Methods of Solving Problems in High-School Mathematics

Table of contents :
Front Page
Title Page
Contents
From the Editor
From the Authors
Chapter 1 Algebraic Equations and Systems of Equations
Chapter 2 Logarithms. Exponential and Logarithmic Equations
Chapter 3 Inequalities. Equations and Inequalities with Parameters
Chapter 4 Trigonometry
Chapter 5 Complex Numbers
Chapter 6 Sequences
Chapter 7 Limit of a Function. Continuity of a Function
Chapter 8 The Derivative and Its Applications
Chapter 9 The Antiderivative and the Integral
Chapter 10 Problems on Deriving Equations
Chapter 11 Plane Geometry
Chapter 12 Solid Geometry
Chapter 13 The Method of Coordinates
Chapter 14 Combinatorics.The Binomial Theorem. Elements of the Theory of Probabilities
Answers and Hints

Citation preview

A.Tsypkin and APinsky t.

M ethods of Solving Problems inHigh-School M athematics

M ethods of Solving Problems in High-School M athematics

A. T. UMIIKHH, A. H. IIHHCKHfl

GnPABOHHOE IIOCOBHE n o METOAAM PEfflEHHH 3AflAH n o MATEMATHKE HsftaTejibCTBo «Hayica» MocKBa

A.G.Tsypkm and A.I.Pinsky M ethods of Solving Problem s in High-School M athem atics Under the editorship of V. I. Blagodatskikh Translated from the Russian by Irene Aleksanova

Mir Publishers Moscow

First published 1986 Revised from the 1983 Russian edition

Ha am/iuucKoM si3biKe

© MaflaTejibCTBo «HayKa» rjiaBHan peAanitHH 0, equa­ tion (3) has two real roots: x' ~

-b + VD 2a

_ ’

-b -Y D 2a

b If D = 0, equation (3) has one real root of multiplicity 2: x — —^ If D < 0, then equation (3) has no real roots. A solution of many rational equations consists in reducing them, by some technique, to equations of form (2) or (3). An introduction of an auxiliary unknown is one of those techniques. Example 1.1. Solve the equation 1 1 _ 1 x (x-\-2) ( £ + 1)2 12 Solution. Designating z = x2 -|- 2x, we write the initial equation in the form 1 1 _ 1 {*) z z+ 1 12 *

1 Rational Equations in One Unknown

13

Simple transformations reduce equation (*) to the equation z2-f-z—12_ 12z (z -f-1)

(**)

which is equivalent to the equation z2 + z — 12 = 0. The equivalence of these equations follows from the fact that the roots z = —4 and z = 3 of the last equation belong to the set of permissible values of equation (**). Thus the initial equation is equivalent to two quad­ ratic equations, x2 -|- 2x — 3 = 0 and x2 -f- 2x -j- 4 = 0. The roots of the first equation are x1 = 1 and x2 = —3. The second equation has no real roots. Answer, x = 1, x = —3. Solve the following equations.

1.4.

^ x2-)-4x = 6. x2—4x + 10 (x2—6x)2—2(x —3)2= 81. 24 15 2+ 2x —3 x2+ 2x —8

,.5.

7 (h 4 ) - 2

1. 6 .

x2-f- 2a;+ 1 2+ 2x + 2

1. 1.

1.3.

1.2.

=

+4 ^

a;2+ 5

=2.

2.

(«■ + ■£-)=»■ 2+ 2x + 2 x2+ 2x+ 3 -48

1.7.

x2—4 x2— 1

=0.

1. 8 . (x2 + 2a;)2 — (x + l )2 = 55.

1.9. \x2 (x2 + x + 1) (x2 + x + 2) -— 12 = 0. 1.10*. + I)2 7)2 — — (x — - 2) I(x — 3) = 0. *. (x2 — — 5x 5x41.11*. (x — 2) (x + 1) (x + 4) (x + 7) = 19. 1.12*. (2x2 + 3x — 2) (5 — 6x — 4x2) = —5 (2x2 + 3 x + 2). 1.13. x4 — 13x2 + 36 = 0. 1.14. 2x8 + x4 — 15 = 0. 1.15. (2x — 1)« + 3 (2x - l )3 = 10. 1.16*. (1 + x)8 + (1 + a;2)4 = 2x4. 1.17. (x — 2)fl - 19 (x — 2)3 = 216.

