Get Your Head Around: Basic Algebra I

Table of contents :
Title Page
1: Equations
Practice Questions 1
Answers 1
2: Addition and Subtraction
Practice Questions 2
Answers 2
3: Squares and Square Roots
Practice Questions 3
Answers 3
4: Brackets (and Terms)
Practice Questions 4
Answers 4
5: Brackets Multiplied by Brackets
Practice Questions 5
Answers 5
6: Factorisation
Practice Questions 6
Answers 6
7: Quadratic Functions and non-whole numbers
Practice Questions 7
Answers 7
8: Deriving an Equation to Solve Quadratics
Practice Questions 8
Answers 8
9: Extending the Formula for Quadratics
Practice Questions 9
Answers 9
10: Cannonballs
Practice Questions 10
Answers 10
Conclusion

Citation preview

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Get Your Head Around: Basic Algebra I

by Austin Hartnell-Jones

© Copyright 2019 Austin Hartnell-Jones.

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Welcome to ‘Get Your Head Around: Basic Algebra I’. You will probably be aware that mathematics is considered to be the language of science and technology. It is also often used in many other important areas of life, including economics and finance. If we think of mathematics as a ‘language’, then we can think of algebra as being the ‘grammar’, making up the basic rules and building blocks of that language. Whilst mathematics is hugely important in scientific applications, I probably won’t use too many scientific examples, at least not in the early stages of the book. Instead, I will try to use more ‘everyday’ examples that can more easily be visualised. This is to encourage you to get a genuinely instinctive ‘feel’ for algebra. Ideally, it should seem like a lot of common sense. If it doesn’t, you might be over-thinking it, or you might have missed something earlier on. 3

It is very important not to move on to a new section until you have thoroughly grasped the earlier sections at a fundamental ‘common-sense’ level. The only other thing I’ll say before we get started is that the book begins with some very simple stuff. Don’t worry if this seems a little rudimentary, it will get more interesting later on.

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1. Equations Let us begin. First of all, we’ll introduce the idea of ‘equations’, as well as some of the symbols and notations used. As you progress with your mathematical education, you will pick up on the fact that much of what is written down is basically a short-hand way of writing statements that could otherwise be written in plainEnglish (but would take up a lot more space). Lets start with a plain-English statement. Imagine we go to a grocer’s store and we want to buy some eggs (four eggs to be precise). We see that the grocer is selling eggs at 25p (or £0.25) each. So, one egg costs £0.25. We wanted four eggs. As the grocer unfortunately doesn’t give any discount for multiple purchases, the cost for four eggs is going to be four times the cost for one egg. So, the total cost will be £1.00. 5

This all seems very straightforward. Lets write down our initial statement as an equation:

1 egg = £ 0.25 Here, the equals symbol ‘=’ is basically saying ‘has a cost equal to’. We can think about the whole equation as being equivalent to saying ‘the cost of one egg equals £0.25’, or in more common English ‘the cost of one egg is £0.25’. As we are talking about money and cost, it can be assumed that this is what the equation is about. All equations have two sides, the left hand side and the right hand side. The left hand side is what is on the left of the equals sign ‘=’, and the right hand side is what is on the right hand side of the equals sign ‘=’. The equation ‘equates’ the two sides. It says that they are equal to each other, that they are the same. In this case, as we are talking about cost and money, the equation is about the cost of the eggs. If we were talking about the weight of the eggs, then the equation would be about weight. In that case, we 6

would obviously have a weight value on the right hand side instead of a money value. The context of an equation will generally be known to us. It is useful to note that there is nothing particularly important about which side of the equation things go on. We could also have written the previous equation as:

£ 0.25 = 1 egg This isn’t perhaps the way we would say it if we were talking, but it is perfectly fine as an equation. We might say it this way if we said ‘£0.25 will buy me one egg’. Now, lets write down the equation for ‘the cost of four eggs is £1.00’:

4 eggs = £ 1.00 So far, so good. Nothing very interesting, but we need to make sure we are building our understanding on very solid foundations. Let’s slow down a little. How did we get from the first equation to the 7

second equation? We basically increased the number of eggs from 1 to 4 (four times as much) and also increased the cost by the same amount (four times as much). Lets see that in steps:

1 egg = £ 0.25 multiply both sides by 4:

4 × 1 egg = 4 × £0.25 rewrite:

4 eggs = £ 1.00 At this point, its probably worth pointing out something fairly obvious. If I write:

4 eggs = 4 × £ 0.25 and also that:

4 × £ 0.25 = £ 1.00 Then it makes absolute sense that: 8

4 eggs = £ 1.00 In plain English, this is like saying ‘4 eggs cost 4 times £0.25, which is £1.00, so that means 4 eggs cost £1.00’. In more general terms, we can say ‘if A equals B and B equals C, then A equals C’. Or in equations: if:

A=B and:

B =C then:

A=C Sometimes in mathematics, you will see two equations written on one line, like:

A=B=C

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This just means ‘A equals B and B equals C’. Its just a shorter way of writing the two separate equations for A = B and B = C. For our equation for 4 eggs, we could have written:

4 × 1 egg = 4 × £ 0.25 = £ 1.00 Now, all of this might seem so intuitively obvious that we probably didn’t even need to think about it. The bit about multiplying both sides by the same number doesn’t just apply to this equation about eggs, we can do the same thing with any equation. If the left hand side and the right hand side are equal, then 4 times the left hand side will also be equal to 4 times the right hand side. In fact, any number times the left hand side will be equal to any number times the right hand side. For instance:

8 × 1 egg = 8 × £ 0.25 which we would rewrite as:

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8 eggs = £ 2.00 If you multiply one side of an equation by a number, the two sides will still be equal if you multiply the other side of the equation by the same number. If we were buying apples, for instance, and each apple cost £0.40, then 3 apples would cost 3 times as much, which is £1.20. In equations this can be written as:

1 apple = £ 0.40 Multiply both sides by 3:

3 × 1 apple = 3 × £ 0.40 Rewrite:

3 apples = £ 1.20 As I said, this is all hopefully so obvious that you don’t really need to think about it. But it does no harm to think about it.

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Lets look at the reverse process. If we know that three apples cost £1.20, we know that one apple will cost £0.40 (assuming no special offers at the grocer’s shop). Lets go through this using equations, very slowly:

3 apples = £ 1.20 Rewrite as:

3 × 1 apple = 3 × £ 0.40 By writing each side as 3 × something, we now see that we can divide each side by three to get what we want. Lets do this step by step. First divide each side of the equation by three:

3 × 1 apple 3 × £ 0.40 = 3 3 Drawing a line under each side, and placing a 3 below is one way to indicate 12

that we are dividing both sides by three. Another way to do this is to put everything on each side into brackets () (by the way, these are generally known as ‘brackets’ in the UK, and ‘parentheses’ in the US, I’ll use ‘brackets’), and put a slash to the right of it and put the 3 to the right of that:

(3 × 1 apple)/ 3 = (3 × £ 0.40) / 3 Brackets in mathematics specify the order which operations are done. You do whatever is inside the brackets first, and then do what is outside the brackets. In this case, we do the ‘3 × 1 apple’ first, to get 3 apples, and then we divide this by 3. We’ll say more about brackets later on. This equation (with brackets) means exactly the same thing as the one above it. Both ways are fine and you will see both ways used in different books. In fact, we don’t actually need the brackets in this case, but it is generally better to use them just in case (if we had any ‘+’ or ‘-’ symbols then the brackets might be required). 13

Now, if we have 3 times something, but we are also dividing by 3 (in other words ‘unmultiplying’ by three) then the multiplication and division just cancel each other out, and we end up with 1 apple. When you are writing the equations out for yourself, a good way to keep track of things that cancel each other out is to put a mark through them, like:

3 × 1 apple 3 × £ 0.40 = 3 3 where all the ‘3’s have a line struck through them, so when we rewrite the equation we simply leave them out:

1 apple = £ 0.40 As I mentioned, division can be thought of as the opposite of multiplication, or ‘unmultiplication’. If we split the things on each side into an equal number of things, we can divide by that number. When we divide £1.20 by 3, what we are doing is

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finding the cost that would equal £1.20 if we had three times that amount. Usually, we won’t go through each of these steps in detail if we don’t feel we need to, we can just divide each side by a certain number:

3 apples = £ 1.20 Divide each side by 3:

1 apple = £ 0.40 Now. A word about symbols. So far, we have used the words ‘apples’ or ‘eggs’ to represent the things we have a certain number of. This has been okay for now, but if the equations get more complicated it will get a bit cumbersome to write all the words down. For instance, if we were talking about artichokes or cauliflowers, we might get fed up of writing them out all the time. It would be easier to use an abbreviation instead.

