SSS), we can rearrange the formulae and use these variations also: 2 + c 2 −a 2 2 +c2 −b2 a= a2 +b2 −c2 = cos A b= cos B cos C 2bc 2ac 2ab

Let’s re-visit the last example using this variation of the Cosine Law... EXAMPLE 15

Solution

Find the missing angle θ.

Using the Cosine Law :

A

2 2 2 cosB = a +2cac−b

20 35

2 2 2 cosθ = 24 +20 −35

θ

2(24)(20) −1225 = 576+ 400 960 249 = −960

B

24

∴

C

(

249 θ = cos −1 −960

)

θ ≈ 105° As you can see, you get the same answer using either version of the Cosine Law! PRACTICE 44.

Determine the measure of x. a.

b.

x

c. 23

x 12

100o

8

39 34

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54 66o 31

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Chapter 2: Trigonometry

45.

85

Determine the value of θ. Round your answer to the nearest degree. a.

b. 38

16

12

θ 26

8.6

θ 18

46.

6.2

c.

θ

11.1

30

In ∆ABC determine the value of: a.

∠A if AB = 12 AC = 18 BC = 13

b.

c.

BC if ∠A = 48° AB = 12

AC if ∠B = 64° BC = 9.1 AB = 11.3

AC = 22

47.

Solve ∆JKL where ∠J = 70o , JK = 11 and JL = 16 (i.e. find all missing measures).

48.

a. Determine the smallest angle of a triangle with sides 7 cm, 12 cm and 15 cm

49.

Calculate the area of quadrilateral ABCD:

b. Determine the largest angle of a triangle with sides 18 cm, 20 cm and 25 cm

A

15

D 100o 11 35o C

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86

2.5

Chapter 2: Trigonometry

APPLICATIONS OF THE SINE AND COSINE LAWS

PRACTICE 50.

To measure the size of Gull Pond, a warden walked from Point P to Q to R along the path shown. Using his sketch of his trail, determine the length of Gull Pond to the nearest meter. P

Q 165 m

100o

123 m

R

Gull Pond

51.

A soccer net is 4.6 m wide. When Beckham gets possession of the ball he immediately notices that he is 12 m from one goal post and 16 m from the other goal post. Within what angle must he shoot in order to score a goal?

52.

You have purchased a house on the end of a cul-de-sac. As a result, your land is irregularly shaped as shown. Calculate the area of land that you own, rounding your answer to one decimal place. B 24 m A o 82

67o

D 53.

20 m

C

Two airplanes depart Gander at the same time. One plane travels at 400 km/h. The other travels at 620 km/h. If the angle of separation between the planes is 37°, how far apart will the planes be in 90 minutes?

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Chapter 2: Trigonometry 54.

87

The straight-line distance between Kelly’s Island and Bell Island is 5.6 km. Glenn and Elsie want to take their boat from Kelly’s Island to the tip of Little Bell Island, then on to Bell Island. How far will they travel in total? Give your answer to the nearest tenth of a kilometre.

Bell Island

5.6 km

64°

Kelly’s Island 81° Little Bell Island

55.

15 m

Chris wishes to pave his driveway with measurements as shown in the diagram. Determine the length of the missing side of his driveway.

84o

18 m

67o

24 m

56.

From central headquarters, the city is able to track the locations of its snowplow equipment at all times to speed up clearing operations. At the moment when one plow is 35 km from headquarters and the other plow is 26 km from headquarters, what is the angle of separation between them when they are 50 km apart??

57.

Dwight is trying to determine the height of a flagpole outside Confederation Building. He decides to take two measures of the angle of elevation from the ground to the top of the pole as shown in the diagram below. Determine the height of the flagpole, correct to the nearest metre. •

24o

36o

20 m Sharpe Mathematics  2017

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88

3

Chapter 3: Quadratic Functions

CHAPTER

QUADRATIC FUNCTIONS Contents.............................................................. 3.0

CHARACTERISTICS OF A QUADRATIC FUNCTION ................................ 89

3.1

VERTEX FORM OF A QUADRATIC FUNCTION ....................................... 91 A. HORIZONTAL TRANSLATION y = (x – p)2 ..................................... 95 B. VERTICAL TRANSLATION y = x2+q ............................................... 98 C. VERTICAL STRETCH y = ax2 ........................................................ 101 D. VERTICAL REFLECTION y= –x2 .................................................... 105 E. OVERVIEW ................................................................................... 109 F. VERTEX FORM OF A QUADRATIC FUNCTION FROM ITS GRAPH ....... 114

3.2

STANDARD FORM OF A QUADRATIC FUNCTION .................................. 119 A. DETERMINING THE NUMBER OF X-INTERCEPTS ............................ 125

3.3

COMPLETING THE SQUARE ................................................................ 126 A. USING COMPLETING THE SQUARE TO CHANGE FORMS ............... 128 B. MAXIMUM/MINIMUM WORD PROBLEMS ..................................... 132

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Chapter 3: Quadratic Functions

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3.0 CHARACTERISTICS OF A QUADRATIC FUNCTION As you will see throughout this chapter, quadratic functions can be expressed in different forms including: • Vertex form  y = a( x − p )2 + q ; a ≠ 0 • Standard form  y = ax 2 + bx + c ; a ≠ 0 We will examine both of these forms along with each of their characteristics. We begin by first examining the vertex form. In the previous courses, we expressed a function in various forms. One of the forms presented was that of a table of values. A table of values describes how to represent a function as sets of ordered pairs. Throughout this section, we will review how we can generate an image graph of a quadratic function under various transformations. The standard base table of values for the quadratic y = x 2 is given by:

x -3 -2 -1 0 1 2 3

When the points in the table of values are graphed we get the base parabola:

y 9 4 1 0 1 4 9

a.

Describe the shape of the function.

b.

Identify the domain and range of the function graphed. DOMAIN = _______________________

RANGE = _______________________

All transformations and image graphs generated in this chapter will be applied to the base quadratic function y = x 2 . As a result, we will examine the change in the ordered pairs which will then be used to graph the image function. Sharpe Mathematics  2017

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90

Chapter 3: Quadratic Functions

The graph of every quadratic function is a curve called a PARABOLA. The characteristics of a parabola can be used to identify what is happening in different situations, particularly with reallife applications. We will now identify some key terms used throughout this chapter. VERTEX

The highest or lowest point of a parabola, also known as the “turning point”. If the graph of the quadratic opens upwards, then the vertex is the lowest point of the graph. If the graph of the quadratic opens downwards, then the vertex is the highest point of the graph.

AXIS OF SYMMETRY A vertical line that intersects the parabola at the vertex. The parabola is symmetrical about this line and thus divides the graph into two equal parts. The axis of symmetry is defined by the x-coordinate of the vertex. MINIMUM VALUE

The lowest y-coordinate of the vertex that opens upward.

MAXIMUM VALUE

The highest y-coordinate of the vertex that opens downward.

Vertex (max point) Axis of Symmetry

Axis of Symmetry

Vertex (min point)

TRY THIS! Complete the table for each quadratic function. Graph A

Graph B

Direction of opening Vertex Max or min value at vertex? Axis of symmetry Domain Range

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Chapter 3: Quadratic Functions

91

3.1 VERTEX FORM OF A QUADRATIC FUNCTION A quadratic function can be expressed in different forms, each with their own advantages as we shall see. Each form allows us to graph the function and identify key components from the equation itself, such as the vertex and the axis of symmetry. The first form we will examine is known as the vertex form of a function: VERTEX FORM y = a ( x − p) 2 + q ; a ≠ 0 With the vertex form, the vertex is ( p, q ) and the axis of symmetry is x = p . The vertex form (VF) of a quadratic function is expressed as a function of x. In the upcoming sections we will focus on changing from one form of an equation to another, and identify the key characteristics of the given function. TRY THIS:

10

To see the benefits of expressing a quadratic function in vertex form, let us examine the following graph of y = 2( x − 3)2 − 10 : a. Use the graph to identify the vertex.

y = 2( x − 3)2 − 10

8

b. How could you obtain the vertex from the equation?

6 4

c. Use the graph to identify the axis of symmetry.

2 2

4

6

8

d. How could you obtain the axis from the equation?

-2 -4

e. Use the graph to identify the y-intercept.

-6 -8

f. How could you obtain the y-intercept from the equation?

-10 -12

g. Use the graph to identify the vertical stretch. h. How could you obtain the vertical stretch from the equation? i. State the Domain and Range of the function.

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Chapter 3: Quadratic Functions

TRY THIS:

1 Let us now examine the graph of y = − ( x + 4) 2 + 7 : 2

1 y= − ( x + 4) 2 + 7 2

a. Use the graph to identify the vertex.

9 8 7

b. How could you obtain the vertex from the equation?

6 5 4

c. Use the graph to identify the axis of symmetry.

3 2 1 -9

-8

-7

-6

-5

-4

-3

-2

-1

-1

1

2

3

d. How could you obtain the axis from the equation?

-2 -3

e. Use the graph to identify the y-intercept.

-4 -5 -6

f. How could you obtain the y-intercept from the equation?

-7 -8 -9

g. Use the graph to identify the vertical stretch.

-10

To identify the signs of the xand y-coordinates of the VERTEX in Vertex Form use “VOS”: V ertex Form O pposite p S ame q

h. How could you obtain the vertical stretch from the equation? i. State the Domain and Range of the function.

EXAMPLE 1 Identify the axis of symmetry and vertex of y = 2( x − 1)2 − 3 . Solution The axis of symmetry is of the form x = p ⇒ x = 1 . Since the Vertex Form of a function y = a ( x − p )2 + q has the vertex located at ( p, q ) , then the function has its vertex located at (1, –3).

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Chapter 3: Quadratic Functions

93

SUMMARY:

y = a ( x − p )2 + q , a ≠ 0

Vertex Form: • • • • • • •

a > 0 will produce a parabola that opens upward a < 0 will produce a parabola that opens downward

the axis of symmetry is x = p the vertex is ( p, q ) the y-intercept is found by letting x = 0 and solve for y the Range of a parabola is { y | y ≥ q , y ∈  } if a > 0, and { y | y ≤ q , y ∈  } if a < 0 the vertical stretch factor is | a |

PRACTICE 1.

Determine the axis of symmetry, the vertex, y-intercept, vertical stretch, direction of opening, domain and range for each. Sketch the graph of each function on the given axes.

8

a. y = 2 ( x − 1)2 − 6

1 b. y = − ( x + 4 )2 + 2 2

c. y = 3 ( x + 2)2 − 4

Axis of Symmetry:

Axis of Symmetry:

Axis of Symmetry:

Vertex:

Vertex:

Vertex:

y-intercept:

y-intercept:

y-intercept:

Vertical Stretch:

Vertical Stretch:

Vertical Stretch:

Direction:

Direction:

Direction:

Domain:

Domain:

Domain:

Range:

Range:

Range:

-6

-4

8

8

8

6

6

6

4

4

4

2

2

2

2

-2

4

6

8

-6

-4

-2

2

4

6

8

-6

-4

-2

2

-2

-2

-4

-4

-4

-6

-6

-6

-8

-8

-8

10

10

10

-2

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6

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Chapter 3: Quadratic Functions

PRACTICE 2.

Express in vertex form. a. y − 2 = ( x + 3)2 d.

3.

1 ( y + 6 ) = ( x − 2) 2 3

d. y = −3 x 2 +

g. y =

1 2

1 ( x + 3)2 − 1 2

1 j. y = − ( x + 4 )2 − 2 5

5.

1 ( y + 1) = ( x − 2)2 5

e. −2( y − 4 ) = ( x + 3)2

c.

2( y − 5) = x2

1 f. − ( y + 3) = ( x − 1)2 2

Identify the axis of symmetry, vertex and vertical stretch of each function. a. y = 7 x 2

4.

b.

b. y = −4 x 2 + 1

c. = y 3 x2 − 2

1 e. y = − x2 + 6 3

f.= y

h. y = 2( x + 1)2 + 3

i. y = −7( x − 3)2 + 8

k. y =

1 ( x − 1)2 + 2 2

l. y =

1 2 x −1 4

2 ( x − 9 )2 − 11 3

Determine the direction of opening and range for each function. a.= y 2 x2 + 7

b. y = −( x + 3)2 − 4

2 d. y = − ( x + 3)2 − 1 3

e. y =

3 ( x − 4 )2 + 1 8

c. y = −4 ( x − 2)2

f. y =

1 ( x − 1)2 − 5 2

Place a “” in the appropriate spaces below to indicate which values can be quickly obtained from the vertex form of the equation. axis of symmetry

vertex

y-intercept

vertical stretch

range

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Chapter 3: Quadratic Functions

95

y ( x − p) 3.1A HORIZONTAL TRANSLATION =

2

Throughout the remainder of this unit, we will examine various transformations of the parabola y = x 2 . We will first examine the effect of a Horizontal Translation. The Horizontal Translation (HT) affects the base graph y = x 2 by translating or shifting the graph left or right. Let’s examine the following functions. Use the given equations to generate the table of values for the image function. Graph the three functions on the same axes. (a)

vertex

y = x2 x -3 -2 -1 0 1 2 3

(b)

y 9 4 1 0 1 4 9

vertex

= y ( x + 3) 2 x -6 -5 -4 -3 -2 -1 0

y 9 4 1 0 1 4 9

(c)

vertex

= y ( x − 5) 2 x 2 3 4 5 6 7 8

y 9 4 1 0 1 4 9

subtract 3 add 5 When there is (x + 3)2 in the equation then there is a HT of _____ units. This will result in a graph that is shifted ______ units to the _________ of y = x 2 . When there is (x – 5)2 in the equation then there is a HT of _____ units. This will result in a graph that is shifted ______ units to the _________ of y = x 2 . The Horizontal Translation is the OPPOSITE of the value added to x in the equation. Equation: • •

= y ( x − p) 2

HT = p

When p>0 the graph is shifted p units to the RIGHT of y = x 2 When p0 the graph is shifted q units UP in relation to the graph of y = x 2

When q 1, the graph is ____________________ in appearance. When 0 < a < 1, the graph is ____________________ in appearance. Sharpe Mathematics  2017

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-4

-3

-2

-1 -1

1

2

3

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Chapter 3: Quadratic Functions

103

PRACTICE 21.

Identify the vertical stretch in each function. a. y = 2 x 2

b. y = 4 x 2

c. = y 3 ( x − 1) 2

d.= y 4 ( x + 2) 2

e.= y 21 ( x + 5) 2

f.= y 51 ( x − 6) 2

g. y = 2 ( x − 3) 2 − 1

h. y = 7 ( x − 1) 2 + 4

y = 61 ( x − 1) 2 + 1

2 x2 +4 i.= y 3

k. y = 3 ( x + 2) 2 + 3

l. y =( x + 2) 2 − 1

n. y = 5 ( x − 2) 2 − 9

o. y = 81 ( x + 3) 2 + 5

j.

3 ( x + 2) 2 m.= y 4

22.

Identify the VS, HT and VT in each function. a. d. g. j.

y =( x − 3)2 + 2 = y x2 − 4 y =( x + 3)2 + 5 y = ( x − 1)2 + 3

b. e. h. k.

= y ( x + 1)2 y = 2( x + 1)2 − 8 y =( x − 4 )2 − 2 = y x2 − 1

c. f. i. l.

n. y = 21 ( x + 2)2 − 1

m. y = 2( x − 3)2 + 1

y =( x − 7)2 − 3 = y x2 − 3 y =( x − 3)2 − 6 = y ( x + 6)2

o. y = 31 ( x − 4 )2 − 21

Express each transformation of y = x 2 as a function in vertex form.

23.

a. VS scale factor of 2, HT of 3 units left, VT of 5 units down b. VS scale factor of 1, HT of 2 units right, VT of 7 unit up c. VS scale factor of 5, HT of 1 unit right, VT of 2 units down d. VS scale factor of 4, HT of 0 units right, VT of 3 units down e. VS scale factor of 21 , HT of 4 units left, VT of 1 unit up 24.

Match each function with the appropriate graph. c. y =

b. y = 2 x 2

a. y = 4 x 2 A

B

-4

-3

-2 -1

C

6

6

6

5

5

5

4

4

4

3

3

3

2

2

2

1

1

1

-1

1

2

3

4

-4

-3

-2 -1

-1

1

2

3

4

-4

-3

-2 -1

-1

-2

-2

-2

-3

-3

-3

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1 2 x 3

1

2

3

4

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Chapter 3: Quadratic Functions Using y = x 2 as the base function, identify the VS, HT, VT and the equation of each graph. a. b. c.

25.

-4

-3

-2 -1

6

6

6

5

5

5

4

4

4

3

3

3

2

2

2

1

1

1

-1

1

2

3

4

-4

-3

-3

-1

1

2

3

4

-4

-1

-3

-3

-3

-4

-4

-4

-5

-5

-5

6

6

6

5

5

5

4

4

4

3

3

3

2

2

2

1

1

1

-1

1

2

3

4

-4

-3

-2 -1

-1

1

2

3

4

-4

-3

-2 -1

-1

-2

-2

-2

-3

-3

-3

-4

-4

-4

-5

-5

-5

6

12

5

5

10

4

4

8

3

3

6

2

2

4

1

1

2

1

2

3

4

-4

-3

-2 -1

-1

1

2

3

4

-4

-3

-2 -1

-2

-2

-2

-4

-3

-3

-6

-4

-4

-8

-5

-10

-5

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4

1

2

3

4

1

2

3

4

i.

6

-1

1

f.

h.

-2 -1

-2 -1

-2

e.

-2 -1

-3

-2

g.

-4

-2 -1

-2

d.

-4

-3

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Chapter 3: Quadratic Functions

3.1D

VERTICAL REFLECTION

105

y = − x2

Thus far, we have explored the effects of various transformations on the base graph of y = x 2 , namely the horizontal translation (p), vertical translation (q) and vertical stretch (a). We will now examine the situation when the image graph is a reflection in the x-axis. A reflection in the x-axis is indicated by the value of a being less than 0. By examining the coordinates we notice that we keep the x-values the same in the image table of values, but we multiply all y-values by –1 in the image table of values. Complete the new table of values for −y = x2:

y=x : 2

x –3 –2 –1 0 1 2 3

y 9 4 1 0 1 4 9

x –3 –2 –1 0 1 2 3

Sketch the graphs of −y = x 2 and y = x 2 :

y

10 9 8 7 6 5 4 3 2

same x-value

1

y-value multiplied by –1

-4

-3

-2

-1-1

1

2

3

4

-2

Another way to think of the “–” is that it results in a parabola opening downward.

-3 -4 -5 -6

+

–

-7 -8 -9 -10

SUMMARY: When a < 0 in the function, the result will be a REFLECTION in the x-axis. Equation:

y = − x2 Sharpe Mathematics  2017

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Chapter 3: Quadratic Functions

It should be noted that when identifying or discussing the Vertical Stretch factor, the value of the vertical stretch is always positive and, as a result, does NOT indicate whether or not there is a reflection in the x-axis.

VS = a For example, (a) in the function y = −3 x 2 , the value of a is a = −3 but the vertical stretch is VS = _____. 1 1 (b) in the function y = − x 2 , the value of a is a = − but the vertical stretch is VS = _____. 6 6 10

Generate the table of values for each function. Graph the three functions on the same axes. (a)

y=x

2

(b)

y = −2 x

2

(c)

VS = ____ x –3 –2 –1 0 1 2 3

y

x –3 –2 –1 0 1 2 3

9 8 7

1 y = − x2 2

6 5 4

VS = ____

y

x –3 –2 –1 0 1 2 3

3 2

y

1 -4

-3

-2

-1-1

1

2

3

4

-2 -3 -4 -5 -6 -7 -8 -9 -10

SUMMARY: Equation:

y = ax 2

VS = a

• •

the Vertical Stretch factor is always positive When a > 1 the graph is “thinner” than y = x 2

•

When 0 < a < 1 the graph is “wider” than y = x 2

•

when a < 1 in the equation, the result will be a REFLECTION in the x-axis.

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107

PRACTICE 26.

Identify the vertical stretch in each function, and determine the direction of opening. a. y = 3 x 2 d. = y 5 ( x + 1)2 g. y = 2 ( x + 1)2 + 3 1 j. y = − ( x + 2)2 − 2 5 1 m. ( y − 1) = ( x + 3)2 2

b. y = −2 x 2

c. y = −4 ( x + 7)2

1 e. y = − ( x − 3)2 2 1 h. y = − ( x − 4 )2 + 5 2 1 k. y = ( x + 1)2 + 4 3

1 f. y = − ( x + 2)2 6

n. −3 ( y + 2) = ( x − 8)2

o.

i. y = −3 ( x − 2)2 + 1 l. y =

3 ( x − 5)2 − 2 4

2 ( y − 4 ) =( x + 2)2 3

27.

Determine the vertex of each function in the question above.

28.