One of the ways to solve an equation of a degree exceeding two is to factor the polynomial appearing on the left-hand side of the equation and thus reduce the solution of the initial equation to the solution of several equations of lower degrees. This method is based on the follow­ ing property of the roots of an nth-degree polynomial. If the number c is a root of the polynomial P (x) = a0xn -f- fljx71-1 an_jx -fthen the polynomial can be written in the form P (x) = (x — c) Q (x),

(4)

where Q (x) is a polynomial of degree n — 1 (i.e. the polynomial P (x) can be divided by the polynomial x — c).

14

Ch. 1 Algebraic Equations and Systems of Equations

Factoring a polynomial is equivalent to finding its roots. Finding the roots of a polynomial is a difficult problem and in a general case it is impossible to indicate a universal method of seeking the roots of an nth-degree polynomial with real coefficients. There is a theorem, however, which makes it possible to find rational roots of an nth-degree polynomial with integral coefficients. The rational roots of the polynomial a0zn + a1xn~1 + . . . + an_ix + an, where a0, a1? . . ., a7l_1, an are integers, can only be numbers mlp (m is integral and p is natural), the number | m | being the divisor of the number | an | and the number p, the divisor o f the number I ao I* Example 1.2. Find the roots of the equation 4X4 + 8r» — 3a;2 — lx + 3 = 0. Solution. The numbers 1, 3 are the divisors of the number 3, and 1, 2, 4 are the divisors of the number 4. The set {1, —1, 3, —3} is the set of values of m, and the set {1. 2, 4} is the set of values of p. The set {±1, ± 3 , ±1/2, ±1/4, ±3/2. ±3/4} is the set of various distinct rational numbers of the form mlp. Substituting these numbers into the equation, we find that the numbers xt = 1/2 and x2 = —3/2 are its roots. According to (4), this means that the given polynomial can be divided by the linear polynomials (x — 1/2) ana (x -|- 3/2) and, consequently, it can also be divided by their product ( * ~ t ) ( x + t ) = x -+ x ~ t Performing a division, we find the polynomial of the quotient: 4xa+ Ax — 4. Solving the quadratic equation 4a;2 -f* 4a; — 4 = 0, - 1 4— - / 5 andj xA= we get two real, roots: a;8= ----------

- 1g— - 1/— 5 .

We have thus completed the solution of the problem, i.e. have found the four roots of the initial equation: 1

xi —~2~ * Answer.

fl

3 —— 2" *

3

,

—1 + / 5 &

-1 4 -/5

Solve the following equations.

2

— 1— Vo *

X*

—1 - / 5 1

g---. ------- 2------ /*

1. 18. 8x* + 6x* — 13** — x + 3 = 0. 1. 19. * * + 6* + 4 * * + 3 = 0. 1. 20. 2*4 — *» - 9** + 13* — 5 = 0. 1. 21*. (* - 1)» + ( 2* + 3)» = 27** + 8. 1. 22. ** — (2a 4- 1) ** 4 - («* + a) * — (a* — a) = 0. 1. 23. a* - 4** - 19** + 106* — 120 = 0.

----- 2------

1 Rational Equations in One Unknown

15

Some equations of a special form. Equations of the fourth degree whose left-hand side is a product of quadratic trinomials, which differ by a constant term, and the right-hand side is a number, can be re­ duced to quadratic equations by introducing an auxiliary unknown which is equal to the common part of the two factors. Example 1.3. Solve the equation * (x + 1) (x + 2) (x + 3) = 0.5625

(*)

Solution. Multiplying separately x (x -f- 3) and (z-f- 1) (x -f 2). we obtain (x2 -f- 3a:) (x2 -)- 3a; -f- 2) = 0.5625 Introducing an auxiliary unknown y = x2 + 3a:, we get, after simple transformations, a quadratic equation y2 -f- 2y — 0.5625 = 0, whose roots are yx = 0.25 and y2 = —2.25. Returning to the initial unknown, we infer that (*) is equivalent to two equations x2 + 3a: — 0.25 = 0 and x2 -f- 3a: -|- 2.25 = 0. The first equation has two different roots, x = x = ----3 plicity

/ I Q ^ an(j the second equation has one root of multi­ 2, x = — 2" •

Answer.