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An obvious choice would be to use the letter ‘a’ for artichoke and the letter ‘c’ for cauliflowers, etc. Such single letter abbreviations are used very often in mathematics. They are often called ‘variables’. That just means that the value of the thing (the egg, the artichoke or whatever) can vary, or might be unknown. Very often, it is the unknown value of something that we are trying to work out from the equation. So, using the letter ‘e’ to represent the value of an egg, another way of writing our equation saying that ‘the cost of four eggs is £ 1.00’ would be:

4 e = £ 1.00 We can also write this without a space between the ‘4’ and the ‘e’:

4 e = £ 1.00 This is the more common way of writing numbers and variables.

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If we are fairly certain that we know what the numbers mean (cost in pounds or dollars or whatever, weight in kilos or ounces or whatever), we can also leave the unit symbol out as well:

4e = 1 This is much simpler and easier to write. When we get to longer and more complicated equations, we will not run out of space on the paper. Now, lets have a few more examples before you do some practice. If three pears cost £1.00, what will 9 pears cost? Lets use the letter ‘p’ to write down an equation for this:

3p = 1

This is for 3 pairs. As 9 pairs is 3 times as many, we need to multiply both sides by 3:

3×3 p = 1×3

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As 3×3 is 9 and 1×3 is 3, we can rewrite this as:

9 p =3 So, 9 pears will cost £3.00 If 5 bananas cost £1.50, how much will one banana cost? If we use the letter ‘b’ to represent the bananas, we can write an equation:

5b = 1.5 To get to 1 banana, we need to divide both sides of the equation by 5:

5b 1.5 = 5 5 The 5s on the left hand side cancel out, and 1.5 divided by 5 is 0.3:

5b = 0.3 5 18

where the lines in the 5s indicate that they cancel each other out. When we then rewrite it without the 5s, we get:

b = 0.3 If you are happy to, you can just go straight from the first equation to the last one:

5b = 1.5 divide both sides by 5:

b = 0.3 But: don’t skip steps unless you are 100% confident about everything. So far, the unknown quantity has been the price of an individual item. What if we know how much one item costs, but we need to work out how many of them we can buy with a certain amount of money? Well, its more-or-less the same approach. For instance, supposing that I had £12 and a bottle of wine cost £4, and I wanted to 19

know how many bottles I could afford to buy with my money. Our initial statement is: ‘a certain number of £4 bottles of wine can be bought with £12’. We can write an equation for this:

(number of bottles)×£4 = £12 We can make this a lot simpler if we use the letter ‘n’ instead of writing ‘number of bottles’ and also if we drop the ‘£’ units:

n×4 = 12 Instead of writing ‘n × 4’, we can just write ‘4n’, it means the same thing:

4n = 12 So, to get the left hand side to just be ‘n’, rather than ‘4n’, we need to dived both sides of the equation by 4:

n=3 This means the number of bottles we can buy with £12 is three. 20

Here’s a very similar, but slightly more complicated example. Suppose we have £30 and want to know how many bottles of brandy we can buy, if each bottle costs £20. The equation we need is:

20n = 30 If we divide each side by 20, we get:

n = 1.5 In other words, we can buy one-and-a-half bottles. As buying half a bottle doesn’t make sense (assuming we can buy only whole numbers of bottles as we don’t want to break a bottle), we can actually only buy one bottle. This will cost £20, so we will have £10 left over at the end. Now, its time for you to do some practice questions.

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Practice questions 1 (for all of these questions, make sure you write equations, even if it seems too straightforward to bother – its important to get used to thinking in equations. The answers are on the following pages. Make sure you try the question before looking at the answer!) a) If one tin of tomatoes costs £0.15, how much will eight tins of tomatoes cost? b) If one tin of peas costs £0.50, how much will 10 tins of peas cost? c) If two loaves of bread cost £2.20, how much will six loaves of bread cost? d) If you bought seven packets of cheeseand-onion crisps and the price came to £2.10, what was the cost of a single packet? 22

e) If eleven bottles of lemonade cost you £22 in total, what does a single bottle cost? f) If sixteen bars of chocolate cost £8, what would four bars of chocolate cost? g) If you have £21 and bottles of milk are £3 each, how many bottles can you afford to buy? h) If you have £43 and boxes of chocolate cost £5 each, how many boxes can you buy?