Express each transformation of y = x 2 as a function in vertex form. a. VS scale factor of 4, HT of 2 units right, VT of 3 units up, reflection in the x-axis b. VS scale factor of 3, HT of 1 unit left, VT of 5 units down, reflection in the x-axis c. VS scale factor of 5, HT of 1 unit right, VT of 1 unit up d. VS scale factor of 2, HT of 0 units left, VT of 2 units up, reflection in the x-axis 2 , HT of 4 units left, VT of 4 units up e. VS scale factor of 3

29.

Match each function with the appropriate graph. a. y = 3 ( x − 2)2 + 1

b. y = −2 ( x − 2)2 + 1

A

B

-4

-3 -2

C

1 c. y = − ( x − 2)2 + 1 2

6

6

6

5

5

5

4

4

4

3

3

3

2

2

2

1

1

1

-1 -1

1

2

3

4

-4 -3

-2

-1 -1

1

2

3

4

-4 -3

-2

-1 -1

-2

-2

-2

-3

-3

-3

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1

2

3

4

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108

Chapter 3: Quadratic Functions

The image graphs of y = x 2 are shown. Identify the HT, VT and VS in each graph, and state whether or not the graph is a reflection in the x-axis.

30.

a.

-4

-3 -2

b.

-1

6

6

6

5

5

5

4

4

4

3

3

3

2

2

2

1

1

1

-1

1

2

3

4

-4

31.

-3 -2

-3 -2

-1

-1

1

2

3

4

-3 -2

-1

-1

-2

-2

-3

-3

-3

-4

-4

-4

-5

-5

-5

e.

-1

-4

-2

d.

-4

c.

6

6

5

5

5

4

4

4

3

3

3

2

2

2

1

1

1

1

2

3

4

-4

-3 -2

2

3

4

1

2

3

4

f.

6

-1

1

-1 -1

1

2

3

4

-4

-3 -2

-1 -1

-2

-2

-2

-3

-3

-3

-4

-4

-4

-5

-5

-5

Write the equation for each transformation of y = x 2 in the question above. a.

Equation: _______________________

b.

Equation: _______________________

c.

Equation: _______________________

d.

Equation: _______________________

e.

Equation: _______________________

f.

Equation: _______________________

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109

3.1E OVERVIEW As you have seen, when graphing quadratic functions in vertex form we can generate a table of values for the function, which can then be graphed. Let us now combine the various transformational components to analyze a quadratic function in vertex form. The vertex form of an equation is of the form:

y = a ( x − p )2 + q where

a = vertical stretch (VS) p = vertical translation (VT) up or down q = horizontal translation (HT) left or right

Notice that the horizontal component is inside the brackets and the vertical components are outside the brackets. Once you realize how the parameters a, p and q transform the graph of y = x 2 , you can then analyze quadratic functions by examining its key characteristics including the Vertex, Axis of Symmetry, Domain, Range, Direction of Opening, and y-intercept. SUMMARY: VERTEX FORM:

y = a ( x − p )2 + q

When a > 0 : • The parabola opens up and has a minimum y-value of q • The range of the function is y ≥ q •

There is a Vertical Stretch about the x-axis by a factor of |a|

When a < 0 : • The parabola opens down and has a maximum y-value of q • The range of the function is y ≤ q • •

There is a Vertical Stretch about the x-axis by a factor of |a| The result will be a REFLECTION in the x-axis.

Compared to y = x 2 , the HT is p units right if p > 0 and |p| units left if p < 0 . Compared to y = x 2 , the VT is q units up if q > 0 and |q| units down if q < 0 . The Domain of the function is the set of all real numbers x ∈  . The equation of the Axis of Symmetry is x = p . The Vertex is ( p, q ) .

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Chapter 3: Quadratic Functions

TRY THESE! 1.

Identify the characteristics of the graph of each function. = y 2 x2 − 5

y= −3 ( x + 1) + 2 2

y= − ( x + 7) − 3 2

2 y = 41 ( x − 2 ) + 8

Direction of opening Vertical stretch Vertex Axis of symmetry y-intercept Domain Range 2.

Identify the characteristics of y = 2 ( x − 1) + 3 and then graph the function below. 2

a) Direction of opening

b) Vertical stretch

c) Vertex

d) Axis of Symmetry

e) y-intercept

f) Domain

g) Range

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111

PRACTICE 32.

Identify the characteristics of y =( x + 3)2 + 1 and then graph the function below. a) Direction of opening b) Vertical stretch c) Vertex d) Axis of Symmetry e) y-intercept f) Domain g) Range

33.

Identify the characteristics of y = −( x − 1)2 − 4 and then graph the function below. a) Direction of opening b) Vertical stretch c) Vertex d) Axis of Symmetry e) y-intercept f) Domain g) Range

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112

34.

Chapter 3: Quadratic Functions

Identify the characteristics of y = 2( x − 2)2 + 3 and then graph the function below. a) Direction of opening b) Vertical stretch c) Vertex d) Axis of Symmetry e) y-intercept f) Domain g) Range

35.

Identify the characteristics of y = −2( x − 4 )2 + 5 and then graph the function below. a) Direction of opening b) Vertical stretch c) Vertex d) Axis of Symmetry e) y-intercept f) Domain g) Range

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Chapter 3: Quadratic Functions 36.

Identify the characteristics of y =

113 1 ( x − 3)2 − 1 and then graph the function below. 2

a) Direction of opening b) Vertical stretch c) Vertex d) Axis of Symmetry e) y-intercept f) Domain g) Range

37.

1 Identify the characteristics of y = − ( x − 1)2 − 3 and then graph the function below. 2

a) Direction of opening b) Vertical stretch c) Vertex d) Axis of Symmetry e) y-intercept f) Domain g) Range

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Chapter 3: Quadratic Functions

3.1F VERTEX FORM OF A QUADRATIC FUNCTION FROM ITS GRAPH Much of this chapter has focused on using the given quadratic equation to sketch the graph of the function by identifying several key characteristics, including the vertex, y-intercept, axis of symmetry and the vertical stretch factor. In this section we will now reverse the process by identifying the vertex form of the quadratic function from the given graph. We will use these same characteristics and work backwards to identify the function graphed. TRY THESE! Examine the following graphs and identify the equation of each function. A.

B.

9

10

8

8

7

6

5

6 4

4

3

2

2 1

2

4

6

8

-9

-2

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

-2

-4

-3 -4

-6

-5

-8

-6 -7

-10

-8 -9

-12

a. Identify the vertex.

-1

-10

 ________

a. Identify the vertex.

 ________

b. Identify the axis of symmetry.  ________

b. Identify the axis of symmetry.  ________

c. Identify the y-intercept.

 ________

c. Identify the y-intercept.

 ________

d. Identify the vertical stretch.

 ________

d. Identify the vertical stretch.

 ________

e. Express the equation in Vertex Form (VF).

e. Express the equation in Vertex Form (VF).

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115

When attempting to find the equation of a quadratic function, perhaps the most challenging characteristic to find is the vertical stretch factor or the value of the parameter “a”. This value can be determined if we know the vertex and one other point. By substituting in these values, the vertical stretch factor can then be determined, as demonstrated in the following example. EXAMPLE 2 A golf ball is hit from the fairway with a high chip shot. It reaches a maximum height of 20 m and lands on the green 10 m away. Determine the equation that describes the golf ball’s path. Express the equation in Vertex Form. Solution

20

From the sketch it is apparent that the vertex is (5, 20) and the x-intercepts are (0, 0) and (10, 0). y = a ( x − h) 2 + k

 VF

0

y = a ( x − 5) + 20

 Plug in the vertex (5, 20)

*Remember VOS*

0 = a (10 − 5)2 + 20

 Plug in either the point (10, 0) or (0,0)

2

10

 Simplify 0 a (5)2 + 20 =  Simplify 0 25a + 20 =  Simplify 25a − 20 = −20  Simplify =a 25 4  Simplify − = a 5 4  y= − ( x − 5)2 + 20  VF 5

TRY THIS! Write an equation for each quadratic function described. Express your answer in Vertex Form. a. vertex (3, 1) and has a y-intercept of 5

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b. vertex (–2, –4) and passing through (–6, 3)

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Chapter 3: Quadratic Functions

PRACTICE 38. Identify the equation for each graph shown in Vertex Form. A.

B.

C. 7

9

8

6

8

7

5

7

6

4

6

5

3

5

4

2

4

3

1

2

−3

1 −3

−2

y

y

y 9

−1

−2

−1

3

x 1

2

3

4

5

6

7

2

8

1

2

−2

3

−3

1 −1

2

4

2

8

8

7

7

6

6

2

1

1

1 -1

2

3

2

-1

1

2

4

3

4

3

-3 -2

3

4

5

4

-4

2

F.

5

-6 -5

x

−1

−3

E.

-7

−2

−1

D.

-9 -8

1

−1

-6

-5

-4

-3

-2

-1

-1 -2

-2

1

2

-1

1

2

3

4

5

6

7

8

9

-1

-3

-3

-4

-4

-5

-5

6

-2 -3

39. Write an equation for each quadratic function described. Express your answer in Vertex Form. a. vertex (2, 5) with y-intercept of 3

b. vertex (6, –2) with y-intercept of –8

c. vertex (4, 3) with x-intercepts 2 and 6

d. vertex (–2, –4) with x-intercepts –4 and 0

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40. Write an equation for each quadratic function described. Express your answer in Vertex Form.

41.

a. vertex (1, –5) and passes through the point (2, –3)

b. vertex (0, 3) and passes through the point (–2, 1)

c. vertex (–1, –2) and passes through the point (–4, –1)

d. vertex (5, –6) and passes through the point (1, 2)

Write an equation for each quadratic function. Express your answer in Vertex Form. a. vertex (2, –3), opens upward, and has the same vertical stretch factor

b. vertex (–1, 4), has the same vertical stretch as y = 31 ( x − 1)2 − 5 , and

as y = 2( x − 1)2 + 3

reflected in the x-axis

c. vertex (5, 2), vertical stretch factor 2 , where the vertex is a of 3

d. vertex (0, 5), vertical stretch factor of 41 , where the vertex is a

e. the graph of y = x 2 is reflected across the x-axis, shifted 2 units right, 3 units down with a

f. the graph of y = x 2 is translated 4 units left and is vertically stretched by a factor of 21 .

minimum point

vertical stretch factor of 2.

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Chapter 3: Quadratic Functions

42.

A rock is thrown into the air from an initial height of 2 meters. After 2 seconds it reaches a maximum height of 10 meters. Determine the equation of the quadratic function that describes the path of the rock. Express your equation in Vertex Form.

43.

A wedding arch is in the shape of a parabola. If the arch is 2 m wide and 3 m tall, determine the equation that describes the shape of the arch. Express your equation in Vertex Form.

44.

An arrow is fired into the air and reaches a maximum height of 30 m at a horizontal distance of 50 m from where it was fired. It stuck in the ground 90 m away from where it was fired. a. Determine the equation of the quadratic function that describes the path of the arrow. Express your equation in standard form.

b. How high is the arrow after travelling a horizontal distance of 80 m?

45.

A football is kicked for a field goal attempt and reaches a maximum height of 25 m at a horizontal distance of 20 m. a. Determine the equation of the quadratic function that describes the path of the football. Express your equation in Vertex Form.

b. If the field goal marker is located 38 m away and at a height of 4 m, would the kick count to score a point?

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3.2 STANDARD FORM OF A QUADRATIC FUNCTION The transformational form has several advantages as we have just seen. It allows us to graph the function and identify key components from the equation itself, such as the vertex ( h, k ) and the axis of symmetry x = h . Similarly, there is another form known as the standard form of a function: STANDARD FORM y = ax 2 + bx + c , a ≠ 0 To see the benefits of expressing a quadratic function in standard form, let us examine the following graph of y = 21 x 2 − 4 x + 5 . a. Identify the values of ‘a’, ‘b’ and ‘c’. 10

a = _____ , b = _____ , c = _____

y = 21 x 2 − 4 x + 5

8

b. Use the graph to identify the vertex.

6

c. Use the graph to identify the axis of symmetry.

4 2

d. Use the graph to identify the y-intercept. -2

2

4

6

8

10

12

-2

e. Evaluate the y-intercept algebraically (let x = 0 ).

-4

f. Identify the Vertical Stretch Factor. g. What effect will ‘a’ have on the graph if: a>0: h. Determine the value of x = −

a 0 will produce a parabola that opens upward

•

a < 0 will produce a parabola that opens downward

•

the y-intercept is y = c

• •

the Axis of Symmetry is x = − 2ba , which is also the x-coordinate of the vertex the Range of a parabola is { y | y ≥ k , y ∈  } if a > 0 , and { y | y ≤ k , y ∈  } if a < 0

•

the Vertical Stretch factor is |a|

PRACTICE 46.

For each function, determine the axis of symmetry, the vertex, y-intercept, direction of opening, domain and range. Sketch the graph of each function on the given axes. a. y = 2 x 2 − 8 x + 3

b. y = −3 x 2 − 12 x − 4

c. y = x 2 − 8 x + 10

Axis of Symmetry:

Axis of Symmetry:

Axis of Symmetry:

Vertex:

Vertex:

Vertex:

y-intercept:

y-intercept:

y-intercept:

Direction:

Direction:

Direction:

Domain:

Domain:

Domain:

Range:

Range:

Range:

8

-8

-6

-4

6

6

6

4

4

4

2

2

2

-2

2

4

-2

47.

8

8

6

8

10

-8

-6

-4

-2

2

4

6

8

1

-8

-6

-4

-2

2

4

6

8

1

-2

-2

-4

-4

-4

-6

-6

-6

-8

-8

-8

10

10

10

Identify the y-intercept, axis of symmetry and vertex of each function. a. b. c. d. e. f.

y = x2 + 2 x + 7 y = 3 x2 − 2 x + 4 y= −2 x 2 + 4 x + 1 y= − x2 + 3 x − 2 y = 3 x2 − 6 x + 5 y= −4 x 2 − 2 x

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y-int = _______ y-int = _______ y-int = _______ y-int = _______ y-int = _______ y-int = _______

axis of sym = _______ axis of sym = _______ axis of sym = _______ axis of sym = _______ axis of sym = _______ axis of sym = _______

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124 48.

Chapter 3: Quadratic Functions Identify the y-intercept, axis of symmetry and vertex of each function. a. y = 2 x 2 − 5 x − 1

y-int = _______

axis of sym = _______

vertex = _______

b. y =−4 x 2 + x + 2

y-int = _______

axis of sym = _______

vertex = _______

y-int = _______

axis of sym = _______

vertex = _______

d. y = − 21 x 2 + 4 x + 2

y-int = _______

axis of sym = _______

vertex = _______

e.

y-int = _______

axis of sym = _______

vertex = _______

y-int = _______

axis of sym = _______

vertex = _______

g. y = 2.4 x 2 + 1.2 x + 1.6

y-int = _______

axis of sym = _______

vertex = _______

h. y =

2 x2 − 2 x − 6

y-int = _______

axis of sym = _______

vertex = _______

3 x2 − 6 x + 5

y-int = _______

axis of sym = _______

vertex = _______

c.

f.

i. 49.

y= 10 x 2 + 5 x + 3 2 x2 + 1 x y= −3 6 2 3 2 x+8 y= −4 x −3

y=

A ball is thrown into the air, and its height (h) above the ground in meters t seconds after it was thrown can be described by the function h( t ) = −5t 2 + 20 t + 1.8 . a. Identify the y-intercept of the function h( t ) .

24

b. What was the initial height of the ball?

22 20

c. Determine the value of h(1.5) .

18

d. What was the height of the ball at 1.5 seconds?

14

e. Determine the axis of symmetry of h( t ) .

12

f. When did the ball reach its maximum height?

10 8

g. Identify the vertex of the function h( t ) .

6

h. What was the maximum height of the ball?

4

i. Assuming the ball hits the ground at 4.1 seconds, sketch the graph of the function h( t ) . 50.

16

2 1

2

3

4

5

6

A toy rocket is fired into the air. Its height (h) above the ground in meters after t seconds is given by the function h( t ) = −4.9 t 2 + 24 t . a. What was rocket’s initial height?

b. What was rocket’s height at 4 seconds?

c. When did the rocket reach its max height?

d. What was the rocket’s max height?

e. Assuming the rocket hits the ground at 4.9 seconds, state the domain of the function. Sharpe Mathematics  2017

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Chapter 3: Quadratic Functions 51.

52.

125

A bat swoops down to catch a moth in mid-flight. Its height (h) above the ground, in meters, after t seconds is given by the function h( t ) = 2t 2 − 10 t + 25 . a. What was the initial height of the bat?

b. What was the bat’s height at 2 seconds?

c. When was the bat closest to the ground?

d. What was the minimum height of the bat?

Jenna is on a diving board, and her height (h) in meters above the water t seconds after she leaves the board is given by the function h( t ) = −4 t 2 + 10 t + 5 . a. What was the height of the diving board?

b. What was her height at 1.5 seconds?

c. When did she reach her maximum height?

d. What was Jenna’s maximum height?

3.2A DETERMINING THE NUMBER OF X-INTERCEPTS When the equation of a quadratic function is presented in Vertex or Standard Form, we can determine the number of x-intercepts by examining the location of the vertex and the direction of opening. Alternatively, we can determine the number of x-intercepts by examining: • the value of a to determine if the graph opens upward or downward • the value of q to determine if the vertex is above, below, or on the x-axis a)

Vertex Form of Equation

b)

= f ( x ) 0.4 x 2 − 2

c)

f ( x) = −3( x + 1)2

d)

f ( x) = − 51 ( x − 4 )2 − 3

e)

Direction of Opening

Sketch the Graph

Number of x-intercepts

How many x-intercepts would the graph of y = a ( x − p ) + q have if: 2

i) a > 0 and q > 0 ? f)

Vertex

ii) a > 0 and q < 0 ?

0? iii) a < 0 and q = iv) a < 0 and q < 0 ?

Determine the number of x-intercepts for each quadratic function:

−0.5 x 2 + 4 i) f ( x ) = Sharpe Mathematics  2017

( x ) 31 ( x − 2)2 ii) f =

− 4 ( x + 6 )2 + 7 iii) f ( x ) =

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126

Chapter 3: Quadratic Functions

3.3 COMPLETING THE SQUARE Finally, the last two conversions involve changing an equation from general form into standard form or transformational form. To do this we need to use the method of “completing the square” in which we create and factor a perfect square trinomial. Let us first do a quick review of factoring perfect square trinomials of the form x 2 + bx + c . When factoring trinomials of this type, we try to find two numbers that multiply to give “c” and add to give “b”. By listing off the possible products to give “c”, we examine the various combinations to find the pair that has a sum of “b”. For example, to factor x 2 + 8 x + 16 , list the different ways to multiply to get 16:

Product is 16 =

1 16 2 8 4 4

Sum is +8

∴ x 2 + 8 x + 16 = ( x + 4 )( x + 4 )

EXAMPLE 4 Factor each perfect square trinomial. a. x2 + 10x + 25

b. x2 – 16x + 64

c. x2 + 20x + 100

b. Product is +64

c. Product is +100

Solution a. Product is +25 1 5

-1 -64 -2 -32 -4 -16 -8 -8 = sum

25 5 = sum

1 2 4 5 10

100 50 25 20 10 = sum

∴ ( x + 5)( x + 5)

∴ ( x – 8)( x – 8)

∴ ( x + 10)( x + 10)

CHECK: ( x + 5)( x + 5) = x2 + 5x + 5x + 25 = x2 + 10x + 25 

CHECK: ( x – 8)( x – 8) = x2 – 8x – 8x + 64 = x2 – 16x + 64 

CHECK: ( x + 10)( x + 10) = x2 + 10x + 10x + 100 = x2 + 20 x + 100 

Look again at the above perfect square trinomials of the form x 2 + bx + c . What do you notice about the numbers that were selected? How are the values of b and c related? Sharpe Mathematics  2017

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Try the following exercises to see how well you recognize the patterns. 1. Expand each binomial: a. ( x + 2)2

b. ( x − 3)2

c. ( x + 6 )2

d. ( x − 7 )2

e. ( x + 9 )2

2. Factor each perfect square trinomial: a.

x2 + 2 x + 1 =(

x 2 + 8 x + 16

b.

)2

=(

)2

x 2 − 14 x + 49

c.

=(

)2

x 2 + 18 x + 81

d.

=(

)2

x 2 − 5 x + 25 4

e.

=(

)2

3. To continue our investigation, complete the following by determining the values that would make each expression a perfect square trinomial: a. x 2 + 4 x + ___ =(

b. x 2 + 12 x + ___

)2

f. x 2 + 6 x + ___ =(

=(

)2

=(

g. x 2 − 20 x + ___

)2

=(

c. x 2 + 10 x + ___

)2

)2

h. x 2 + 3 x + ___ =(

)2

d. x 2 − 14 x + ___ =(

)2

i. x 2 + 7 x + ___ =(

e. x 2 + 16 x + ___ =(

)2

j. x 2 − 1.2 x + ___

)2

=(

)2

From the previous examples, there is a very distinct pattern that emerges when factoring perfect square trinomials of the form x 2 + bx + c : 2

b To determine the value of c, divide the value of b by 2 and square the result. Thus c =   .  2

This process of identifying a number that produces a perfect square trinomial is referred to as “completing the square”, since we are completing the trinomial to produce a perfect square. 4. Fill in the blanks below by completing the square.

x2 + 4 x

a.

d.

x 2 + 12 x

b.

x 2 − 10 x

c.