-3 + /1 0 x = ------—^------ ,

x=

- 3 - Y iO ^

, a: =

3 —.

Find the roots of the following equations. 1.24. (x + a) (a: -f- 2a) (x — 3a) ( x — 4a) = 64. 1.25. (x — 4)(x — 5) (x — 6) (a: — 7) = 1680. 1.26. (6* + 5)2 (3* + 2) (* + 1) = 35. 1.27. x4 - 2x* + * — 132 = 0. 1.28. (x ~ 1) (x + 1) ( x + 2) x = 24. 1.29. (x - 4) (x + 2) (x + 8) (x + 14) = 354. 1.30*. (x2 -f- x + 1) (2x2 + 2x + 3) = 3 (1 — x — x2). The equation of degree n a0xn

aiXn * -f- . . . -f- &n~ix

— 0

(3)

is said to be symmetric if ak = an„h for allk = 0, . . ., n. If n = 21, then we can divide both parts of equation (5) by xl and pass to an equivalent equation + . . . + an-i

1

= 0,

16

Ch. 1 Algebraic Equations and Systems of Equations

and equation (5) can be reduced to an equation of degree I by introducing a new unknown z = x -]— . If n = 21 4- 1, then a direct veri­ fication shows that x = —1 is a root of the equation. The division by (x -f- 1) reduces equation (5) to a symmetric equation of degree n = 21. Equation (5), where ah = (—1)han-k is called skew-symmetric. The arguments presented above are applicable to it with due account of the following changes: for n = 2Z, it is necessary to introduce z = x — 1/x as a new unknown, and for n = 21 -f 1, x = 1 is one of the roots of the equation. Example 1.4. Solve the equation x7 2xe — 5x5 — 13x4 — 13a;3 — 5a;2 -|- 2a; -f- 1 = 0. (*) Solution. The given equation is symmetric and n = 1 and x = —1 are its roots. Consequently, equation (*) can be represented as (x + 1) (a;6 + xB— 6a;4 — 7a;3 — 6a;2 -J- x -f- 1) = 0, and its solution reduces to the solution of an equation of an even degree x6 -j- xb — 6a;4 — 7a;3 — 6a;2 -f- £ + 1 = 0. Dividing both sides of the equation by a;3, we get

We introduce the designation z = x + 1/^ and, taking into account that X2-J----jr*= Z 2—2, X3-j----q“= z3—3z, X

2

X

8

we get an equation z3 -)- z2 —9z — 9 = 0 which is equivalent to the equation (z + 1) (z2 - 9) = 0. Consequently, the solution of the initial equation reduces to the solu­ tion of the following three equations: 1 1 1 ^ + “1" ^ 3, x-f- — = —3, X X X which are respectively equivalent to the quadratic equations x2 + x -j- 1 = 0, x2 — 3x 1 = 0, x2 -f- 3x + 1 = 0. The first equation has no real roots, and the roots of the second and third equations can be calculated by the formula for the roots of a buadratic equation. Ansirer. x

x = —1,

—3 — / 5 2

3 + /§

3 -/5 2

—3-f2

1 Rational Equations in One Unknown Solve 1.31. 1.32. 1.33. 1.34. 1.35.

17

the following equations. x4 + 5*3 + 2x2 + 5x + 1 = 0. 2x4 4- 3 r, - 4 x 2 - 3 i 4 - 2 = 0. 15xB-f- 34x4 + 15x3 — 15a:2 — 34r — 15 = 0. 6z3 — a:2 — 20a: + 12 = 0. x4 + 1 = 2 (1 + x)4.

Hint. To solve equations 1.34 and 1.35, it is first necessary to make a change y = ax + b. As a result of a requisite choice of the numbers a and b, the equations become symmetric (or skew-symmetric) with respect to the unknown y. A rational algebraic equation of the form P{x) 0 (6)