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Answers 1: a)

t = £ 0.15 8t = £ 1.20

b)

t = £ 0.50 10t = £ 5.00

c)

2l = £ 2.20 6l = £ 6.60

(multiply each side by 3) d)

7 p = £ 2.10 p = £ 0.30

(divide each side by 7) 24

e)

11b = £ 22 b=£ 2

(divide each side by 11) f)

16b = £ 8 4b = £ 2

(divide each side by 4) g)

3n = 21

(n = number of bottles)

n=7 (divide each side by 3) 25

h)

5n = 43

(n = number of boxes)

n = 8.6 (divide both sides by 5) As you probably won’t be able to buy 0.6 of a box, you can buy 8 boxes for £40 and you will have £3 left over.

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2: Bringing subtraction

in

addition

and

So far, we have only looked at very simple equations involving multiplication or division. Now we will add some addition and subtraction. Imagine the following scenario: I am working in a factory making chairs. I get paid a fixed daily amount of £100, and on top of that I also get £30 for each chair I make in that day.

pay = £ 30 × (numberof chairs) + £ 100 To make this simpler, let’s use the letter ‘p’ to represent ‘pay’, and ‘n’ for the number of chairs, and let’s also drop the £ units:

p = 30n + 100 Suppose I make 3 chairs on one particular day. My pay for that day will be:

p =(30 × 3) + 100 27

which is 90 + 100 which is £190. Now, supposing I wanted to earn £250 instead. How many chairs would I have to make? Lets write the equation. In this case, we know the value of p, which is 250:

250 = 30n + 100 If we only had the ‘30n’ on the right hand side of the equation, we would be able to divide both sides by 30 to get n on its own. At the moment, we also have the ‘+ 100’ as well. Lets remove this by taking away 100 from each side:

250 − 100 = 30n + 100 − 100 Now, we know that 250 – 100 is equal to 150. So the left hand side will just become 150. On the right hand side, the 100 – 100 is equal to 0, so we will just have 30n + 0:

150 = 30n + 0

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But as 30n plus nothing is just 30n, we can write this as:

150 = 30n (We didn’t really need to bother writing the equation with ‘30n + 0’, we could have just skipped to the one with ‘30n’, but I wanted this to be very clear). We can now easily complete our work because we can just divide both sides by 30 to get:

5=n So, n equals 5. This means I can earn £250 for my day’s work if I make 5 chairs. Here is another example: A barber charges £10 for each hair cut. The rent for his shop is £50 per day. How many haircuts must he do in a day to earn £100 after the rent is taken out of his earnings? So, the amount of money he wants to earn is £100, but the actual amount earned will 29

be equal to £10 for each haircut (so £10 times the number of haircuts), minus the £50 for rent. We can write this as the following equation:

(numberof haircuts)× 10 − £50 = £100 In simpler form, this can be written as:

10n − 50 = 100 where we have used the letter ‘n’ to represent the number of haircuts, and have removed the £ units symbol. At the moment, we don’t have the ‘10n’ on its own on the left hand side. If we did, we could just divide everything by 10 to work out what n is. To get the ‘10n’ on its own, lets add 50 to each side:

10n − 50 + 50 = 100 + 50 On the left hand side, we are taking away 50 then adding 50, which is the same as doing nothing (they cancel each other 30

out). On the right hand side, 100 + 50 equals 150, so we can write:

10n = 150 Now we can just divide both sides by 10 and we will get:

n = 15 So, the barber needs to do 15 haircuts a day if he is to earn £100 after rent. Here is another example: suppose an egg farmer sells eggs in 50-egg boxes and gets £12 for each box. In order to make £50 profit in a day, how many boxes must she sell if the daily cost of chicken-feed is £33? In this case, the actual money made will be £12 times the number of boxes, minus the £33 for feed. We want this to be £50 (or higher), so we can write the following equation:

12b − 33 = 50

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Where we have used ‘b’ to represent the number of boxes, and we have also left out the £ units. To get the ‘12b’ on its own, we need to add 33 to both sides. This gives us:

12b − 33 + 33 = 50 + 33 Which can be rewritten as:

12b = 83 We could have just gone straight from ‘12b – 33 = 50’ to ‘12b = 82’ without the step in-between, but at this stage I want to keep it all very clear. Now, to get ‘b’ rather than ‘12b’, we just divide both sides by 12:

b = 6.92 (Its actually 6.916666 recurring, but I’ve rounded it to 2 decimal places). If we assume the egg farmer only sells in whole boxes, then she will actually have to sell 7 32

boxes and this will give her 12 × 7 – 33, which is equal to £51 rather than £50. Now, its time to do the next set of practice questions to see how much you have learned.