=( x 2 + 4 x + ___ ) – ___

=( x 2 + 12 x + ___ ) – ___

=( x 2 − 10 x + ___ ) – ___

=(

=(

=(

)2 – ___

2 x2 + 4 x

e.

)2 – ___

3 x 2 − 18 x

f.

)2 – ___

−2 x 2 + 7 x

= 2( x 2 + 2 x + ___ ) – ___

= 3( x 2 − ___ x + ___ ) – ___

= –2( x 2 − ___ x + ___ ) – ___

= ___(

= ___(

= ___(

)2 – ___

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Chapter 3: Quadratic Functions

3.3A USING COMPLETING THE SQUARE TO CHANGE FORMS Now that we know how to identify the number needed to complete the square, we can now investigate converting a quadratic equation from Standard Form to Vertex Form.

SF  VF To convert from Standard Form to Vertex Form, we need to complete the square. Let’s examine a few examples to see how the process can be applied. EXAMPLE 5 Convert the function y =x 2 + 10 x − 7 from Standard Form to Vertex Form. Solution y =x 2 + 10 x − 7

 SF

y= ( x + 10 x + ____) − 7 + ____ 2

( )

2

y= ( x 2 + 10 x + ____) 25 − 7 + ____ –25  Add and balance out the value 10 2 = 25 y =( x + 5)2 − 32

 VF

EXAMPLE 6 Convert the function y = 2 x 2 + 12 x + 23 from Standard Form to Vertex Form. Solution y = 2 x 2 + 12 x + 23

 SF

y= 2( x 2 + 6 x + ____) + 23 + ____  Factor out "2" from the first two terms

( )

2

y= 2( x 2 + 6 x + ____) 9 + 23 + ____ –18  Add and balance out the value 62 = 9 y = 2( x + 3)2 + 5

 Simplify to put in VF

Why is this number “–18” and not “–9”?

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PRACTICE 53.

54.

55.

Determine the value of ‘b’ or ‘c’ needed to make each expression a perfect square trinomial. a. x 2 + 4 x + c

b. x 2 + bx + 4

c. x 2 − 12 x + c

d. x 2 + bx + 25

e. x 2 + 14 x + c

f. x 2 + bx + 8

g. x 2 + x + c

h. x 2 + bx + 49 4

i. 2 x 2 + 4 x + c

j. x 2 + bx + 11.56

k. 3 x 2 + 12 x + c

l. x 2 + bx + 2

Use the Complete the Square method to convert from Standard Form to Vertex Form. a. y = x 2 + 2 x − 5

b. = y x2 − 6 x

c. y =x 2 − 14 x + 45

d. y = x 2 − 4 x + 9

e. y = x 2 − 3 x − 1

f. y = x 2 + x − 12

g. y = x 2 + 9 x + 14

h. y = x 2 − 5 x + 3

i. y =− x 2 + x + 6

Use the Complete the Square method to convert from Standard Form to Vertex Form. a. y = 2 x 2 − 12 x + 16

b. y = −3 x 2 + 24 x − 42

c. y = 21 x 2 − 4 x + 4

d. y = −2 x 2 + 14 x − 13

e. = y 3 x2 − 6 x

f. y = 3 x 2 − 8 x + 2

g. y = 2 x 2 + 9 x + 18

h. y = 2 x 2 − 5 x + 2

i. y = 3 x 2 + 8 x − 3

j. y = −5 x 2 + 13 x − 6

k. y = 2 x 2 − 11 x + 5

l. y= x ( x + 5) + 2 ( x + 6 )

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Chapter 3: Quadratic Functions

In the previous examples we learned how to use the method of Completing the Square to change an equation from Standard Form to Vertex Form. There is another way we can do the same thing but using a completely different method. This alternate method requires using your knowledge of the characteristics of a quadratic function to determine the answer.

 Alternate Method to convert from Standard to Vertex Form  STEP 1: Determine the x-coordinate of the vertex (ie. the axis of symmetry) of the parabola using b x= − 2a STEP 2: Plug this value back into the original function to determine the y-coordinate of the vertex. STEP 3: Identify the “a” value from the original function in Standard Form. This will be the same value of “a” for the function when expressed in Vertex Form. To see this alternate method of conversion in action, let’s re-visit the previous example… EXAMPLE 7 Convert the function y = 2 x 2 + 12 x + 23 from Standard Form to Vertex Form. Solution STEP 1: Determine the x-coordinate of the vertex: b 12 x= − = − = −3 2a 2 ( 2) STEP 2 : Determine the y-coordinate of the vertex: y = 2( −3)2 + 12 ( −3) + 23 y = 18 − 36 + 23 y=5 Thus, the vertex is ( −3, 5). STEP 3 : Identify the a value of y = 2 x 2 + 12 x + 23. When compared to y = ax 2 + bx + c we have a = 2. Thus, the Vertex Form of the equation is y = 2( x + 3)2 + 5.

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PRACTICE 56.

57.

Convert each function from Standard Form to Vertex Form using the Alternate Method. a. y = x 2 + 6 x + 7

b. y = x 2 + 8 x + 12

c. y = x 2 − 2 x − 5

d. y = x 2 + 4 x + 6

e. y =x 2 − 10 x + 20

f. y =x 2 + 12 x − 30

g. y =x 2 − 14 x − 38

h. y = x 2 + 3 x + 4

i. y = x 2 + 5 x + 4

Convert each function from Standard Form to Vertex Form using the Alternate Method. a.

y = 2 x 2 + 12 x + 15

d. y = 5 x 2 + 10 x − 13

58.

b. y = 2 x 2 − 16 x + 29

−3 x 2 + 18 x − 20 c. y =

−2 x 2 + 10 x − 17 e. y =

f. y = 21 x 2 − 4 x + 10

A rock is thrown up in the air and its height, h, in meters, is indicated by the function h( t ) = −2t 2 + 12t , where t is the time in seconds. 24

a. Express the function in Vertex Form.

22 20

b. What was the intial height of the rock?

18 16 14

c. When did the rock reach its maximum height?

12 10 8

d. What was the maximum height?

6 4 2

e. Sketch the path of the rock on the grid. Sharpe Mathematics  2017

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2

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Chapter 3: Quadratic Functions

3.3B MAXIMUM / MINIMUM WORD PROBLEMS EXAMPLE 8 A farmer wants to enclose a rectangular pen that is to have a fence on three sides and a river on the fourth side. If he has 1000 m of fencing, what are the dimensions of the largest possible pen that can be constructed? Solution

River

STEP 1: Draw and label a diagram to represent the situation. w

w

STEP 2: Identify two equations using the given information.

l

Equation 1: Since there are 1000 m of fencing, then P=1000, where P=2w + l Thus: 2 w + l = 1000  Equation 2: Since we want to construct the largest possible pen, we want the pen with the largest area. Thus: Area = w x l



STEP 3: Solve the perimeter equation  for one of the dimensions and substitute into the Area equation  (i.e plug  into ). 2w + l = 1000  l = 1000 – 2w

Area = w x l = w x (1000 – 2w) = 1000 w – 2w 2 = –2w 2 + 1000w 

STEP 4: Determine the axis of symmetry of  to identify when the maximum will occur. b 1000 1000 w= − = − = − = 250 −4 2a 2( −2) Thus, the maximum pen will occur when the width, w, is 250 m. STEP 5: Use the value from the previous step to determine the other dimension. l =1000 – 2w  = 1000 – 2(250) = 1000 – 500 = 500 Thus, the maximum pen will occur with dimensions w = 250 m and l = 500 m. Sharpe Mathematics  2017

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133

PRACTICE 59.

Two numbers have a sum of 64. Identify these numbers if their product is a maximum.

60.

Determine two numbers that differ by 28 and whose product is a minimum.

61.

Determine the maximum product of two numbers whose sum is 18.

62.

Find two non-negative numbers whose sum is 9 and the sum of their squares is a minimum.

63.

Two numbers have a sum of 14. Find the numbers if the sum of their squares is a minimum.

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Chapter 3: Quadratic Functions

64.

A farmer wants to enclose a rectangular pen that is to have a fence on three sides and a barn on the fourth side. If he has 500 m of fencing, what are the dimensions of the largest possible pen that can be constructed?

65.

You have 50 m of fencing and a large field with which to construct a rectangular playground area. What are the dimensions of the largest such yard?

66.

Joshua has a 400 m roll of fencing to enclose his rectangular garden. If he wishes to build a fence around the outside, as well as build an interior fence to split his garden in half, what are the dimensions of the largest such enclosure?

67.

A farmer has to build three rectangular pastures of the same size for his cow, his horse, and his sheep and he only has 192 m of fencing. Determine the dimensions of his pastures so that the area is a maximum.

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135

68.

The owner of a ranch decides to enclose a rectangular region with 140 m of fencing. To help the fencing cover more land, he plans to use one side of his barn as part of the enclosed region. What is the maximum area the rancher can enclose?

69.

A rectangular lot is bounded on one side by a river, and one the other three sides by 80 m of fencing. Determine the dimensions of the largest possible lot.

70.

A farmer wishes to enclose a rectangular region bordering a river using 600 m of fencing. He wants to divide the region into two equal parts using some of the fence material. What is the maximum area that can be enclosed with the fencing?

71.

A local grocery store has plans to construct a rectangular parking lot on land that is bordered on one side by a highway. There are 1280 m of fencing available to enclose the other three sides. Determine the dimensions that will maximize the area of the parking lot.

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136 72.

Chapter 3: Quadratic Functions A rancher has 1200 m of fencing to enclose two adjacent rectangular regions of equal lengths and widths as shown in the figure. What is the maximum area that can be enclosed in the fencing?

x

x y

73.

Dave wishes to build a rectangular pen for his dogs with three parallel partitions as shown in the diagram. If he has 500 m of fencing, what dimensions will maximize the total area of the pen?

74.

A rectangular parcel with a square base is to be sent in the mail. The sum of all the edges of the parcel measures 140 cm. What dimensions will produce a parcel with the largest possible surface area?

y

y

x

75.

x

A Norman window is a window in the shape of a rectangle with a semicircle on top. If the window has a perimeter of 6 metres, what dimensions will produce the largest window?

y

x

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MAXIMUM REVENUE PROBLEM! EXAMPLE 9 Mr. Lewis is collecting names of students who are interested in buying a school t-shirt. He has been told by the manufacturer that the t-shirts will each cost $20 if he places an order for 300 shirts. If he is unable to generate enough interest, however, the price of each t-shirt will increase by $5 for every 30 fewer students willing to buy a shirt. What selling price will produce the maximum revenue? What will be the maximum revenue? Solution STEP 1: When solving Revenue problems, you need to use the Revenue equation: REVENUE = (SALE PRICE) x (NUMBER SOLD) When the price of each shirt is $20, he is able to sell 300 units. Initial Revenue = (20) x (300) Let x = number of shirts For every 30 students who do not buy a shirt (–30 x), the unit price of each shirt will increase by $5 (+5x).

( SALE PRICE ) = ( 20 + 5 x )

Revenue = =

× ×

( NUMBER SOLD ) ( 300 − 30 x )

6000 − 600 x + 1500 x − 150 x 2 = − 150 x 2 + 900 x + 6000

STEP 2: Identify the axis of symmetry to determine when the maximum will occur. 900 900 b x= − = − = − = 3 2a 2 ( −150 ) −300 Thus, the maximum revenue will occur when x = 3 . STEP 3: Use the value x = 3 to determine the resulting sale price and number sold. SALE PRICE = 20 + 5x = 20 + 5(3) = 20 + 15 = 35

NUMBER SOLD = 300 – 30x = 300 – 30(3) = 300 – 90 = 210

Thus, the Maximum Revenue will occur with sales of 210 shirts at $35 each and the Maximum Revenue generated will be 210 x $35 = $7350.

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Chapter 3: Quadratic Functions

76.

A school auditorium can seat 600 people at $10 per ticket. The principal estimates that 50 less people will attend if prices are increased by $1. What ticket price would result in the greatest revenue? What is the maximum revenue possible?

77.

A ferry service carries about 2000 passengers per day for a cost of $0.90. A survey indicates that for every $0.05 decrease in the ticket price, the number of passengers will increase by 200 per day. What ticket price will produce the greatest revenue? What is the maximum profit?

78.

A company sells 750 games each month at a price of $34.50 each. They believe that sales will increase by 125 each month for every $2.50 decrease in price. What price will maximize the revenue generated?

79.

A chocolate bar sells for $1.10 with monthly sales averaging 140 bars. The distributor estimates that for every 5 cent increase in price, sales will decline by 5 bars. Determine the maximum revenue.

80.

The local bakery sells an average of 1000 muffins each week at 50 cents each. The owner of the bakery would like to increase revenue, and has determined that for each 10-cent increase in price, the bakery will sell 40 fewer muffins each week. At what price will the muffins bring in maximum revenue?

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CHAPTER

4

139

QUADRATIC EQUATIONS Contents..............................................................

4.1

SOLVING QUADRATIC EQUATIONS BY GRAPHING ............................... 140

4.2

SOLVING QUADRATIC EQUATIONS BY FACTORING ............................. 148 A. FACTORING REVIEW .................................................................... 148 B. SOLVING QUADRATIC EQUATIONS BY FACTORING ........................ 154

4.3

SOLVING QUAD EQUATIONS BY COMPLETING THE SQUARE ................ 156

4.4

A. THE QUADRATIC FORMULA .......................................................... 159 B. THE NATURE OF ROOTS OF A QUADRATIC EQUATION .................. 164 C. WORD PROBLEMS USING QUADRATIC EQUATIONS ....................... 169 D. THE SUM AND PRODUCT OF ROOTS *ENRICHMENT* ................... 174

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Chapter 4: Quadratic Equations

4.1 SOLVING QUADRATIC EQUATIONS BY GRAPHING Much of the last chapter focused on using the given quadratic equation to sketch the graph of the function by identifying several key characteristics, including the vertex, y-intercept, axis of symmetry and the vertical stretch factor. In this section we will now reverse the process by identifying the equation from the given graph. We will use these same characteristics and work backwards to identify the function graphed. As we saw in the previous unit, quadratic equations of the form y = ax 2 + bx + c are said to be in STANDARD FORM. If we wish to solve the equation, however, we would have to examine ax 2 + bx + c = 0 to determine the values of x that result in y = 0 . The values that satisfy this relationship are the solutions or roots or zeros of the equation. You already know of several methods in which to solve a quadratic equation which shall be reviewed here. We will also introduce you to other methods which will be new to you. A quadratic equation of the form y = ax 2 + bx + c is said to be in STANDARD FORM. The x-intercepts of a function are also referred to as the roots or zeros of the function. EXAMPLE: A.

Examine the following graphs and identify the equation of each function. B.

10

9 8 7

8

6

6

5 4

4

3

2

2 1

2

4

6

8

-9

-2

-8

-7

-6

-5

-4

-3

-2

-1

1

2

3

-2

-4

-3 -4

-6

-5

-8

-6 -7

-10

-8 -9

-12

a. Identify the vertex.

-1

-10

 ________

a. Identify the vertex.

 ________

b. Identify the axis of symmetry.  ________

b. Identify the axis of symmetry.  ________

c. Identify the y-intercept.

 ________

c. Identify the y-intercept.

 ________

d. Identify the vertical stretch.

 ________

d. Identify the vertical stretch.

 ________

e. Express the equation of each quadratic function in Vertex Form. SF: Sharpe Mathematics  2017

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Chapter 4: Quadratic Equations

141

PRACTICE 1. Identify the equation for each graph shown in Vertex Form. A.

B.

C.

D.

E.

F.

G.

H.

I.

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Chapter 4: Quadratic Equations

We will be primarily focused on an algebraic approach to determine the roots of a quadratic function. Let us first begin by examining the graphs of several functions and analysing the results that follow. EXAMPLE: Sketch the graphs of each function by identifying the characteristics listed. a. y = x 2 − 2 x − 3

-3

b. y = x 2 − 6 x + 9

c. y = x 2 + 2 x + 2

i. vertex

 ________

i. vertex

 ________

i. vertex

 ________

ii. y-int

 ________

ii. y-int

 ________

ii. y-int

 ________

-2

-1

11

11

11

10

10

10

9

9

9

8

8

8

7

7

7

6

6

6

5

5

5

4

4

4

3

3

3

2

2

2

1

1

1

-1

1

2

3

4

5

6

-3

-2

-1

-1

1

2

3

4

5

6

-3

-2

-1

-1

-2

-3

-3

-3

-4

-4

-4

-5

-5

-5

 ________

iii. How many roots does the function have?  ________

2

3

4

5

6

-2

-2

iii. How many roots does the function have?

1

iii. How many roots does the function have?  ________

iv. x-intercepts  ________

iv. x-intercepts  ________

iv. x-intercepts  ________

v. Based on the graph above, solve x 2 − 2 x − 3 = 0 .

v. Based on the graph above, solve x 2 − 6 x + 9 = 0 .

v. Based on the graph above, solve x 2 + 2 x + 2 = 0 .

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143

USING TECHNOLOGY You could use the graphing calculator to determine the roots of a quadratic equation of the form ax 2 + bx + c = 0 . To do so, we must enter the function as Y1 and graph it. For example, if we wish to use the graphing calculator to determine the x-intercepts for the previous example x 2 + 3 x − 10 = 0: •

Press Y = and enter the function x 2 + 3 x − 10

•

Adjust the window so that you can see the x-intercepts displayed on the graph.

As you can see, there are two x-intercepts displayed. We will first obtain the value of the left-most intercept. •

To determine the left x-intercept, press 2nd TRACE 2

•

The calculator will respond with “Left Bound?”. Use the   arrows to position the cursor so that it is anywhere to the left of the x-intercept. Press ENTER .

•

The calculator will now respond with “Right Bound?”. Use the   arrows to position the cursor so that it is to the right of the x-intercept. Press ENTER .

•

The calculator will now respond with “Guess?”. Press ENTER to proceed to the answer.

As a result, we have determined that x = −5 is one of the xintercepts. To find the remaining x-intercept, we repeat the above procedure by selecting points on the left and right of the other x-intercept, namely x = 2 . Thus, the x-intercepts of the function y = x 2 + 3 x − 10 can be determined by finding the roots of the equation x 2 + 3 x − 10 = 0 are x = −5 and x = 2 . Sharpe Mathematics  2017

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Chapter 4: Quadratic Equations

If the quadratic function is given in the form y = ax 2 + bx + c , the x-intercepts can be determined by letting y = 0 and then solving the equation ax 2 + bx + c = 0.

EXAMPLE 1 Solve each quadratic equation by graphing. x2 − 2 x − 8 = 0

a.

x2 + 4 x + 4 = 0

b.

Solution a. Sketch the graph of y = x 2 − 2 x − 8

b. Sketch the graph of y = x 2 + 4 x + 4

8 6 4 2

8 6 4 2

-8 -6 -4 -2-2

2 4

-8 -6 -4 -2-2 -4 -6 -8

6 8

-4 -6 -8

The graph crosses the x-axis at –2 and 4. ∴ the solutions are x=–2 and x=4.

2 4

6 8

The graph crosses the x-axis at –2. ∴ the solution is x=–2 (double root).

PRACTICE 2. Identify the roots of the quadratic equations represented by the graphs of each function. A.

B.

C.

D.

6

6

6

6

5

5 4

5 4

5

3

3

3

4 3 2 1 -4 -3 -2 -1 -1

2 3

4

-4 -3 -2 -1 -1

2 1

2 1

2 1 1

4

1

2 3

4

-4 -3 -2 -1 -1

1 2 3

4

-4 -3 -2 -1

-1

-2

-2

-3

-3

-3

-2 -3

-4

-4

-4

-4

-5

-5

-5

-5

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Chapter 4: Quadratic Equations 3.

145

Solve each equation by graphing the corresponding function. a. 0 = x 2 + 9 x + 14

b. = 0 3 x2 − 6 x

c. 0 = − p2 + 6 p − 5

d. 0 = m 2 + 4 − 4 m

e. − 21 x 2 − 5 x − 8 = 0

f. 0 =x 2 − 14 x + 49

g. x 2 = 9

h. − x 2 + x + 6 = 0

i. v 2 + 4 v − 5 = 0

j. −k 2 − 6 k = 0

2 k. 2r= 2r + 12

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l. − 1 t 2 − 4 t = 0 2

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146 4.

Chapter 4: Quadratic Equations Determine the roots of each equation. a. 9d 2 + 12d + 4 = 0

b. 2 x 2 − 3 x − 5 = 0

c. 7 x − 6 x 2 = 0

d. 2 x 2= x + 6

e. 3w 2 − w = 2

f. 6 x − 16 = − 41 x 2

g. 5b 2= 6 − 7b

h. 4 x 2 + 4 x − 3 = 0

i.

j. 2 x 2 + 7 x − 4 = 0

k. 0 = 12k − 2k 2 − 16

l. x 2 + 43 x − 41 = 0

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− 31 n2=

9 − 4n

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147

A rock is thrown from the top of a house. The height of the rock, h metres above ground, t seconds after it was thrown, is modelled by the equation = h 20 − 5t 2 . a. Graph the quadratic function. b. When did the rock hit the ground?

c. What is the domain?

d. What does the domain represent?

6.

A football is kicked from ground level and its height, h metres above ground, t seconds after it was kicked, is modelled by the equation h = −3t 2 + 18t . a. Graph the quadratic function. b. When did the football hit the ground?

c. What is the domain?

d. What does the domain represent?

7.

Determine whether each quadratic function has two real roots, a double root, or no real roots. a. = b. y = c. y = d. y = −2 x 2 + 5 x + 3 − x2 − 4 x + 4 y x2 + 2 x − x2 − 4

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Chapter 4: Quadratic Equations

4.2 SOLVING QUADRATIC EQUATIONS BY FACTORING A. FACTORING REVIEW Your skills with factoring are essential for this unit. Many of the problems encountered involve applying the various types of factoring to simplify the given expression. There are four main types of factoring with which you should be familiar: (i) Extracting the greatest common factor (GCF) (ii) Difference of Squares ( x 2 − a 2 ) (iii) Trinomials of the form x 2 + bx + c (iv) Trinomials of the form ax 2 + bx + c We will first briefly review the types of factoring to assist with solving quadratic equations. (i)

EXTRACTING THE GREATEST COMMON FACTOR (GCF) With this type of factoring we can remove a common monomial factor. The common factor may be just a number or contain a variable(s). By removing the Greatest Common Factor, no other factors can be removed from the polynomial, other than 1 or –1. EXAMPLE 2 Factor each of the following. a. 3 x − 12

b. 5 x 2 y 5 + 6 x 3 y 4 − 8 x 6 y 3

c. 15 x 4 –   12 x 3 + 18 x 2

b. 5 x 2 y 5 + 6 x 3 y 4 − 8 x 6 y 3

c.

Solution a.

3 x − 12

GCF = x 2 y 3

GCF = 3

=3( x − 4)

15 x 4 –   12 x 3 + 18 x 2

= x 2 y 3 ( 5 y 2 + 6 xy − 8 x 4 )

GCF = 3 x 2

= 3 x2 ( 5 x2 − 4 x + 6)

PRACTICE 8.

Factor. a. 18 x − 12

b. 24 y + 8

c. 5m − 20

d. −3w − 15

e. 3 x 2 + 2 x

f. 2r 2 − 9 r

g. 3m − 15n

h. 5 x 4 − 40 x 3

i. 24 x 4 − 6 x 3 + 12 x 2

j. 3 y 5 − 9 y 6 + 12 y 7

k. 5 p3 − 10 p 2 + 35 p

l. 16a 3 b 2 − 12a 2 b3 + 20 ab 4

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149

DIFFERENCE OF SQUARES ( x 2 − a 2 ) This type of factoring occurs when we can factor with two binomial products of the form (x + a)( x – a). Using the FOIL method we get: (x + a)( x – a) = x 2 – ax + ax – a2 = x 2 – a2 For example: (x + 5)( x – 5) = x 2 – 5x + 5x – 52 = x 2 – 25 To factor a difference of squares, you can think of taking the square root of each term. For example:

4 x 2 − 49 = ( 4 x 2 + 49 )( 4 x 2 − 49 ) = (2 x + 7)(2 x − 7)

EXAMPLE 3 Factor each of the following. a. x 2 – 36

b. 9x2 – 16y2

c. 8x – 32x3

b. (3x + 4y)(3x – 4y)

c.

Solution a. (x + 6)(x – 6)

8x (1 – 4x2) = 8x (1 + 2x)(1 – 2x)

PRACTICE 9.

Factor completely. a. x2 – 9

b. a2 – 36

c. p2 – 100

d. 4x2 – 1

e. 49m2 – 16

f. 25x2 – 9y2

g. x2 – 16y2

h. 63 – 7t2

i. 48x2y – 12yz2

j. 25 – 100k2

k. 36u2 – 9v4

l. f 4 – 16g4

m. 7x2y2 – 28x2z2

n. 3p2 – 27q2

o. 81m2 – 9n2

p. 16a2 – 64b2

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Chapter 4: Quadratic Equations TRINOMIALS OF THE FORM x 2 + bx + c When factoring trinomials of this type, we try to find two numbers that multiply to give “c” and add to give “b”. By listing off the possible products to give “c”, we examine the various combinations to find the pair that has a sum of “b”. For example, to factor x 2 + 11 x + 24 , list the different ways to multiply to get 24:

Product is 24 =

1 24 2 12 3 8 4 6

Sum is +11 ∴ x 2 + 11 x + 24 =( x + 3)( x + 8)

EXAMPLE 4 Factor each of the following. a. x 2 + 9 x + 18

b. x 2 − 14 x + 40

c. x 2 + 16 x − 36

b. Product is +40

c. Product is –36

Solution a. Product is +18 1 18 2 9 3 6 = sum

-1 -2 -4 -5

-40 -20 -10 = sum -8

-1 36 -2 18 = sum -3 12 -4 9 -6 6

∴ (x + 3)(x + 6)

∴ (x – 4)(x – 10)

∴ (x – 2)(x + 18)

CHECK: (x + 3)(x + 6) = x 2+6x +3x +18 = x 2+9x +18 

CHECK: (x – 4)(x – 10) = x 2–10x –4x +40 = x 2 – 14x +40 

CHECK: (x – 2)(x + 18) = x 2+18x –2x –36 = x 2 + 16x – 36 

PRACTICE 10.

Factor completely. a. x 2 + 12x + 20

b. m2 + 9m + 20

c. x 2 + 6x – 27

d. p 2 – 8p + 7

e. x 2 + x – 42

f. y2 – 11y + 28

g. x 2 + 10x + 24

h. w2 – 4w – 12

i. t2 – 16t – 36

j. 2x 2 + 4x – 160

k. n4 – 4n 3 – 32n 2

l. 3x 3– 36x 2 + 105x

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151

TRINOMIALS OF THE FORM ax 2 + bx + c Factoring of this type of trinomial can be thought of in a similar way as above. To see how to factor these, let’s work backwards by using the FOIL method: (2x+3)(x – 5) = (2x)(x) + (2x)( –5) + (3)(x) + (3)( –5) = 2x 2 – 10x + 3x – 15 = 2x 2 – 7x – 15 –10 and +3 have a:

sum = –7

and

product = –30

Thus, working backwards to factor 2x 2 – 7x – 15: Product = –30 2 x 2 – 7x – 15 Sum = –7

Product = –30 Sum = –7

= 2x 2 – 10x + 3x – 15 = (2x 2 – 10x) + (3x – 15) = 2x(x – 5) + 3(x – 5) = (x – 5)(2x +3)

–10 and +3

“Decompose” –7x Group each pair of terms Factor each group Factor out (x – 5)

This is known as the method of DECOMPOSITION, where we examine the possible numbers that multiply to give “ac” and add to give “b”. 12 For example, to factor 2x 2 + 7x + 6 , list the different ways to multiply to get 2x6 =12:

Product is 12 =

1 2 3

12 6 4

Sum is +7

+4 and +3

∴ 2x 2+7x + 6 = 2x 2+ 4x + 3x +6 = (2x 2 + 4x) + (3x + 6) = 2x(x + 2) + 3(x + 2) = (x + 2)(2x + 3) There is another useful method of factoring known as the “AUSTRALIAN METHOD” that can also be used. The method below is similar to decomposition, except when factoring the expression ax 2 + bx + c , the value of “ax” is added to each individual value, then divided by the value of “a”. Sharpe Mathematics  2017

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Chapter 4: Quadratic Equations 12

For example, to factor 2x2+7x+6 , again list the different ways to find the possible combination of numbers that multiply to give +12 and add to give +7: +4 and +3 This time, however, start by listing the two numbers: Place the term "ax" in front of each number: Divide the expression by the value of "a" (=2): Simplify where possible:

 +4 +3  (2 x + 4 )(2 x + 3) (2 x + 4 )(2 x + 3)  2  ( x + 2)(2 x + 3)

EXAMPLE 5 Factor each of the following. a. 2x2+7x+3

b. 4x2 + x – 3

c. 3x2 – 17x – 6

b. Product = –12 Sum = +1 ∴ +4 and –3

c. Product = –18 Sum = –17 ∴ –18 and +1

Solution Using Decomposition: a. Product = +6 Sum = +7 ∴ +6 and +1

                             2 x2 + 7 x + 3 4 x 2 + x – 3                            = 2 x 2 + 6 x + 1 x + 3                   = 4 x 2 + 4 x – 3 x – 3                   = =

(2x

2

+ 6 x ) + ( 1 x + 3 )            

=

(4x

2

+ 4 x ) + ( –3= x             –3  )

= 4 x ( x + 1) – 3 ( x + 1             ) 2 x ( x + 3 ) + 1 ( x + 3 )               =( x + 3 )( 2 x + 1)                      =( x + 1)( 4 x – 3 )                       Solution Using Australian Method:

=

3 x 2 – 17 x –    6 2 3 x – 18 x + 1 x – 6

(3x

2

– 18 x ) + ( 1 x – 6 )

= 3 x ( x – 6 ) + 1( x – 6 )  – =( x 6 )( 3 x + 1)

                                                         a. As before, use +6 & +1 b. As before, use +4 & –3 c. As before, use –18 & +1

(

)(

)

(

)(

)

2x + 2x + 4x + 4x + = =                     2 4

(

)(

)

(

)(

)

( 3 x + )( 3 x + ) (

3

)(

2x + 6 2x + 1 4 x + 4 4 x + −3 3 x + −18 3 x + 1 =                            = 2 4 3 4 ( x + 1)( 4 x − 3 ) 3 ( x − 6 )( 3 x + 1) 2 ( x + 3 ) ( 2 x + 1) = =                                     = 2 4 3  =( x + 3 )( 2 x + 1) =( x + 1)( 4 x − 3 ) =( x − 6 )( 3 x + 1)

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PRACTICE 11.

Factor completely. a. 3x2 – 16x + 20

b. 5m2 – 7m + 2

c. 2x2 + 5x – 7

d. 3p2 – 10p + 3

e. 6x2 + 11x + 5

f. 2y2 – 11y + 12

g. 4x2 + 8x + 3

h. 6w2 – w – 40

i. 10t2 + 51t + 27

j. 14x2 + 49x + 42

k. 36k4 – 48k3 – 20k2

l. 10x3y – 55x2y + 60xy

We will now combine the various types of factoring to ensure that you can first identify the type of factoring involved, and then choose an appropriate method to factor. Some questions may involve several types of factoring, so make sure that your answer is expressed in simplest form.

PRACTICE 12.

Factor completely. a. 5 x 2 –   20

b. m 4 + 5m 2

c.

y 2 + 2 y – 15  

d. 4 x 2 + 20 x + 16  

e.

25a 2 –  1

f.

6 t 2 − 5t – 21  

g.

x 2 – 49

h.

r 2 + 16 r + 64  

i.

4 x3 + 8 x 2 + 4 x

j.

  y2 – 5 y – 6

k.

–12n2 – 34 n – 10  

l.

x 4 – 81

m. 24 x 3 + 68 x 2 + 48 x

n.

48c 2 – 200 c + 200

o. x 2 – 3 x + 2

2a 2 + 17a + 21

q.

16c 2 – 100 d 2

r. 30 x 2 + 21 x – 36

p.

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Chapter 4: Quadratic Equations

B. SOLVING QUADRATIC EQUATIONS BY FACTORING As we have seen, if we wish to solve the equation we would have to examine ax 2 + bx + c = 0 to determine the values of x that result in y = 0 . The values that satisfy this relationship are the solutions or roots or zeros of the equation. Our skills with factoring can be used to determine the roots of a quadratic equation. A useful property when using this method is known as the “Zero Product Property”, which states that if the product of two numbers is zero, then at least one of them must be zero. ZERO PRODUCT PROPERTY If a× b=0, then a=0 or b=0, or both, where a and b are real numbers. EXAMPLE 6 Determine the roots of each quadratic equation. a. x 2 − 10 x + 16 = 0

b. 2 x 2 − 5 x − 3 = 0

c. x 2 + 3 x = 10

Solution a.

x 2 − 10 x + 16 = 0 ( x − 8)( x − 2) = 0 x= − 8 0 , x= −2 0 = x 8= , x 2

b.

2 x2 − 5 x − 3 = 0 (2 x + 1)( x − 3) = 0 2 x= + 1 0 , x= −3 0 2x = 3 −1 , x = x= −

c.

x2 + 3 x = 10

x 2 + 3 x − 10 = 0 ( x − 2)( x + 5) = 0 x= − 2 0 , x= +5 0

1 2

x = 2 , x = −5

PRACTICE 13.

14.

Use factoring to determine the roots of each quadratic equation. a. ( x − 4 )( x + 7 ) = 0

b. ( x − 2)( x + 2) = 0

c. 2 x ( x + 8) = 0

d. 3( x − 1)( x + 5) = 0

e. (2 x + 1)( x − 3) = 0

f. 5(3 x − 2)( x − 6 ) = 0

g. x 2 + 9 x + 14 = 0

h. 3 x 2 − 6 x = 0

i. x 2 + x = 12

j. x 2 + 2 x − 63 = 0

k. x 2 = 25

l. x 2= x + 6

Use factoring to determine the zeros of each quadratic function. a. y = 2 x 2 + 9 x − 18

b. y = 2 x 2 − 5 x + 2

c. y = 3 x 2 + 8 x − 3

d. y = −5 x 2 + 13 x + 6

e. y = 2 x 2 − 11 x + 5

f. y = x ( x + 5) + 2( x + 6)

g. x 2 − 3 x + y = 70

h. x 2 + 4 = 4 x + y

i. x 2 =+ y 14 x − 49

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In the previous examples, we have been given a quadratic equation and asked to determine its roots. Using our knowledge of functions, we are also able to determine an equation if the roots are known. By working backwards, we can create an equation using only its roots or zeros by using the fact that if x = r is a root, then ( x − r ) is a factor of the equation. Hence, if 0. an equation has roots x = r1 and x = r2 , then a resulting equation is ( x − r1 )( x − r2 ) = EXAMPLE 7 1 2 b. 2 + 3 and 2 − 3

a. 3 and −

Find a quadratic equation with the following roots:

SOLUTION

(

)

a. Since the roots are x = 3 and x = − 21 , then the factors are ( x − 3) and x + 21 . Thus we have the equation:

(

)

( x − 3) x + 21 = 0 ( x − 3)(2 x + 1) = 0  Multiply by 2 2 x2 − 5 x − 3 = 0

b. Since the roots are then the factors are

x= 2 + 3

and

( x − 2 − 3)

Thus we have the equation:

and

x= 2 − 3 ,

( x − 2 + 3) .

( x − 2 − 3 )( x − 2 + 3 ) = 0

x2 − 2 x + x 3 − 2 x + 4 − 2 3 − x 3 + 2 3 − 3 = 0  Distributive property x2 − 2 x + x 3 − 2 x + 4 − 2 3 − x 3 + 2 3 − 3 = 0  Simplify x2 − 4 x + 1 = 0

PRACTICE 15.

Determine a quadratic equation with the roots indicated. a. 2 and 3

b. 5 and –1

c. –2 and –4

d. 2 and 21

e. –3 and − 31

2 f. 21 and − 3

g. ± 2

h. ±2 3

i. −2 ± 6

j. 3 ± 5

k.

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l.

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Chapter 4: Quadratic Equations

4.3 SOLVING QUAD EQUATIONS BY COMPLETING THE SQUARE Earlier you had learned how to convert a quadratic expression into a perfect square trinomial. We can apply the same technique to solve a quadratic equation where we solve for x. EXAMPLE 8 Determine the zeros of the function y = x 2 + 6 x − 10 . Solution The zeros or x-intercepts of a function occur when y=0. Thus: 0 x 2 + 6 x − 10 =

 Let y=0

10 ( x 2 + 6 x + ____) + ____ =

 Move 10 to the RHS of the equation

( x2 + 6 x + 9 ) + − 9

( x + 3 ) =10 + 9 2 19 ( x + 3) = 2

x + 3 =± 19 x =−3 ± 19

()

2

 Add and balance out the value 62 10= 9 

Move 9 to the RHS of the equation

 Simplify  Take the square root of both sides  Move 3 to the RHS of the equation

PRACTICE 16.

17.

Solve for x. a. ( x + 5)2 = 2

b. ( x + 6 )2 = 11

c. ( x − 3)2 = 8

d. 2( x + 4 )2 = 10

e. 3( x − 1)2 = 12

f. 5( x − 2)2 = 3

Use completing the square to determine the x-intercepts of each quadratic function. a. y = x 2 + 9 x + 14

b. = y 3 x2 − 6 x

c. y = x 2 + x − 12

d. y = x 2 − 3 x − 1

e. y = x 2 + 4 − 4 x

f. y =x 2 − 14 x + 49

g. y = x 2 + 2 x − 5

2 h. x= 25 + y

i. x 2 + y = x + 6

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Let us examine another example but this time with a leading coefficient where a ≠ 1. EXAMPLE 9 Use completing the square to determine the x-intercepts of the function y = 2 x 2 + 7 x − 4 . Solution The zeros or x-intercepts of a function occur when y=0. Thus we have: 2 x2 + 7 x − 4 0 =  Let y 0 = 7 x2 + x − 2 = 0  Divide each term by leading coefficient 2 2 7 ( x 2 + x + ____) + ____ = 2  Move 2 to the RHS of the equation 2 2 49  49  2 7 7 ÷2 = 7 x x 2  Add and balance out the value + + − =  2 4 2 16  16 

(

2

49 7   x + 4  = 2 + 16  



Move

) ( )= 2

49 16

49 to the RHS of the equation 16

2

7 32 49   x + 4  =16 + 16   

Get a common denominator

2

7 81  x+ 4 = 16   7 81 = ± 4 16 7 9 x+ = ± 4 4 7 9 x= − ± 4 4 −7 + 9 −7 − 9 , x x = = 4 4 −16 2 , x x = = 4 4 1 , x = −4 x= 2 x+

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 Simplify  Take the square root of both sides  Simplify  Move

7 to the RHS of the equation 4

 Simplify

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Chapter 4: Quadratic Equations

PRACTICE 18.

Use completing the square to determine the zeros of each quadratic function. a. y = 2 x 2 + 9 x − 18

b. y = 2 x 2 − 5 x + 2

c. y = 3 x 2 + 8 x − 3

d. y = −5 x 2 + 13 x + 6

e. y = 2 x 2 − 11 x + 5

f. y = x ( x + 5) + 2( x + 6)

g. x 2 − 3 x + y = 70

h. x 2 + 4 = 4 x + y

i. x 2 =+ y 14 x − 49

19.

Compare your answers above with those of Question 14. Which method do you prefer?

20.

Use completing the square to solve each quadratic function. 1 2 x − 4x + 4 2

a. y = 2 x 2 − 12 x + 16

b. y = −3 x 2 + 24 x − 42

c. y =

d. y = −2 x 2 + 14 x − 13

e. y = 2 x 2 − 9 x + 2

f. 3 x 2 − 8 x + 2 = 0

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4.4A THE QUADRATIC FORMULA In this section we continue to examine another method for solving quadratic equations. As we have seen, to solve an equation is to identify the x-intercepts of the function, also known as its roots or zeros. Using the method of completing the square is a very efficient way to do this, but can at times be very laborious with some very nasty results. Let us now use the method of completing the square on any general quadratic function y = ax 2 + bx + c and examine the results. ax 2 + bx + c b x2 + x+ a ( x2 +  2 x + 

0 =  Let y 0  Divide each term by leading coefficient " a "

= 0

c x + ____) + ____ = − a x+

c t o the RHS of the equation a

  Add and balance out the value  

c = − a

 − 

 Move

2

2

   ÷ 2 =   =   

2

b  c  − +  x + 2a  = a  



Move

to the RHS of the equation



Get a common denominator

2

 x+ 

 −  = 

 x+ 

  = 

+

2

 Simplify

x+

= ±

 Take the square root of both sides

x+

= ±

 Simplify

x= −

±

−

±

x=

 Move

to the RHS of the equation

 Get a common denominator

Look back at your final answer above. What you have essentially done is to create a formula that will allow us to find the roots for any quadratic equation. The resulting formula is known as the QUADRATIC FORMULA, and is a very powerful and useful tool in solving equations as you shall soon discover.

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Chapter 4: Quadratic Equations

THE QUADRATIC FORMULA Given any quadratic equation ax 2 + bx + c = 0 with coefficients a, b and c, the roots of the equation are: x=

− b ± b 2 − 4 ac 2a

The quadratic formula can be used on any quadratic equation, and is especially useful when dealing with equations that have irrational roots.

EXAMPLE 10 Solve each equation using the quadratic formula. a.

2 x2 + 5 x − 3 = 0

b.

2 x2 − x + 4 = 0

Solution b. = a 2= , b –1= , c 4

a. = , b 5= , c –3 a 2=

− b ± b 2 − 4 ac x= 2a

− b ± b2 − 4 ac x= 2a

−5 ± 52 − 4 (2)( −3) x= 2(2)

x=

−( −1) ± ( −1)2 − 4 ( −2)( 4 ) 2( −2)

x=

1 ± 1 + 32 −4

x= x= x= x = x = x=

−5 ± 25 + 24 4 −5 ± 49 4 −5 ± 7 4 −5 + 7 −5 − 7 , x = 4 4 −12 2 x = 4 4 1 x = −3 2

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TRY THESE! Use the quadratic formula to solve each of the following equations. Express any irrational roots in simplest radical form. a. x 2 + 3 x − 10 = 0

b. 5 x 2 − 4 x − 2 = 0

c. 2 x − 8 =− x 2

d. 3 x 2 − 8 x = −2

Try solving the problems above using completing the square. Which method do you prefer? Sharpe Mathematics  2017

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Chapter 4: Quadratic Equations

PRACTICE 21.

Solve each quadratic equation using the quadratic formula. Express any irrational roots in simplest EXACT radical form. a. x 2 − 2 x − 2 = 0

b. x 2 − 4 x − 3 = 0

c. x 2 + 6 x + 7 = 0

d. x 2 + 1 = 4x

e. x 2 − 4 x + 2 = 0

2 f. 2 x= 8x − 5

g. 2 x 2 + 3 x − 4 = 0

h. 2 x 2 − 2 x − 3 = 0

i. 6 x 2 − 8 x = 0

k. x 2 − 2 2 x + 2 = 0

l.

j.

1 2 x + 11 x + 12 = 0 2

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22.

163

A toy rocket is fired into the air from the top of a house. Its height, h, above the ground in meters after t seconds is given by the function h( t ) = −5t 2 + 10 t + 20 .

23.

a. What was the initial height of the rocket?

a. What was the initial height of the diving board?

b. When did the rocket reach its max height?

c. What was the maximum height of the rocket?

24.

Joshua is on a diving board where his height, h, in meters above the water t seconds after he leaves the board is given by the function h( t ) = −4.9 t 2 + 8t + 5 .

b. When did Joshua reach his maximum height? Round to the nearest hundredth. c. What was Joshua’s maximum height? Round to the nearest hundredth.

d. How long was the rocket in the air before hitting the ground? Round to the nearest hundredth.

d. How long was Joshua in the air? Round to the nearest hundredth.

e. At what time(s) will the rocket be at a height of 22 m? Round to the nearest hundredth.

e. State the domain and range of the function.

A bottle rocket is launched from a patio. Its trajectory can be closely represented by the −0.01d 2 + 3d + 1 , where h is the rocket’s height in meters and d is the function h ( d ) = rocket’s horizontal distance in meters. What is the maximum height reached by the rocket?

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4.4B THE NATURE OF ROOTS OF A QUADRATIC EQUATION All of the examples encountered so far have involved the set of Real numbers ( x ∈ R ) . More specifically, when using the quadratic formula the number underneath the square root ( b2 − 4 ac ) has always been a non-negative Real number. In other words, the examples to date have always had b2 − 4 ac ≥ 0 . What would an equation look like otherwise, and how would we deal with it? EXAMPLE: Use the quadratic formula to solve the quadratic equation x 2 − 2 x + 5 = 0

a. What do you notice about your answer? 7

b. Identify the vertex and y-intercept.

6 5 4 3 2

c. The quadratic formula identifies the x-intercepts of a function. Graph the function y = x 2 − 2 x + 5 on the grid provided. What does this tell you?

1

1

-1

2

3

-1

2

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Let us now examine the graphs of several functions and analyse the results that follow. EXAMPLE: Sketch the graphs of each function by identifying the characteristics listed. y = x2 − 2 x − 3

a.

-3

y = x2 − 6 x + 9

b.

y = x2 + 2 x + 2

c.

i. vertex

 ________

i. vertex

 ________

i. vertex

 ________

ii. y-int

 ________

ii. y-int

 ________

ii. y-int

 ________

iii. x-int

 ________

iii. x-int

 ________

iii. x-int

 ________

-2

-1

11

11

11

10

10

10

9

9

9

8

8

8

7

7

7

6

6

6

5

5

5

4

4

4

3

3

3

2

2

2

1

1

1

-1

1

2

3

4

5

6

-3

-2

-1

-1

1

2

3

4

5

6

-3

-2

-1

-1

-2

-2

-2

-3

-3

-3

-4

-4

-4

-5

-5

-5

iv. How many roots does the function have?  ________ v. Calculate the value of b2 − 4 ac .  ________

iv. How many roots does the function have?  ________ v. Calculate the value of b2 − 4 ac .  ________

1

2

3

4

5

iv. How many roots does the function have?  ________ v. Calculate the value of b2 − 4 ac .  ________

Look at the roots of each function and the value of b2 − 4 ac . What conclusion can you make?

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166

Chapter 4: Quadratic Equations

In the quadratic formula, the expression b2 − 4 ac is known as the DISCRIMINANT. By calculating the value of the discrminant, we can deduce certain information about the roots of any quadratic equation of the form ax 2 + bx + c = 0. If we let D = b2 − 4 ac , we can make the following discoveries:

When D = 0

The roots are real and distinct.

The roots are real and equal.

EXAMPLE:

b. 2 x 2 − 8 x + 8 = 0

D=

9

-2

-1

The roots are non-real and distinct.

Calculate the discriminant of each equation, and compare the value of D with the graph of the corresponding function.

a. 2 x 2 − 8 x + 6 = 0

-3

When D < 0

When D > 0

c. 2 x 2 − 8 x + 10 = 0

D=

y = 2 x2 − 8 x + 6

9

D=

y = 2 x2 − 8 x + 8

9

8

8

8

7

7

7

6

6

6

5

5

5

4

4

4

3

3

3

2

2

2

1

1

1

-1

1

2

3

4

5

6

-3

-2

-1

-1

1

2

3

4

5

6

-3

-2

-1

-1

-2

-2

-2

-3

-3

-3

-4

-4

-4

-5

-5

-5

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y = 2 x 2 − 8 x + 10

1

2

3

4

5

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Chapter 4: Quadratic Equations

167

PRACTICE 25.

If a quadratic equation has the discriminant value below, determine if the equation will have two distinct roots, two equal roots, or two non-real distinct roots. a. 6

26.

c. –2

d. –7

e. 21

f. 0.24

Given the function y = ax 2 + bx + c , how many times will the graph intersect the x-axis if the discrminant has the value: a. 3

27.

b. 0

b. –5

c. 0

d. 21

e. –8

f. 0.6

Determine the value of the discriminant for each equation. Use it to describe the roots as real or non-real, and equal or distinct. a. x 2 + 2 x − 1 = 0

b. x 2 − 4 x + 4 = 0

c.

x2 + 2 x + 3 = 0

d. 4 x 2 − 4 x + 1 = 0

e. x 2 + 3 x + 9 = 0

f.

2 x2 + 3 x − 4 = 0

g. 9 x 2 + 12 x + 4 = 0

h. 2 x 2 − 4 x + 5 = 0

i.

6 x2 − 7 x = 0

k. 3 x 2 − x = 2

l.

11 x + 12 = − 21 x 2

2 m. 5 x= 8x − 6

n. −4 x 2 + 6 x + 3 = 0

o.

− 31 x 2 − 15 = 7x

p. x 2 + 6 x = 0

q. x 2 + 8 x + 1 = 0

r.

j.

s.

2 x 2= x + 6

2 x2 + 4 x − 8 = 0

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3 x2 + 5 x − 3 = 0

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2 x2 − 2 x + 2 = 0

u. − 3 x 2 + 6 x − 12 = 0

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168 28.

Chapter 4: Quadratic Equations Determine the values of c in the equation x 2 − 8 x + c = 0 such that the equation has: a. two distinct real roots

29.

b. two equal real roots

c. two non-real roots

Determine the values of d in the equation x 2 + dx + d − 1 = 0 such that the equation has: a. two distinct real roots

b. two equal real roots

c. two non-real roots

30.

For what values of k does the equation 4 x 2 + kx + 9 = 0 have non-real roots?

31.

Determine the value of h so that the quadratic equation hx 2 + (2h − 1) x + h = 0 has two equal real roots.

32.

If the equation ( t + 2) x 2 − tx − 2 = 0 has two equal real roots, determine the value of t.

33.

For what value(s) of k does the equation x 2 + ( k − 8) x + 9 = 0 have real and equal roots?

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169

4.4C WORD PROBLEMS USING QUADRATIC EQUATIONS EXAMPLE 11 Karen’s rectangular flower bed measures 10 m by 15 m in size. She plans on doubling its area by adding a strip of uniform width around the flower bed. Determine the width of the strip. Solution x

STEP 1: Draw and label a diagram to represent the situation.

x x

x

10 + 2x

10 x

x x

15

x

15 + 2x

STEP 2: Write an equation to represent the situation using the given information. Since we are considering doubling the area, we want the area of the garden. Thus: new area = double old area (15 + 2 x )(10 + 2 x ) =2 × 15 × 10 150 + 30 x + 20 x + 4 x 2 = 300 4 x 2 + 50 x + 150 = 300 4 x 2 + 50 x − 150 = 0 2 x 2 + 25 x − 75 = 0 0 (2 x − 5)( x + 15) = 2= + 15 0 x − 5 0 or x= 2x = 5 x = −15 ( reject ) 5 x= 2 x = 2.5 STEP 3: Interpret your results. Since we cannot have a negative length for the strip, we reject the solution x=–15. Thus the width of the strip is 2.5 m. NOTE: You should always check your solution!

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Chapter 4: Quadratic Equations

PRACTICE 34.

Eric’s carrot garden is 20 m by 30 m. She wishes to double its present area by adding a strip of uniform width around the entire garden. Determine the width of the strip.

35.

Joshua’s baby barn is 10 m by 15 m. He wishes to add on an extension which would double its present area by adding a strip of equal width on one side and on the back of the barn. Determine the width of the extension needed.

36.

A rectangular picture measuring 20 cm by 40 cm is to be matted with a border of uniform width. If the border is to be exactly half the area of the picture, determine the border’s width.

37.

A square sheet of cardboard will be made into a box by cutting equal-sized squares measuring 4 cm from each corner and folding up the four sides. If the volume of the box is 1024 cm3, what is the original length of each side?

4

4

4

4

x 4

4 4

4 x–8

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Chapter 4: Quadratic Equations

171

38.

The length of Debbie’s driveway is 5 m longer than the width. If the area of the driveway is 300 m2, what are its length and width?

39.

How wide a strip must be painted around a square wall with side length 4 m so that one half of the wall is painted?

40.

A field has dimensions 120 m x 50 m. Christian begins mowing from the outside perimeter of the field until he has mowed half of the entire field. How wide is this border of uniform width that he has mowed?

41.

A deck of uniform width is to be constrtucted partially around a pool as shown. The pool has dimensions 7 m by 3.95 m. What is the width of the deck if the total combined area is 42.9 m2?

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pool

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Chapter 4: Quadratic Equations

42.

Sarah wants to landscape her rectangular back garden by planting shrubs and flowers along a border of uniform width as shown in the diagram. Determine the width of the border if the outside fence has dimensions 28 x 25 m and the remaining garden is to be ¾ of the original size.

43.

A rectangle is four centimetres longer than twice its width. If its area is 96 cm2, determine its dimensions.

44.

Find two numbers whose product is 32 and whose difference is 14.

45.

Find two consecutive integers whose product is 72.

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173

46.

Find two consecutive even integers whose product is 440.

47.

The sum of the squares of two consecutive integers is 145. What are the integers?

48.

A group of students wish to travel to Toronto at a total cost of $810. However, after the trip was booked three students were unable to go, increasing the cost by $3 per person. Determine the number of students who originally booked the trip.

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174



Chapter 4: Quadratic Equations

*** ENRICHMENT TOPIC *** 4.4D THE SUM AND PRODUCT OF ROOTS In the previous exercise, we discovered how to work backwards to determine an equation using the given roots. Similarly, we can create an equation from the sum and product of its roots as well. Any quadratic equation ax 2 + bx + c = 0 can be expressed in the form 0 x 2 − ( sum of the roots ) x + ( product of the roots ) = where:

sum of the roots b = − a

&

product of the roots c = a

EXAMPLE 12 Calculate the sum and product to determine a quadratic equation with roots x= 2 ± 3 . Solution The roots are 2 + 3 and 2 − 3 . Thus the sum of the roots is:

(2 + 3 ) + (2 − 3 ) = 4

and the product of the roots is: (2 + 3 )(2 − 3 ) = 4 − 2 3 + 2 3 − 3 = 1 ∴ the quadratic equation is x 2 − 4 x + 1 = 0.

PRACTICE 49.

For the quadratic equation ax 2 + bx + c = 0 , let us refer to the two roots as r1 and r2 . Hence, let

r1 =

− b + b 2 − 4 ac 2a

b a. Show that r1 + r2 = − . a

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and

r2 =

− b − b 2 − 4 ac . 2a c b. Show that r1 × r2 =. a

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Mathematics 2200

Chapter 4: Quadratic Equations 50.

Without solving, determine the sum and product of the roots of each quadratic equation. a. x 2 − 2 x + 1 = 0

d. x 2 +

g.

51.

52.

175

3 1 x− = 0 4 4

1 2 0 x + 4 x + 15 = 3

b. x 2 − 4 x + 4 = 0

c. x 2 + 3 x + 2 = 0

2 e. x= 6 x + 40

f. px 2 − qx + r = 0

h.

3 x 2 − 4 x + 12 = 0

i.

3 x2 − 2 x + 2 3 = 0

Calculate the sum and product of the given roots to determine a quadratic equation. a. 2 and 3

b. 5 and –1

c. –2 and –4

d. 2 and 21

e. –3 and − 31

2 f. 21 and − 3

g. ± 2

h. ±2 3

i. −2 ± 6

j. 3 ± 5

k.

5±3 2 2

l.

−3 ± 7 2

Compare your answers above with those of Practice Question #15 on Page 155. Which method do you prefer?

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176

5

Chapter 5: Radical Expressions and Equations

CHAPTER

RADICAL EXPRESSIONS & EQUATIONS Contents.............................................................. 5.1

WORKING WITH RADICALS ................................................................ 177 A. ORDERING RADICALS .................................................................. 182 B. RESTRICTIONS ON VARIABLES .................................................... 183 C. ADDING AND SUBTRACTING RADICALS .......................................... 184

5.2

A. MULTIPLYING RADICAL EXPRESSIONS .......................................... 186 B. DIVIDING RADICAL EXPRESSIONS .................................................. 189 C. DIVIDING RADICAL EXPRESSIONS USING CONJUGATES ................. 192

5.3

SOLVING RADICAL EQUATIONS ........................................................... 195

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177

5.1 WORKING WITH RADICALS You have encountered square roots in previous mathematics courses. The initial exercises that follow will serve as a partial review to refresh your skills for use in the upcoming concepts, followed by new material that will enhance your understanding of square roots and their properties. PARTS OF A RADICAL radical symbol

8

3

index

radicand

radical

*Note that in a square root an index of 2 is understood and generally not written. It is important that we also understand the relationship between the exponent of a power and the index of the radical: You should notice that the index of the radical is the same as the exponent!!!

25 5 since 5 × 5 25 = or 52 25 = = 3 8 2 since =

4 81 3 since =

2 or 23 8 = ×2×2 8 = 3 ×= 3 × 3 × 3 81 = or 34 81

5 32 2 since =

2 × 2 ×= 2 × 2 × 2 32 = or 25 32

TRY THESE! 1.

Identify the index and radicand for each of the following. a.

2.

3

5

b.

7

72

c.

28

d.

6

43

e. 2 3 6

f.

3

−64

Without using a calculator, evaluate each of the following where possible. 1 a. 3 27 b. 5 −32 c. 4 d. 6 4 3 e. 2 3 125 f. 3 −64 81

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1 2

g.

g.

3

27 1000

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178

Chapter 5: Radical Expressions and Equations

To understand the many uses of radicals let us begin by exploring some basic properties. The first property is the multiplication property. For basic square roots we have:

= a⋅ b

ab, where a ≥ 0, b ≥ 0

4 ⋅ 9 = 2×3 = 6

For example,

4 ⋅ 9=

same!

36= 6

This property can be extended to radicals with any index: Multiplication Property of Radicals n

n a ⋅ nb= ab

where n ∈ , and a, b ∈ .

For example,

3

8 ⋅ 3 125 = 2 ⋅ 5 = 10

3

8 ⋅ 125=

3

same!

1000= 10

When working with radicals, they can be expressed as entire radicals or as mixed radicals.

20

2 5

entire radical

mixed radical

You can sometimes express entire radicals as mixed radicals… EXAMPLE 1 Express as a mixed radical:

a.

b.

20

18

c.

50 x 3 y 8

This method of reducing is referred to as simplifying by extracting the largest perfect square.

Solution

a.

20

b.

18

50 x 3 y 8

c.

= 4×5

= 9×2

= 25 x 2 y 8 × 2 x

= 4× 5

= 9× 2

= 25 x 2 y 8 × 2 x

= 2× 5

= 3× 2

= 5 xy 4 × 2 x

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Chapter 5: Radical Expressions and Equations

179

… and mixed radicals can also be expressed as entire radicals. EXAMPLE 2 Express as an entire radical:

a.

5 2

b.

3 7

c.

2 x3 y 5 x

2 x3 y 5 x

c.

Solution a.

b.

5 2

3 7

2 × ( 5 × 5)

=

7 × (3 × 3)

=

2 × ( 25 ) =

7 × (9) =

50 =

( 5 x ) ( 2 x 3 y )( 2 x 3 y )

=

( 5 × 2 × 2 ) ( xx 3 x 3 ) ( yy )

=

63

20 x 7 y 2

=

This idea can be extended for any value of n, not just square roots where the index is 2. EXAMPLE 3 Express as an entire radical: a.

3

40

b.

4

162

c.

3

54 x y z 11

6

To be a perfect square, the prime factor must occur twice. To be a perfect cube, the prime factor must occur three times.

4

Solution

a.

3

40

b.

4

162

c.

3 8×5 =

4 81 × 2 =

3 8×35 =

4 81 × 4 2 =

2× 3 5 =

3

54 x 11 y 6 z 4

3 27 x 9 y 6 z 3 × 2 x 2 z 1 = 3 27 x 9 y 6 z 3 × 3 2 x 2 z 1 =

3 x3 y2 z1 × 3 2 x 2 z1 =

3× 4 2 =

3 4 2= 5 3= 2 3 x3 y2 z

3

2 x2 z

TRY THESE! 1.

Determine the exact value of each radical. a.

3

125

b.

d.

4

10000

e.

g.

3

8 27

h.

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343

c.

3

512

3

1

f.

4

81

3

64 125

i.

4

0.0016 0.0081

3

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Mathematics 2200

180 2.

3.

Chapter 5: Radical Expressions and Equations Determine which radicals can be evaluated. Evaluate those that are possible. a.

3

−125

b.

d.

5

−32

e.

g.

3

−

3

54

6.

3

−1

−1

f.

4

−81

i.

3

−

4

−

16 81

c.

3

0.008 0.027

b.

4

32

250

d.

5

486

Convert to an entire radical: a. 2 3 6

5.

h.

c.

Convert to a mixed radical: a.

4.

8 125

4

−27

b. 3 3 5

c. 2 5 3

d. 5 4 2

Express as a mixed radical in simplest form. a.

12 x 2

b.

25 y 3

c.

e.

16a 4 b9

f.

36 p7 q 5

g.

3

18k 5

d.

16 m 3 n8

h.

27 x 6

4

81 x 13 y 10 z 7

Express as an entire radical in simplest form. a. 3 x 2

b. 5m 3 m

c. 2k 5k 3

d. 8 x 5 2 x 3

e. 2 x 3 y 2 4 3 x 2 y

f. 5h4 k 2h3 k 7

g. 2 x 2 3 5 x 8

h. 3a 2 bc 4 3 2ab 2 c

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Chapter 5: Radical Expressions and Equations

181

PRACTICE 1.

2.

Express as a mixed radical in simplest form by extracting the largest perfect square. a.

12

b.

24

c.

50

d.

32

e.

200

f.

54

g.

27

h.

96

i.

108

j. 2 80

a.

3

16

b.

3

24

c.

4

810

d.

3

250

e.

3

−135

f.

4

16

g.

4

1250

h.

3

−48

40 000

j.

5

64

k.

3

1 16

l.

4

3

24 54

Express each entire radical as a mixed radical in simplest form. a. e.

i.

4.

l. 4 45

Express each entire radical as a mixed radical in simplest form.

i.

3.

k. −3 28

3

12a 4

b.

54 c 4 d 3

f.

a4 20 b3

j.

3

4

27 m 6

c.

80 x 5

d.

3

8 x 9 y3

200 g 7

g.

a3 b4 c5

h.

3

4 j5 k 8

16 m n 6

12

k.

3

48 x 10

l.

27 p5 20 q8

Express each mixed radical as an entire radical. a. 3 5

b. 2 x 3 x

c. 4 m 3 2m 2

d. 2 x 2 y 3 3 xy

e. 2a 3 5a 2

f. 3 x 2 x

g. 5abc ac

h. 2d 2 c 3 d 2

i. 3k 3 2k

j. 2 g 2 h3 4 2 g 3

k.

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1 x 3

3

9x

l.

2a 3 3b 2

3

a2 2b

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182

Chapter 5: Radical Expressions and Equations

5.1A ORDERING RADICALS Most radicals are not from perfect numbers that result in an integral value. As a result, radical values often result in non-terminating, non-repeating decimals when evaluated on a calculator. How can we compare radicals without the use of a calculator? To do so, we can use our skills with radicals by rewriting mixed radicals as entire radicals. EXAMPLE 4 Arrange the following real numbers in ascending order:

163 , 13 , 4 10 , 2 42 , 3 19

Solution To compare each value, first express each number as an entire radical. Thus,

163 13 =

13 × 13 =

160 ,

∴

169

4 10 =

4 × 4 × 10 =

160

2 42 =

2 × 2 × 42 =

168

3 19 =

3 × 3 × 19 =

163 ,

168 ,

169 ,

171

Thus, in ascending order we have:  4 10 ,

171

163 , 2 42 , 13 , 3 19

TRY THESE! 1.

Complete each statement to find two whole numbers between which the radical lies. a. Since 12 is between 9 and 16, then

12 is between

b. Since 50 is between 27 and 64, then c. Since 85 is between 64 and 125, then 2.

3

.

50 is between

and

.

85 is between

and

.

Arrange the following radicals on the number line below by placing the letter in the correct box (try without using a calculator!). (A)

3.

3

and

5

200

(B)

20

(C)

3

45

(D)

4

0.32

(E)

3

5

(F)

4

100

Arrange in order from least to greatest (without the use of a calculator!). a.

2 9 , 5 3, 8 2, 4 5, 3 7

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Chapter 5: Radical Expressions and Equations

183

5.1B RESTRICTIONS ON VARIABLES What happens when you square root a negative number? What message does your calculator respond with? If a radical represents a real number and has an even index, the radicand must be non-negative. For example, the radical 2 − x has an even index. Thus, the radicand must be greater than or equal to zero. Hence: 2− x ≥ 0 − x ≥ −2 x≤2 \ The radical

2 − x is only defined as a real number if x is less than or equal to two.

What happens when you cube root a negative number? Since the index is an odd number, the variable can be any real number. Why can a radical with an odd index have a radicand that is positive, negative or zero? To fully answer this question we need to discuss a new number system, namely imaginary numbers. For now, however, we will simply say that an even index requires a radicand derived from a number multiplied by itself an even number of times. For example, −4 is not permitted since the only options are either 2 or –2, yet neither will produce a negative product. That is, 2 × 2 ≠ −4 and − 2 × − 2 ≠ −4 . On the other hand,

3

−8 works since − 2 × − 2 × − 2 = −8 . Thus

3

−8 =−2 .

In summary, when a radical has an EVEN index, the radicand must be non-negative. When a radical has an ODD index, the radicand can be positive or negative. x−6

Some other examples are: (a)

12 − 3 x has an even index \ 12 − 3 x ≥ 0 ⇒ − 3 x ≥ −12 ⇒ x ≤ 4

(b) (c)

has an even index \ x − 6 ≥ 0 ⇒ x ≥ 6

3

x−7

has an odd index so there are NO restrictions on the variable.

(d)

\ x∈.

x 2 + 1 has an even index but x 2 + 1 is always greater than zero. Thus there are NO restrictions on the variable. \ x ∈  .

(e)

1 x+3

has an even index but

x + 3 is in the denominator and

thus cannot equal zero. \ x + 3 > 0 ⇒ x > −3

TRY THESE! Identify the restrictions on the values of the variables in each radical expression. a)

x

b)

1 x2

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c)

1 3x − 1

d)

3

x−1

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e)

2 x − 10

f)

1 3

4 − 2x

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184

Chapter 5: Radical Expressions and Equations

5.1C ADDING AND SUBTRACTING RADICALS Many new concepts are easier to remember when compared to information previously learned. unlike radicals :

unlike algebraic terms : 3x

− 2 x2

x

3 x

5x

2 3

2

like algebraic terms : 2x

5x

2x 3

2x

−23 2

5

3 7 2

−54 6

like radicals : −x

2

5 2

2 2 3

−3 2

− 21 2

same

same

Adding and subtracting radicals is very similar to adding and subtracting algebraic expressions. With algebra, like terms can be added or subtracted. Similarly, like radicals can be added or subtracted. EXAMPLE 5 Simplify each expression. a.

6 2+5 2−4 2

b.

3 3 24 + 3 250 − 5 8 − 3 54

b.

3 3 24 + 3 250 − 5 8 − 3 54

Solution a.

6 12 + 5 27 − 4 32 = 6( 4 × 3 ) + 5( 9 × 3 ) − 4 ( 16 × 2 )

= 3 3 8 × 3 + 3 125 × 2 − 5 4 × 2 − 3 27 × 2

= 6( 4 × 3 ) + 5( 9 × 3 ) − 4 ( 16 × 3 )

= 3(2 3 3 ) + 5 3 2 − 5(2 2 ) − 3 3 2

= 6(2 × 3 ) + 5(3 × 3 ) − 4 ( 4 × 3 )

= 6 3 3 + 5 3 2 − 10 2 − 3 3 2

= 12 3 + 15 3 − 16 3

= 6 3 3 + 2 3 2 − 10 2

= 27 3 − 16 3

TRY THESE!

Remember: like radicals have the same index and radicand.

Simplify. a) 2 27 + 5 12

b) 3 8 + 4 3 − 2 18

c) 3 3 16 + 4 3 54

d) 4 27 x − 2 50 x

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Chapter 5: Radical Expressions and Equations

185

PRACTICE 5.

6.

Simplify. Identify any restrictions on the expressions containing variables. a. 3 2 + 5 2

b. 6 3 − 2 3

c. −2 3 5 + 3 3 5

d. 12 6 x + 3 6 x

e. 8 3 7 + 3 3 7

f. −9 3 2n − 4 3 2n

g. 6 3 3 + 2 3 3 − 4 3 3

h. −8 4 5a − 3 4 5a + 2 4 5a

2 7+ 1 7 i. 21 7 − 3 4

Simplify. Identify any restrictions on the expressions containing variables. a.

e.

7.

4

12 + 5 3

b. 2 10 − 3 40

48 + 4 243

f.

48 x − 3 3 x

c. −2 3 7 − 3 3 56

d. −6 20 y + 7 80 y

g. 5 2 + 3 8

h. 3 3 54 x + 2 3 250 x

Simplify. Identify any restrictions on the expressions containing variables. a. 5 2 + 2 3 + 6 3

b. −3 5 − 4 3 7 + 6 3 7 + 2 5

c. 3 x 3 16 x +

d.

3

128 x 4 − 2 3 54 x 4 + x 3 250 x

3 48 + 1 20 − 2 27 f. 31 45 − 4 2 3

e. 2 45k 3 − 2k 27 k + 3 20 k 3 − k 48k

8.

150 1 24 2 − 54 − + 6 3 2 2 3

Determine the perimeter of each figure. Express your answer in simplest form. a.

3 8 + 2 27

5 12 − 3 2

3 3 16 − 2 3

b.

48 − 3 54 5 3 −23 2

2 12 − 3 128 Sharpe Mathematics  2017

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186

Chapter 5: Radical Expressions and Equations

5.2A MULTIPLYING RADICAL EXPRESSIONS As we have seen in the previous section, = a× b To further generalize:

ab , where a ≥ 0 , b ≥ 0 .

= a b × c d (ac ) bd , where b ≥ 0 , d ≥ 0

In other words, when multiplying radicals the numbers on the outside of the radical sign can be combined, as well as the numbers under the radical sign (the radicands). EXAMPLE 6 Simplify each expression. a.

3 5×4 2

Think of algebra: (3x)(4y)=12xy

b.

(2 12 )(5 2 )

Solution a.

b. (2 12 )(5 2 )

3 5×4 2 =× (3 4 ) 5 × 2

=× (2 5) 12 × 2

= 12 10

= 10 24 = 10 4 × 6 = 10 (2 6 ) = 20 6

The distributive law and FOIL method are also useful when multiplying radicals. EXAMPLE 7 Simplify each expression. a.

2 3 (3 5 + 4 7 )

b.

( 3 + 5 2 )(3 2 − 4 3 )

b.

( 3 + 5 2 )(3 2 − 4 3 )

Solution a.

2 3 (3 5 + 4 7 )

= 6 15 + 8 21

F

O

I

L

= 3 6 − 4 9 + 15 4 − 20 6 = 3 6 − 4 (3) + 15(2) − 20 6 = 3 6 − 12 + 30 − 20 6

PRACTICE Sharpe Mathematics  2017

= −17 6 + 18

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Mathematics 2200

Chapter 5: Radical Expressions and Equations

187

PRACTICE 9.

Simplify. a.

10.

3× 2

b.

5× 6

c. 2 5 × 3

d. (3 2 ) × (2 3 )

g. (3 2n )2

h. (−2 5 )2

e. (−4 a )(3 a )

f. 3 3 7 × 2 3 3

i. (2 32 )2

j.

20 × 12

k. 2 18 × 75

l. 2 3 × 5 6 × 3 2

m. (5 8 )(−3 27 )

n. 3 2k × 2 5k

o. 2 6 × 3 18

p. 3 6 × 2 18 × 2 15

Simplify. a. 5 (2 + 3 )

d. 2 x ( x +

b.

y)

3 (4 + 2 )

c.

3

7 (3 5 − 3 2)

e. 3 7 ( 3 − 5 )

f. 4 2 (2 3 + 3 6 )

g. 3 5 (2 2 − 3 3 )

h. −2 a (3 b − 3 a )

i. 3 3 (2 8 + 2 18 )

j. −2 2 (3 12 − 2 27 )

k. 2 3 (2 48 − 3 32 )

l. 4 3 9 w (3 3 6 − 3 3 12w + 2 3 3w 2 )

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Mathematics 2200

188 11.

Chapter 5: Radical Expressions and Equations Simplify. a. (2 + 3 )(4 + 3 )

b. ( 3 + 1)( 3 − 5)

c. ( 2 + 3 )(2 2 + 3 )

d. (6 − 3 2 )(2 − 5 2 )

e. (2 x + 1)( x − 6)

f. ( 7 − 3 )(4 7 + 3 )

g. ( 2d − c )( 2d + c )

h. ( 3 − 2 6 )(5 6 − 3 )

i. (5 10 t + 7 2t )( 10 t + 8t )

j. (3 5 − 2 2 )( 5 − 4 2 ) k. (3 2 − 8 )(2 2 + 8 )

12.

13.

l. ( 2d + 4 3d )(2 4 d + 9d 2 )

Simplify. a. (5 + 3 )2

b. (2 + 3 )2

c. ( a − b )2

d. (2 3 − 3 5 )2

e. (2 3 + 2 )2

f. (3 6 p − 2 3 p )2

Simplify. What do you notice about your answers? a. (3 + 2 )(3 − 2 )

b. (1 + 2 2 y )(1 − 2 2 y )

c. (2 3 + 6 )(2 3 − 6 )

d. ( m + 5)( m − 5)

e. ( 3 + 7 )( 3 − 7 )

f. (2 2 + 4 3 )(2 2 − 4 3 )

Here we are multiplying

conjugates.

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Chapter 5: Radical Expressions and Equations

189

5.2B DIVIDING RADICAL EXPRESSIONS Similar to the rule for multiplying radicals, the rule for dividing radicals can be generalized as: n a = n b

n

a , where a ≥ 0 , b > 0 , n ∈  b

In other words, the quotient of the radicals is equivalent to the radical of a quotient.

100 10 = = 2 and 5 25

For example,

100 = 25

= 4 2.

For a radical to be in simplest form, the denominator in the final answer must not be a radical (i.e. must not be irrational). This is NOT 12 2 3 3 expressed in = = For example, simplest form. 20 2 5 5 In such instances, we can change the denominator to a rational number by multiplying the entire expression by 1 and obtaining the same value.

3

For example,

3

=

5 = =

×

irrational

5 5

15 25 15 5

rational

15 are equivalent to each other. Verify with your calculator! 5 5 The process outlined above is referred to as RATIONALIZING THE DENOMINATOR, and involves expressing the denominator as a rational number as opposed to a radical (irrational) number. To further demonstrate this strategy, examine the following solutions using the two different forms of the quotient rule for radicals. As a result, both

3

× 1

5

and

36 = 2 Evidently, both

6 2

6

and

2

36 = 2

= 18 3 2

and 3 2 must be equivalent. By using the same process as before:

6 2 Sharpe Mathematics  2017

× = 1

6

2 6 2 6 2 × = = = 3 2 2 2 2 4 DO NOT COPY!

 Mathematics 2200

190

Chapter 5: Radical Expressions and Equations

EXAMPLE 8 Simplify by rationalizing the denominator:

a.

3 5

4 3 x2

b.

6

5 2x

, x>0

c.

3 7 m32

, m≠0

Solution a.

3 5 3 5 6 = × 6 6 6

4 3 x2 4 3 x 2 = × 5 2x 5 2 2

b.

3 30 6 30 = 2 =

=

c.

4 6x

3 7 3 7 ( 3 2 )2 = × m 3 2 m 3 2 ( 3 2 )2 =

5 4

=

4 6x 10

=

=

2 6x 5

=

(3 7 )( 3 2 )( 3 2 ) m ( 3 2 )( 3 2 )( 3 2 ) (3 7 )( 3 4 ) m (3 8) (3 7 )( 3 4 ) 2m

PRACTICE 14.

Simplify each expression. State any restrictions on the values for the variables. a.

e.

15.

24 3

2b 5

1

b.

f.

5n

−4 3

2r

c.

g.

−2 6

18 5 3 2

d.

h.

6 3

3

12 3 x 2 5 7x

Simplify. State any restrictions on the values for the variables. a.

e.

5 20

12 7d 5 3 12d 3

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b.

f.

33 6 3

12

7 3t 4 28

c.

g.

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−9 12 18

12 24 3 20

d.

h.

15 3 3 6y

12 3 18 x 7 4 3 8x

Mathematics 2200

Chapter 5: Radical Expressions and Equations

16.

Express in simplest form. State any restrictions on the values for the variables. a.

e.

17.

191

1 2

+

1

b.

3

1+ 2

f.

3

1 3

−

1 6

2 − 3n 6n

c.

g.

2 2

+

6 3

2 20 + 3 10 2 5

d.

h.

8− 2 6a

2 5+3 2 2 3

Debbie simplified the following expression. Identify, explain, and correct any errors in her work.

5m + 6 m 5m + 6 m 3m = × 3m 3m 3m = = = =

(

)( 5m + 6m ) ( 3m )( 3m )

3m

5m 3m + 9 m (3m )2 5m 3m + 3m 3m (5m + 1) 3m , m≠0 3m

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192

Chapter 5: Radical Expressions and Equations

5.2C DIVIDING RADICAL EXPRESSIONS USING CONJUGATES As you have just seen, to simplify radicals we must ensure that we have rationalized the denominator to eliminate the radical in the denominator. All of the previous examples, however, involved only a single radical in each denominator. How can we apply our knowledge to radical fractions involving binomial radical expressions? For example, what could you multiply the following radical expressions by to rationalize each denominator? a.

1 1+ 2

×

7

b.

2+3

×

c.

5 3− 2

×

d.

2 5+7 3 2+2 6

×

When correctly chosen, the new denominator will become a rational number when multiplied by its conjugate. For example, (3 + 2 ) and (3 − 2 ) are conjugates with a product of 7. EXAMPLE 9 Solution Simplify by rationalizing the denominator:

2 3 2+ 6

×

2− 6 2− 6

Conjugate of

(

2+ 6

) is (

)

2− 6 .

2 3 2+ 6

2 3

=

2+ 6

=

×

2− 6 2− 6

2 6 − 2 18 4−

12 +

2 6 − 2(3 2 ) 2−6 2 6 −6 2 = −4 − 6 +3 2 = 2 =

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12 − 36

Distribute and FOIL.

Simplify and reduce. Simplify and reduce. Simplify and reduce. Divide numerator and denominator by -2.

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Chapter 5: Radical Expressions and Equations

193

PRACTICE 18.

Determine the conjugate of each expression. a.

2+2

e. x − 3 2

19.

20.

b.

c. 3 +

3−4

f. 3 2 + 7

x

3−4 5

g.

d. 5 − 2 3

h. 2 3 + 5 6

Simplify. a. ( 3 + 1)( 3 − 1)

b. (2 + 5 x )(2 − 5 x )

c. (5 − 2 )(5 + 2 )

d. ( 2 + 3 )( 2 − 3 )

e. (2 5 − 6 )(2 5 + 6 )

f. (3 2 + 2 6 )(3 2 − 2 6 )

Simplify. a.

2 1+ 5

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b.

−3 3−2

c.

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5 2+

x

d.

− 2 10 + 2

Mathematics 2200

194

21.

Chapter 5: Radical Expressions and Equations

Simplify. a.

22.

6

b.

2+ 6

−2 3 3 2− 3

c.

4 − 2x 2 3 + 2x

d.

2 2+5 6 3 2−2 6

Eric simplified the following expression. Identify, explain, and correct any errors in his work.

2y 6 2y 6 5 + 2y = × 5 − 2y 5 − 2y 5 + 2y =

=

= = = =

( 2 y 6 )( 5 + 2 y ) ( 5 − 2 y )( 5 + 2 y ) 10 y 6 + 2 y 12 y 25 + 5 2 y − 5 2 y − (2 y )2

12 y 18 y 25 − 4 y

(

12 y 3 2 y

)

25 − 4 y 36 y 2 y 21 y 12 2 y , y≥0 7

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Chapter 5: Radical Expressions and Equations

195

5.3 SOLVING RADICAL EQUATIONS In the previous sections, we have been examining ways in which we can SIMPLIFY radical expressions. We now turn our attention to SOLVING radical equations. To solve a radical equation, our primary goal is to isolate the radical and then eliminate it, usually by squaring the expressions on both sides of the equation. The following steps can be used when solving radical equations: STEP 1: State any restrictions on the variable(s). STEP 2: Isolate the radical on one side of the equation. STEP 3: Square each side of the equation (LHS & RHS). STEP 4: Solve for the variable that remains. STEP 5: Check/verify the roots to identify and reject any that do not work. EXAMPLE 10 Solve for x:

−1 + 5 x + 5 =x

Solution 5x + 5 ≥ 0 5 ( x + 1) ≥ 0

STEP 1: State any restrictions on the variable(s). Restrictions come from the radical part of the expression. More specifically, the radicand must be a non-negative real number.

x+1≥ 0 x ≥ −1

STEP 2: Isolate the radical on one side of the equation.

−1 + 5 x + 5 =x 5x + 5 = x + 1

( 5 x + 5 ) =( x + 1) ( 5 x + 5 )( 5 x + 5 ) = ( x + 1)( x + 1)

STEP 3: Square both sides of the equation (LHS & RHS).

2

2

( 5 x + 5 ) = ( x 2 + 2 x + 1)

STEP 4: Solve for the variable that remains.

x 2 + 2 x + 1 − 5 x − 5 =0 0 x2 − 3 x − 4 = 0 ( x − 4 )( x + 1) = 4 0 , x= x −= +1 0 x = 4 , x = −1

STEP 5: Check/verify the roots by substituting the possible solution(s) into the original equation to identify the value(s) that work and and reject any that do not. x= 4 ⇒ − 1 + 5 x + 5= x

x =−1 ⇒ − 1 + 5 x + 5 =x

− 1 + 5( 4 ) + 5 =( 4 ) 

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− 1 + 5( −1) + 5 = ( −1) 

Mathematics 2200

196

Chapter 5: Radical Expressions and Equations

The solutions to the radical equation need to be verified to ensure they satisfy the ORIGINAL equation. The reason for checking is due to the fact that we actually changed the original equation when we squared both sides. It is possible that by squaring the equation, you may have obtained an answer that does not actually work. Such a result is called an EXTRANEOUS ROOT and must be REJECTED as it is not a solution to the original radical equation. This is perhaps more easily explained by viewing the graphical as well as the algebraic solution. When solving a radical equation, sometimes we have NO extraneous roots (ie. two solutions) … EXAMPLE 11 Determine the roots of

3x  3  x  1 .

Solution Restrictions:

Solve for x:

3x + 3 ≥ 0 3 ( x + 1) ≥ 0

3x  3  x  1 3x  3  x  1

x+1≥ 0 x ≥ −1



3x  3



2

y1

2

  x  1

3 x  3  x2  2 x  1 0  x2  x  2

y  3x  3  x

0   x  2 x  1 x  2,  1 Verify solutions:

3x + 3 − x = 1

3x + 3 − x = 1 x= 2

⇒

?

) 1 3(2) + 3 − (2=

x =−1

?

3( −1) + 3 − ( −1) =1 ?

? 6 + 3 − ( 2) = 1

−3 + 3 + (1) = 1 ? 0 + (1) = 1 ? 0 + (1) = 1

?1 9 − ( 2) = ? 3 − ( 2) = 1

1= 1

⇒

1= 1





As you can see, by inserting both x  2 and x  1 into the original equation that both values work. The graph also shows that both x  2 and x  1 are solutions.

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Chapter 5: Radical Expressions and Equations

197

… sometimes one value is rejected, so we have ONE extraneous root (and only one solution) … EXAMPLE 12 Determine the roots of

2x 1  2  x .

Solution Restrictions:

yx

Solve for x:

2x − 1 ≥ 0 2x ≥ 1

y  2x 1  2

2x 1  2  x 2x 1  x  2

1 x≥ 2



2x 1



2

2

  x  2

2 x  1  x2  4 x  4 0  x2  6 x  5 0   x  5 x  1 x  5, 1 Verify solutions: 2 x − 1 + 2 =x

2 x − 1 + 2 =x x 5 =

⇒

?

x= 1

2(5) − 1 + = 2 5 ?

?

2(1) − 1 + 2= 1 ?

2 − 1 + ( 2) = 1

10 − 1 + (2) = 5

?

?

1 + ( 2) = 1 ? 1 + ( 2) = 1

9 + ( 2) = 5 ? 3 + ( 2) = 5 5=5

⇒

3≠1





As you can see, by inserting both x  5 and x  1 into the original equation that only x  5 is valid while x  1 does not work. The graph also shows that only x  5 is a solution.

We must take note of any restrictions in the equation, and reject any that do not appear in the solution set. These are known as EXTRANEOUS ROOTS.

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198

Chapter 5: Radical Expressions and Equations

… and sometimes both values are rejected, so we have TWO extraneous roots (ie. no solution). EXAMPLE 13 Solve for x:

2x  4  1  x  1

Solution Restrictions: 2x − 4 ≥ 0 2x ≥ 4 x≥2

Solve for x:

y  x 1

2x  4  1  x  1 2x  4  x  1  1



2x  4



2

y  2x  4  1

2

  x

2 x  4  x2 0  x2  2 x  4 Notice that

D  b 2  4 ac  (2)2  4 (1)( 4 )  4  16  12  0

Thus, from our previous chapter we know that when the discriminant is less than 0 (D 16

29. a. d ≠ 2

b. d = 2

c. ∅

30.

{k / 0 < k < 36, k ∈ R}

31. h =

f. 2 c. -8, imaginary & unequal f. 41, real & unequal i. 49, real & unequal l. 97, real & unequal o. 29, real & unequal r. -4, imaginary & unequal u. 12, real & unequal

32. t = −4

1 4

4.4C: WORD PROBLEMS USING QUADRATIC EQUATIONS

33. k = 2,14 (PAGES 170–173)

34. 5 m

35. 5 m

36. approx. 3 cm

37. 24 m

38. 20 m, 15 m

39. 2 − 2 ≈ 0.59m

40. 10 m

41. 1.25 m

42. approx. 2.4 m

43. 16cm, 6 cm

44. -16, -2 or 2, 16

45. -8, -9 or 8, 9

46. 20, 22 or -20, -22

47. -8, -9 or 8, 9

48. 30

4.4D: THE SUM AND PRODUCT OF ROOTS

(PAGES 174–175)

49. Work it out! 50. a. sum = 2 , prod = 1 d. sum = − 34 , prod = − 14 g. sum = -12 , prod = 45 51. a. x 2 − 5 x + 6 = 0 2 e. 3 x + 10 x + 3 = 0 2 i. x + 4 x − 2 = 0

b. sum = 4 , prod = 4 e. sum = 6 , prod = -40

c. sum = -3 , prod = 2 f. sum = qp , prod = rp

h. sum =

i. sum =

b. x 2 − 4 x − 5 = 0 2 f. 6 x + x − 2 = 0 2 j. x − 6 x + 4 = 0

4 3 3

, prod = 2 c. x 2 + 6 x + 8 = 0 2 g. x − 2 = 0 2 k. 4 x − 20 x + 7 = 0

2 3 3

, prod = 2

d. 2 x 2 − 5 x + 2 = 0 2 h. x − 12 = 0 2 l. 2 x + 6 x + 1 = 0

52. Answers are the same as #15. Sharpe Mathematics  2017

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Mathematics 2200

362

SOLUTIONS

5.1: WORKING WITH RADICALS

(PAGE 181)

1. a. 2 3 g. 3 3

b. 2 6 h. 4 6

c. 5 2 i. 6 3

d. 4 2 j. 8 5

e. 10 2 k. −6 7

f. 3 6 l. 12 5

2. a. 2 3 2

b. 2 3 3

c. 3 3 10

d. 5 3 2

e. −3 3 5

f. 2

h. −2 3 6

i. 10 4 4

j. 2 5 2

k. 2 3

b. 3m 2

c. 4 x 2 5 x

d. 2x3 y

e. 3cd 3 2c

f. 10 g 3 2 g

2 k. 3 x

3 p2 l. 2q 4

g. 5 4 2 3. a. 2a 2 3 2 2

g. ab c 4. a. g.

ac

h. jk

2 3

2

4j k

2

a2 1 i. 2b 5b

j. 2mn

45

b. 12x 3

c. 3 128m5

d.

25a 3 b 2 c3

h. 3 8c3 d 8

i. 18k 7

j.

m

2

24x 7 y 4

e.

32g 11 h12

k.

3

4

34

3

3

1 2

3

l.

6 x

40a 5

f.

9 x4 27

l.

5.1C: ADDING AND SUBTRACTING RADICALS 5. a. 8 2 f. −13 3 2n , n ≥ 0 6. a. 7 3 f.

3x , x ≥ 0

3 2

3p 5

18x3 3

8a11 54b 7

(PAGE 185)

c.

g. 4 3 3

h. −9 4 5a , a ≥ 0

1 7 i. 12

b. −4 10

c. −8 3 7

d. 16 5 y , y ≥ 0

e. 5 4 3

g. 11 2

h. 19 3 2 x , x ≥ 0 c. 9 x 3 2 x , x ≥ 0

d. − 16 6

3

d. 15 6 x , x ≥ 0

5

b. − 5 + 2 3 7

e. 12k 5k − 10k 3k , k ≥ 0 8. a. 6 2 + 32 3

e. 11 3 7

f. 2 5 − 5 3 b. 11 3 − 3 3 2

5.2A: MULTIPLYING RADICAL EXPRESSIONS 6

3

b. 4 3

7. a. 5 2 + 8 3

9. a.

2 3

b.

30

(PAGE 187)

c. 2 15

d. 6 6

e. −12a , a ≥ 0

f. 6 3 21

g. 18n , n ≥ 0

h. 20

i. 128

j. 4 15

k. 30 6

l. 180

m. −90 6

n. 6k 10 , k ≥ 0

o. 36 3

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p. 216 5

Mathematics 2200

SOLUTIONS

363

5.2A: MULTIPLYING RADICAL EXPRESSIONS 10. a. 10 + 5 3

(PAGES 187-188)

(continued) …

b. 4 3 + 6

c.

e. 3 21 − 3 35

f. 8 6 + 24 3

g. 6 10 − 9 15

h. 6a − 6 ab , a, b ≥ 0

i. 30 6

j. 0

k. 48 − 24 6

l. 36 3 2 w − 36 3 4 w2 − 24 w , w ≥ 0

b. −2 − 4 3

c. 7 + 3 6

d. 42 − 36 2

e. 2 x − 11 x − 6

f. 25 − 3 21

g. 2d − c

h. 21 2 − 63

i. 78t + 34t 5

j. 31 − 14 10

k. 8

b. 7 + 4 3

c. a + b − 2 ab , a, b ≥ 0

e. 14 + 4 6

f. 66 p − 36 p 2 , p ≥ 0

11. a. 11 + 6 3

12. a. 28 + 10 3 d. 57 − 12 15 13. a. 7

b. 1 − 8 y , y ≥ 0

c. 6

3

d. 2 x + 2 xy , x, y ≥ 0

35 − 3 14

l. 4d 2 + 3d 2d + 16d 3 + 12d 3d , d ≥ 0

d. m − 25 , m ≥ 0

5.2B: DIVIDING RADICAL EXPRESSIONS 14. a. 2 2 e.

15. a. e.

10b ,b≥0 5 5 2 2d 21, d > 0 3

3 2+2 3 6 3+ 6 e. 3

16. a.

5n ,n>0 5n −2 3 4r 2 f. ,r>0 r

b.

e. −4

f. −40

(PAGES 190-191) c. − g.

1 6 3

d. 2 3 9

3 10 2

h.

12 21x , x > 0 35

b.

33 4 2

c. −3 6

d.

5 2y , y>0 2y

f.

1 21t , t ≥ 0 8

g.

4 30 5

h.

3x 2 3 18 , x > 0 2

2 3− 6 6 2 3n − 3n 2 f. ,n>0 6n

b.

c. g.

2+2 3

4+3 2 2

4 6a − 3a ,a>0 3a 2 15 + 3 6 h. 6

d.

2 17. First error occurs in Step 3. Steps should be: = 5m 3m + 18m 2 ( 3m )

5m 3m + 3m 2 3m 5 3m + 3 2 ,m>0 = 3 =

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364

SOLUTIONS

5.2C: DIVIDING RADICAL EXPRESSIONS USING CONJUGATES (PAGES 193-194) 18. a. 2 − 2 e. x + 3 2 19. a. 2

b. 3 + 4 f. 3 2 − 7 b. 4 − 5 x , x ≥ 0

20. a.

−1 + 5 2

b. 3 3 + 6

21. a.

− 3+3 2

b.

c. 3 − x g. 3 + 4 5 c. 23

−2 6 − 2 5

d. 5 + 2 3 h. 2 3 − 5 6

d. −1

e. 14

c.

10 − 5 x , x>0 4− x

c.

4 3 − 2 2x − 6x + 2x , x≥0, x≠6 6− x

=

d.

e. = x 6 , x ≥1

f. x = {∅} , x ≥

25. = a. x 99 , x ≥ −1

b.= x 4, x≥2

i. g = 5, 0 , g ≥ −

1 3

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−36 − 19 3 3

2

(PAGES 200-201)

b. x 23 , x ≥ −2 =

e.= r 7 , −3 , r ≤ 15

d.

10 y 6 + 4 y 3 y 25 , y≥0, y≠ 25 − 2 y 2

23. In the function g(x), the denominator cannot equal 0. Hence x >

1 2

(2y)

25 + 5 2 y − 5 2 y −

5.3: SOLVING RADICAL EQUATIONS

26. a. y = 5 , 1 , y ≥

− 5 +1 4

10 y 6 + 2 y 12 y

22. First error occurs in Step 4. Steps should be: =

24. a. x 100 , x ≥ 0 =

f. −6

7 4

3 2

c. x = {∅} , x ≥ −2

d.= x 6, x≥

g. = x 1 , −2 , x ≤ 2

h. x 84 , x ≥ 3 =

c.= x 4, x≥0

b. c = 2 , 16 , c ≥ − f. n = 6 , 2 , n ≥

4 . 3

3 2

j. = x 1 , −4 , w ≤ 4

2 3

c. k = −1 , −4 , k ≥ −5

d. m = 2 , 5 , m ≤ 6

19 54 h. d 9, 10 , d ≥ = 6 7 3 6 7 k. t = l. a 1 , , a ≥ − −1 , −9 , t ≤ = 4 19 2 g. = p 5 , −3 , p ≥ −

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Mathematics 2200

SOLUTIONS

365

5.3: SOLVING RADICAL EQUATIONS 27. a. x = 0 , (no restrictions ) 28. a.= u 1, u ≥

1 4

d.= y 2, y≥−

(PAGES 202-204)

b. x =−2 , 0 , −

3 5

(

5 5 ≤x≤ 2 2

b. x 20 , x ≥ =

c. = x 2 , −8 , x ∈ −∞, − 72  

58 3

c.= h 4,h≥

e. x 22 , x ≥ 2 =

29. a. x 0 , 8 , x ≥ 0 =

b. x = 1 , 9 , x ≥ 1

30. a.= x 2, x≥2

b. x = 1 , 8 , x ≥ −

31. a. x = −8 , 2 , x ≥ 1

32. Solving gives the one solution x =−13 ⇒ Therefore there are no solutions.

∪

7

,∞  2

)

3 2

f. x 10 , x ≥ 3 = c. x 0 , 3 , x ≥ 0 =

1 3 3 e. x = 2 , 114 , x ≥ 2 7 b. x = −9 , 4 , x ≥ 2

d. x = −1 , 3 , x ≥ −1

(continued) …

c. x = 2 , 10 , x ≥ 2 f. x = −

2 1 5 ,2,− ≤x≤ 9 4 2

c. x = 0 , 2 , x ≥ 0

−17 − −17 which are non-real radical values.

45 43 33. Solving P = 2l + 2 w ⇒ 2 14 p − 45 + 2(2 p ) = 54 gives p= . , p 9 where p ≥ = 14 2 Therefore the side lengths are 14 p − 45 ⇒ 2 p ⇒ 2(9) = 18 cm .

14(9) − 45 =

126 − 45 =

6.1: RATIONAL EXPRESSIONS 1. a. x ≠ −4 g. x ≠ ± 2 y

b. x ≠ 6

1 , x ≠ −5 x+5 1 , y ≠ −7, −1 y+7 x+4 i. , x≠4 2 n−4 m. , n ≠ 2,5 n−2

2. a.

Sharpe Mathematics  2017

(PAGES 207-209) c. x ≠

1 h. x ≠ − , −2 2

1 2

1 a , a ≠ 0, 2a − 1 2 1 f. , x ≠ −2, 7 x−7 m−3 j. , m ≠ −3 2 2 1 n. , x ≠ −2, x+2 2

b.

81 = 9 cm and

d. x ≠

2y 3

c. 3, x ≠ −

e. x ≠ 8, −3

f. x ≠ −2, −1

1 2

d.

t +3 ,t≠3 t −3 x −1 k. , x ≠ ±2 x+2 1 x−6 o. , x ≠ −3, 2x − 1 2

g.

DO NOT COPY!

1 , x ≠ ±2 x+2 2x − 1 2 h. , x≠− 3x + 2 3 x +1 l. , x ≠ −5, −3 x+5 x−3 2 p. , x ≠ −3, − x+3 3 Mathematics 2200

366

SOLUTIONS

6.2: MULTIPLYING RATIONAL EXPRESSIONS 3. a.

x , x ≠ 0, ± 3 2( x − 3)

e. (t + 2)(t + 1), t ≠ −3,1 i.

5y , y ≠ −5, 3, 6 8( y − 6)

2 , y ≠ −3, −2, 0 5y x +1 f. , x ≠ ± 3, ± 1 x −1 2x − 1 j. , x ≠ −5, −3, 4 6( x + 5) b.

(PAGE 211) x−5 , x ≠ −5, ± 3 x+3 x+2 g. , x ≠ 0, ± 3 6

c.

b.

1 , x ≠ −1 x +1

x+5 3 , x ≠ − ,5 6 2

h. 1, p ≠ −5, −4, −3, 2

k. a + 3, a ≠ ± 3

6.3: DIVIDING RATIONAL EXPRESSIONS 4. Answers may vary. 1 a. , x≠2 x−2

d.

l. 1, m ≠ 0, ± 1, 5 (PAGE 213)

c.

1 , x≠0 x

d.

1 , x ≠ 1, 3 ( x − 3)( x − 1)

5. False. They are not the same since the NPV’s are different. 6. a. 3( x − 2) , x ≠ 0, −2 1 , x ≠ −4, ± 5 3 4 x( x − 3) g. , x ≠ −2,3,5 3 6 x( x + 7) j. , x ≠ −7 , −1, 2, 4 x−4

d.

b.

a+5 , a ≠ 5, ± 6 a+6

e. 5( y + 5), y ≠ −5, 0 (n − 3)(n + 2) , n ≠ −4, −3,− 2, 1 3(n + 4) 3( x − 1)(2 x + 3) 3 k. , x ≠ −2, −1, 0,1, 2 x( x + 2)( x + 1) 2 h. −

6.4A: ADDING & SUBTRACTING RATIONAL EXPRESSIONS 7. a. d. g. j.

LCD = 15 LCD =(t − 3)(t − 5) LCD = ( x − 4)( x + 4)

LCD =( x − 5)( x − 2)( x + 3)

2x + 5 , x≠0 6x 17 x − 4 d. , no NPV ' s 6

8. a.

g.

3n − 11 , n ≠ 2,3 (n − 2)(n − 3)

Sharpe Mathematics  2017

b. e. h. k.

LCD = 6m LCD = x( x + 2) LCD = 2 y ( y + 2)

LCD = 9 x 2 ( x − 4)(2 x + 3)

−5 , x≠2 x−2 1 e. − , y≠0 12 y b.

h.

x + 15 , x ≠ −3, 0 2 x( x + 3)

DO NOT COPY!

2( x − 7) , x ≠ −2, ± 7 x+7 2 1 f. , x ≠ ± ,3, 4 2x − 1 2 3(t − 2) 1 i. , t ≠ −7 , − , 0, 4 5(t + 7) 2 x+7 3 l. , x ≠ −5, −3, 2x − 3 2

c.

(PAGE 215)

c. LCD =( x + 1)( x + 4) f. LCD =( x + 3)( x + 3)( x − 2) i. LCD = ( x + 7)( x + 2) l. LCD =2( x − 6)( x + 6)( x + 2) a+2 1 ,a≠− 2a + 1 2 7 x + 17 f. , x ≠ −3, −2 ( x + 2)( x + 3) c.

i.

5 x 2 − 3 x − 15 , x ≠ −4,3 ( x + 4) 2 ( x − 3)

Mathematics 2200

SOLUTIONS

367

6.4A: ADDING & SUBTRACTING RATIONAL EXPRESSIONS 9. a. d. g. j.

(PAGE 216)

2 x 2 + 8 x − 15 8x2 + 9 x − 6 3 b. , x ≠ ±4 , x ≠ 0, 5( x − 4)( x + 4) 2 x(4 x − 3) 4 2 4r − 9r + 3 −4 x − 25 e. , r ≠ −2, 0 , x ≠± 7 3r (r + 2) ( x − 7)( x + 7) 1 1 h. , x ≠ 3, 4, 6 , x ≠ ±2 ( x − 3)( x − 4) ( x + 2) 3 7 k + 22 2 k. , x ≠ −2, 0 , k ≠ −1, − ,8 2x (3k + 2)(k + 1)(k − 8) 3

y −1 , y ≠ ±1 y +1 x 2 + 8 x + 14 f. , x ≠ −8, −3 ( x + 8)( x + 3) 5 i. , k ≠ −3, 2 (k − 2)(k + 3) x 2 − x − 10 l. , x ≠ −4, 0 x( x + 4) c.

6.4B: COMPLEX FRACTIONS

(PAGES 218 – 219)

10. a.

x 1 , x ≠ ± ,0 2x − 1 2

11. a.

5x − 1 , x ≠ ±1, −4 2x + 8

b.

4x − 8 11 , x ≠ −3, − , −2,1 4 x + 11 4

c. −

2x + 8 1 , x ≠ −2, −1, 2x − 1 2

e.

3x − 1 1 , x ≠ − ,0 3x + 1 3

f.

d.

b.

−x , x ≠ −2, −1 x+2

c.

3x + 2 , x ≠±2 6

6.5A: SOLVING RATIONAL EQUATIONS 12. a. = x 2, x ≠ −2, −1 d. x 2, 6, x ≠ 0 =

13. a. x = −3,1, x ≠ ± 2 9 d. x = − ,1, x ≠ −3, −1 5

14. a. x = 1 , x ≠ ±1∴ no solution

Sharpe Mathematics  2017

b. x 1, 2, x ≠ 0, = e. x =

d.

x −1 , x ≠ 4, 6 x−6

3 xy , x ≠ ± y, x ≠ 0, y ≠ 0 x+ y

xy 4y , x≠± x ≠ 0, y ≠ 0 3x − 4 y 3 (PAGES 221 – 222)

2 3

c. x = −2, x ≠ −5,1

65 , x≠0 2

f.= x 5, x ≠ 0

1 b. x = − ,1, x ≠ −1, 0 2

7 c. x = − , 2, x ≠ −2, 4 2

e. x = −5 , x ≠ −5 ∴ no solution

f. x = −3 ,9, x ≠ ± 3

5 b. x = , 3 , x ≠ −2,3 3

DO NOT COPY!

c. = x 5, x ≠ ± 2

d. = x 7, x ≠ −3, 2

Mathematics 2200

368

SOLUTIONS

6.5B: APPLICATIONS OF RATIONAL EQUATIONS 15. a.

(PAGES 224 – 226)

16. D

S

T

Glenn

700

s

700 s

Elsie

700

s – 30

700 s − 30

b. Glenn =100 km/h

T

Fraction

Nancy

x+7

1 x+7

1 1 1 + = x + 7 x 12 24 x + 84 = x 2 + 7 x

Debbie

x

1 x

x 2 − 17 x − 84 = 0 ( x − 21)( x + 4) = 0

Together

12

1 12

x = 21 , x = −4 ∴it took Nancy 28 hrs

Elsie = 70 km/h 17.

18.

19. D

Total

#

Cost

Orig

810

x

810 x

Norm

After

810

x–3

810 x−3

Faster

810 810 − = 3 x−3 x ( x − 30)( x + 27) = 0

S

T

24

s

24 s

24

s+1

24 s +1

24 24 −2= s s +1 ( s − 3)( s + 4) = 0

x = 30 , x = −27 ∴ originally 30 students booked

20. S

T

Toby

10

s+1

10 s +1

Mich

10

s

10 s

10 10 9 + = s +1 s 2 5 s= 4, x= − 9 ∴ Toby = 5 km / h Michelle = 4 km / h Sharpe Mathematics  2017

S

T

Norm

200

s

200 s

Faster

200

s + 10

200 s + 10

200 200 − = 1 s s + 10 ( s − 40)( s + 50) = 0

s = 3 , s = −4

s = 40 , s = −50 ∴ normal speed is 40 km / hr

∴ normal speed is 3 km / hr

21. D

D

22.

2x

1000

2x + 200 2

2

(2 x) + (2 x + 200) = (1000) x = 300 , x = −400 ∴ slower speed is 300 km / hr faster speed is 400 km / hr

DO NOT COPY!

2

D

S

T

Down

40

s+5

40 s+5

Up

40

s-5

40 s−5

40 40 9 + = s s−5 5 5 s= 4,s= − 9 ∴ in still water the speed is 4 km / hr Mathematics 2200

SOLUTIONS

369

7.1: ABSOLUTE VALUE 1. a. 11

b. 0.38

2. a. x = ± 5

(PAGE 229)

c. –0.24

d.

5 7

e. 8

b. x = ± 2.6

3. −3 , − 2.5 , −1.6 , 2.5 ,

−5.28 2

, −2

f. 4

g. –16.8

c. x = ± 3

h. 17

i. 2

j. –6

d. no solution

2 11 , 2.71 , − 3 4

7.2: ABSOLUTE VALUE FUNCTIONS

(PAGES 238 – 239)

4. a.

b.

c.

5. a.

b.

c.

 x + 1 , x ≥ −1 6. a. f ( x) =  − x − 1 , x < −1

 2 x + 1 , x ≥ − 12 b. f ( x) =  1 −2 x − 1 , x < − 2

 12 x + 12 , x ≥ −1 c. f ( x) =  1 1 − 2 x − 2 , x < −1

2  x +1 , −1 < x < 1 7. a. f ( x) = −  2

2  ( x + 3) + 4 , − 5 < x < −1 b. f ( x) = −  2

2  ( x − 1) + 4 , − 1 < x < 3 c. f ( x) = −  2

 x − 1 , x ≤ −1 , x ≥ 1

8. a. x = −1 g. x = − 12 m. x = 72

( x + 3) − 4 , x ≤ −5 , x ≥ −1

b. x = 3

c. x = 5

d. x = 2

h. x = − 23 n. x = 6

i. x = 54

j. x = −2

Sharpe Mathematics  2017

( x − 1) − 4 , x ≤ −1 , x ≥ 3

e. x = − 12 k. x = −12

f. x = 12

l. x = 18

o. x = − 43

DO NOT COPY!

Mathematics 2200

370

SOLUTIONS

7.2: ABSOLUTE VALUE FUNCTIONS (PAGES 239 – 242) 9. a. x = ± 1

b. x = ± 3 h. x = −6,1 n. no x − int

g. x =−5, −3 m. x = −1,3

c. x = ± 5 i. x = −3, 4 o. x = − 12 , 4

 x + 1 , x ≥ −1 10. a. f ( x) =  − x − 1 , x < −1  x−2 , x≥ 2 d. f ( x) =  − x + 2 , x < 2

 2 x + 1 , x ≥ − 12 g. f ( x) =  1 −2 x − 1 , x < − 2  1 x + 1 , x ≥ −2 j. f ( x) =  2 1 − 2 x − 1 , x < −2  2( x + 1) − 9, x ≥ m. f ( x) =  −2( x + 1) + 9, x
0

b. k = 0

c. k < 0

7.4: RECIPROCAL FUNCTIONS 26. a. x-intercept

b. 1,– 1

c. 0

(PAGE 265) d. ∞

e. 0

f. ∞

27.

a.

b.

c.

d.

28.

a.

b.

c.

d.

Sharpe Mathematics  2017

DO NOT COPY!

g. 1

Mathematics 2200

SOLUTIONS

373

7.4: RECIPROCAL FUNCTIONS

(PAGE 266)

29.

a.

b.

c.

d.

30.

a.

b.

c.

d.

31. a.

b. As the value of a gets larger, the value of 1a gets smaller. As the value of a gets smaller, the value of 1a gets larger. As the value of a gets larger, the “vertex” of the graph of 1 gets closer to the origin in Quadrants 1 and 3. y = ax 1 As the value of a gets larger, the graph of y = ax approaches the x- and y- axes faster.

c. If a < 0 , the graphs of y = ax would instead in Quadrants 2 and 4, passing through the origin. Likewise, the reciprocal 1 would also occur in Quadrants 2 and 4. graphs y = ax

Sharpe Mathematics  2017

DO NOT COPY!

Mathematics 2200

374

SOLUTIONS

7.4: RECIPROCAL FUNCTIONS 32.

(PAGES 267 – 269)

Function: = y 2x − 4 x-intercept: (2, 0) y-intercept: ( 0, −4 ) Reciprocal Function: y = NPVs: x ≠ 2 HA: y = 0 VA: x = 12

(

1 2x − 4

) (

)

Invariant Points: 32 , −1 , 52 , 1 x-intercept: none y-intercept: 0, − 14

(

33.

)

Function: y = −3 x + 4

(

x-intercept: 43 , 0

)

y-intercept: ( 0, 4 )

Reciprocal Function: y = NPVs: x ≠ 43 HA: y = 0 VA: x = 43

1 −3 x + 4

(

)

Invariant Points: (1, 1) , 53 , −1 x-intercept: none y-intercept: 0, 14

(

34.

)

Function: = y 12 x + 2 x-intercept: (–4, 0) y-intercept: ( 0, 2 ) 1 Reciprocal Function: y = 1 2x+2

NPVs: x ≠ −4 HA: y = 0 VA: x = −4 Invariant Points: ( −6, −1) , ( −2, 1) x-intercept: none y-intercept: 0, 12

(

Sharpe Mathematics  2017

DO NOT COPY!

)

Mathematics 2200

SOLUTIONS

375

7.4: RECIPROCAL FUNCTIONS 35.

(PAGES 270 – 272) Function: y = x 2 + 3 x − 4 x-intercepts: ( −4, 0 ) , (1, 0) y-intercept: ( 0, −4 )

(

vertex: − 32 , − 25 4

)

Reciprocal Function: y =

1 x + 3x − 4 2

NPVs: x ≠ −4, 1 HA: y = 0 VA: x = −4 , x = 1

(

invariant points: −3 ±2 x-intercept: none y-intercept: 0, − 14

21

)(

, −1 , −3 ±2

(

) 4 reciprocal vertex: ( − 32 , − 25 ) 36.

Function: y= 4 − x 2 x-intercepts: ( −2, 0 ) , (2, 0) y-intercept: ( 0, 4 ) vertex: ( 0, 4 ) Reciprocal Function: y =

1 4 − x2

NPVs: x ≠ −2, 2 HA: y = 0 VA: x = −2 , x = 2

(

) (

)

invariant points: ± 5, −1 , ± 3, 1 x-intercept: none y-intercept: 0, 14

(

)

(

reciprocal vertex: 0, 14

Sharpe Mathematics  2017

DO NOT COPY!

)

Mathematics 2200

29

)

,1

376

37.

SOLUTIONS

Function: y =( x − 3) 2 + 2 x-intercept: none y-intercept: ( 0, 11) vertex: ( 3, 2 ) Reciprocal Function: y =

1 ( x − 3) 2 + 2

NPVs: none HA: y = 0 VA: none invariant points: none x-intercept: none 1 y-intercept: 0, 11

(

) reciprocal vertex: ( 0, 14 )

38.

Function: y = − x 2 + 8 x − 16 x-intercept: ( 4, 0 ) y-intercept: ( 0, −16 ) vertex: ( 4, 0 ) Reciprocal Function: y =

1 − x + 8 x − 16 2

NPVs: x ≠ 4 HA: y = 0 VA: x = 4 invariant points: ( 3, −1) , (5, −1) x-intercept: none 1 y-intercept: 0, − 16

(

)

reciprocal vertex: none

Sharpe Mathematics  2017

DO NOT COPY!

Mathematics 2200

SOLUTIONS

377

8.01: SOLVING A LINEAR-LINEAR SYSTEM GRAPHICALLY 1. a.

b.

(PAGE 278)

c.

d.

e.

f.

g.

h.

i.

8.02: USING SUBSTITUTION TO SOLVE A LINEAR-LINEAR SYSTEM 2.= a. x 25 = , y 21

b.= x 2= , y 2

24 c. x 16 , y 19 = = 19

8.03: USING ELIMINATION TO SOLVE A LINEAR-LINEAR SYSTEM 3. a. x = 2 , y = −6

Sharpe Mathematics  2017

b. x = 1 , y = −1

c. x = −2 , y = 3

DO NOT COPY!

(PAGE 281) d.= x 3= , y 2

(PAGE 284) d. x = − 23 , y = −5

Mathematics 2200

378

SOLUTIONS

8.1: SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY 4. a. YES 5. a. (2, −1) e. (−3, 1) , (1, 1)

b. NO

c. NO

b. (−3, 4) , (1, 0) f. none

d. YES c. (−2, 5) , (2, 1) g. (−3, −3) , (5, 5)

6.

7.

8.

9.

Sharpe Mathematics  2017

(PAGE 294 - 296)

DO NOT COPY!

d. (1, 0) h. (2, 1)

Mathematics 2200

SOLUTIONS

379

8.1: SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY

(PAGE 297)

10. a.

b.

c.

d.

e.

f.

g.

h.

i.

8.2: SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY

(

)

(PAGES 304 - 305)

11. a. (0, 1) , (1, 2)

b. − 12 , 2 , (1, 5)

c. (−4, −3) , (2, 3)

d. (−3, 0) , (0, −3)

12. a. (0, −9) , (1, −12)

b. (1, 5)

c. (−2, 27) , (1, 9)

d. (−4, 6) , (4, −2)

Sharpe Mathematics  2017

DO NOT COPY!

Mathematics 2200

380

SOLUTIONS

8.2: SOLVING SYSTEMS OF EQUATIONS ALGEBRAICALLY

(PAGES 306 - 312)

(

)

13. a. (−2, −1)

b. (−5, 8) , (−1, 8)

c. (2, 1)

d. 14 , 11 8 , (4, 7)

14. a. (2, 0)

b. (−2, 5) , (2, 1)

c. (−1, 3) , (3, − 37)

d. (0, 7) , (2, 3)

c. (−1, −5) , (0, −2)

d. (−5, 22) , (1, 4)

(

15. a. (0, 1) , 23 , 19

)

(

) (

b. 32 , − 14 , 52 , 34

16. a. (−5, 4) , (−3, 4)

)

(

c. (−2, 12) , 13 , −2

b. (4, 4)

)

d. (−1, 2) , (3, 10)

17. The line is tangent at (−1, −2) ⇒ b =−4 2

18. Parabola equations are y = 2( x − 3) − 1 . Thus

y= −1( x − 1) 2 + 3

2( x − 3) 2 − 1 =−1( x − 1) 2 + 3 2( x − 3)( x − 3) − 1 =−1( x − 1)( x − 1) + 3 (2 x 2 − 12 x + 18) − 1 = (− x 2 + 2 x − 1) + 3 3 x 2 − 14 x + 15 = 0 (3 x − 5)( x − 3) = 0

(

5, x = x= 3 ⇒ intersect at (3, −1) , 53 , 23 3 9

19. a. t = 6 , t = −4 (reject ) ⇒ t = 6 seconds

b. h(6) = 156 m

)

c. t = 45 seconds

20. t = 6,t= 1 ⇒ David has two chances to hit the skeet: at t = 1 second and again at 6 seconds . 21. a. t = 2 seconds

b. h(2) = 2.4 m

22. x = 18 ⇒ h(18) =3.6 meters

∴ he will catch the ball when it is 18m from

−2.1 meters (reject ) x= 21 ⇒ h(18) = 23.

x + 3y = 24   x − 5 y =y − 13 2

24.

2x − 5 = y   2 y+x = 115

25.

x − 12 = y   x − 30 y = 360  2

26. x + 2 y = 46   2 x − 3y = 93

home plate and at a a height of 3.6 m.

x = 3 , x = 53 (reject since 53 ∉ W) ⇒ The numbers are x = 3 , y = 7.

x= −12 , x = 10 ⇒ The numbers are x = −12, y = −29 and x = 10 , y = 15.

x= 0, x= 30 ⇒ The numbers are x = 0,y = −12 and x = 30 , y = 18.

27 x= 12 , x = − 27 12 , y = 17. 2 ( reject since − 2 ∉ I) ⇒ The numbers are x =

Sharpe Mathematics  2017

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Mathematics 2200

SOLUTIONS

381

9.1 LINEAR INEQUALITIES IN TWO VARIABLES b. YES

1.

a. NO

2.

a. x > 3

b. y ≤ 1

c. y > –2

d. y < x + 3

e. y ≥ – 12 x + 2

f. y < 2x – 4

g. 2x + y > 5

h. 2x – 4y ≤ –12

i. 3x + 2y ≤ 4

Sharpe Mathematics  2017

c. YES

d. NO

e. YES

(PAGE 320)

DO NOT COPY!

f. YES

g. YES

h. YES

Mathematics 2200

382

SOLUTIONS

9.1 LINEAR INEQUALITIES IN TWO VARIABLES 3.

(PAGE 321)

a. 3x – 2y ≤ 6

b.

2y – 3x + 4 > 0

c.

d. – 3y ≥ 15 – 5x

e.

3x < – y + 4

f. – 6 + x ≤ 2y

x+3≤0

i. – y + 1 > 0

g.

y–2>0

h.

j.

y – 2x ≥ 3

k. x < –2

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DO NOT COPY!

2x + 5y – 10 < 0

l. –4y ≤ 3x – 12

Mathematics 2200

SOLUTIONS

383

9.1 LINEAR INEQUALITIES IN TWO VARIABLES 4.

(PAGES 322 - 323)

Assign variables: Let x = number of cars manufactured Let y = number of trucks manufactured Write the inequality:

x + y ≤ 100

State the restrictions:

x≥0 y≥0

Test point check:

Test (40, 20): (40) + (20) = 60 < 100 Thus (40, 20) is in the shaded region. So the correct region has been shaded.

5.

Assign variables: Let x = luggage weight for economy passengers Let y = luggage weight for first-class passengers Write the inequality:

40x + 100y ≤ 5000

State the restrictions:

x≥0 y≥0

Test point check: Test (50, 10): 40(50) + 100(10) = 3000 < 5000 Thus (50, 10) is in the shaded region. So the correct region has been shaded.

6.

Assign variables: Let x = number of pairs of sneakers sold Let y = number of jerseys sold Write the inequality: 60x +75 y ≥ 500 State the restrictions: x ≥ 0 y≥0 Test point check:

Test (6, 8): 60(6) + 75(8) = 960 > 500 Thus (6, 8) is in the shaded region. So the correct region has been shaded.

Sharpe Mathematics  2017

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Mathematics 2200

384

SOLUTIONS

9.1 LINEAR INEQUALITIES IN TWO VARIABLES 7.

(PAGE 323) (cont’d)…

Assign variables: Let x = number of emerald cedars sold Let y = number of blue spruce sold Write the inequality: 8x + 9y ≤ 120 State the restrictions: x ≥ 0 y≥0 Test point check:

Test (4, 8): 8(4) + 9(8) = 104 < 120 Thus (4, 8) is in the shaded region. So the correct region has been shaded.

9.2 QUADRATIC INEQUALITIES IN ONE VARIABLE 8. a. x = −2, 5 9. a.

+

b. x < −2, x > 5

–

+

–3

b.

–

c. −2 < x < 5

+

1

–

–2

11. a.

+

– 3

e.

[ 3, 4 ]

+

+

+

4

– –2

b.

–

4

[ − 2, 4 ]

Sharpe Mathematics  2017

+ − 32

–

+

1

+

c.

3

d.

–

–

5

–

+

+

–

–3

+

d.

+

8

–3

g.

+

− 12

( −∞, − 32 ) ∪ ( − 12 , ∞ )

DO NOT COPY!

– –3

( −3, 52 )

+ 2 5

– 5

( −∞, −3] ∪ [8, ∞ )

( −∞, 2 ) ∪ ( 3, ∞ ) +

+

e. −2 ≤ x ≤ 5

b. {x | − 6 ≤ x ≤ 2 , x ∈ R}

2

f.

d. x ≤ −2, x ≥ 5

2

a. {x | x < − 2 , x > 4 , x ∈ R}

10.

c.

(PAGES 329 – 330)

h.

( 5, 8)

+

+ 8

– − 13

+ 3

 − 13 , 3 

Mathematics 2200

SOLUTIONS

385

9.2 QUADRATIC INEQUALITIES IN ONE VARIABLE 12. a.

+

–

+

–3

b.

+

3

1

[ −3, 3 ] e.

–

+

+ 4

[ − 2, 4 ]

f.

( −∞, 1 −

+

7

+

–

− 23

–

6  ∪ 1 + 6, ∞

g.

+

− 12

)

d.

(

( −3, 5)

+

b.

+

d.

5

–3

– –8

( −∞, − 32 ) ∪ ( − 12 , ∞ )

13. a. 1 − 3, 1 + 3  c.

+

( −∞, 1) ∪ ( 7, ∞ )

– –2

c.

(PAGES 331 – 334)

1 2

( −∞, 6 − 2 5 ) ∪ ( 6 + 2

( −42− 6 , −42+ 6 )

5, ∞

+

3 2 −∞, − 12  ∪  32 , ∞

(

( −8, 12 )

– − 12

h.

+

+

+

– − 72

)

+

9 2 7 9 −∞, − 2  ∪  2 , ∞

)

)

14. t ∈ (1, 2 ) ⇒ The football was at a height greater than 10 m between 1 and 2 seconds 15. x = 0 , 8 ⇒ x ≥ 8 ⇒ The minimum leg lengths are 6 and 8 cm. 16. x = −5 , 10 ⇒ Since x ≥ 0 is a restriction we have 0 ≤ x ≤ 10

⇒ Ticket prices can be $10, $11, $12, ... , $20. 17. 100 − 20 10 ≤ x ≤ 100 + 20 10 ⇒ 36.75 m ≤ x ≤ 163.2 m 18. t ∈ ( 2, 10 ) ⇒ The arrow was at a height greater than 100 m between 2 and 10 seconds. 19. x = −20 , 3 ⇒ Therefore x < −20 or x > 3. Since x > 0 the increase must be at least 3 m. 20. x = −30 , 5 ⇒ Therefore x ≤ −30 or x ≥ 5. Since x > 0 the mat's width must be at least 5 m.

Sharpe Mathematics  2017

DO NOT COPY!

Mathematics 2200

386

SOLUTIONS

9.3 QUADRATIC INEQUALITIES IN TWO VARIABLES

(PAGES 337 – 339)

21. a. NO

b. YES

c. NO

d. YES

e. NO

f. NO

22. a. YES

b. YES

c. NO

d. YES

e. NO

f. YES

23. a.

b.

c.

d.

e.

f.

g.

h.

b. y < −( x − 2) 2

c. y ≥ ( x + 2)( x − 1)

d. y ≥ ( x + 1) 2 + 2

24. a. y < x( x + 3) e. y ≤ − x( x + 3)

f. y ≤ ( x + 1)( x + 3)

g. y > − 12 x 2 + 1

25. a.

b.

c.

d.

e.

f.

Sharpe Mathematics  2017

DO NOT COPY!

h. y > 2( x − 1) 2 − 2

Mathematics 2200

SOLUTIONS

387

9.3 QUADRATIC INEQUALITIES IN TWO VARIABLES

(PAGES 340 – 342)

26. a.

b.

c.

d.

e.

f.

27.

a.

28. a/b.

b. Between 20 and 40 students are needed.

Sharpe Mathematics  2017

c. Between 10 and 60 seconds (50 seconds total).

DO NOT COPY!

Mathematics 2200