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Practice questions 2: a) If a gardener charges £25 per hour for his gardening service, but his daily costs (transport, tool maintenance etc.) come to £34, then how many hours work must he do each day to break even? b) If a basket maker gets paid £30 per day and also gets a bonus of £16 for each basket she makes, how many baskets must she make in order to get a total daily pay of £125? c) Work out the value of x from the following equation:

6x − 22 = 18 d) Work out the value of y from the following equation:

3 y − 6 = 21 34

e) Work out the value of p from the following equation:

25 − 4 p = 13

f) Work out the value of q from the following equation:

4 = 25 − 3q g) Work out the value of m from the following equation:

48 = 5m + 13

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Answers 2: a) Breaking even means his total pay, for his £25 per hour, minus the £34 costs, will equal zero:

25h − 34 = 0 add 34 to both sides:

25h = 34 divide by 25:

h = 1.36 So, he will have to work for 1.36 hours (just over 1h 21min) to break even. b)

30 + 16b = 125

(where b = number of baskets). Take 30 from both sides:

16b = 95 36

Divide both sides by 16:

b = 5.94 So, 6 baskets actually needed (which will make 6×16 + 30, which is equal to £126). c) add 22 to both sides:

6x = 40 Divide both sides by 6:

x = 6.67 (rounded to 2 decimal places) d) Add 6 to both sides:

3 y = 27 Divide both sides by 3:

y =9 37

e) Add 4p to both sides:

25 = 13 + 4 p Take 13 away from both sides:

12 = 4 p Divide both sides by 4:

3= p f) Add 3q to both sides:

4 + 3q = 25 Take 4 from both sides:

3q = 21 Divide both sides by 3:

q=7

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g) Take 13 from both sides:

35 = 5m Divide both sides by 5:

7=m

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3. Squares and Square Roots So far, the example situations we have been thinking about have been fairly simple and perhaps not too interesting. Later in this chapter, we will be using algebra (with some simple mechanics) to work out when a thrown object will land again. To be able to do that, we need to know a little about squares and square roots. These are actually very straightforward. Suppose I had 6 boxes of eggs, and in each box were 6 eggs. How many eggs would I have in total. Well, this is just 6 times 6, which is 36. Another way of saying 6 times 6 is ‘6 squared’. So, ‘6 squared’ is equal to 36. Squaring just means that you multiply the number by itself. 36 can then be called the ‘square’ of 6. It is called the ‘square’, because it is the area you would get if you had a square with each side equal to 6 units of length. To convince yourself of this, draw a square and split it into 6 40

slices in one direction and 6 slices in the other direction, then count the number of squares it has been split into. You will get 36 (see the picture below):

In the form of an equation, we could write:

6 × 6 = 36 There is another, more common way to write this: 2

6 = 36 41

Where the little superscript ‘2’ written just to the upper right of the ‘6’ tells us that the 6 is being multiplied by itself (it is a ‘2’ because in 6×6 we have two sixes). In the other direction, we can say that 6 is the ‘square root’ of 36. This means it is the number that you can multiply by itself to get 36. We can write it like:

√ 36 = 6 The strange elongated ‘tick’ symbol is called the ‘radical’ symbol in mathematics. So, the square root of 36 is 6. Well, although this is true, there is something else we need to think about. 6 isn’t the only square root of 36. There is one more square root, and this is -6. If we multiply this -6 by itself we get -6 × -6 which is 36. By the way, -6 can be referred to as ‘minus 6’ (which is probably more common in the UK) or ‘negative 6’ (which is probably more common in the US).

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In any case, a minus times a minus is a plus. It should also be obvious that -6 × 6 is -36 (a minus times a plus is a minus). So, all numbers (except zero) have two square roots: a positive one and a negative one. We can write ‘the positive square root of 36 equals 6’ like: +

√ 36 = 6 and we can write ‘the negative square root of 36 equals minus 6’ like: −

√ 36 =−6

If we don’t want to specify whether we are talking about the positive or negative square root, and want to consider both, we can use the following symbol: ±

√ 36

Okay, let’s consider the following example: I am making a square solar panel to go on a boat. I want the solar panel to have a large area (to get the most light), but its weight cannot be greater than 75 43

kilograms (or it will be too heavy for the boat to stay afloat). If the solar panelling material has a weight of 3 kilograms per square metre, how long can the sides of the square panel be? If we use the symbol ‘l’ to represent the length of each side in metres, then the area, in square metres, will be the square of this (which we will write as l2). The weight of this, in kilograms, will be 3 times the area (as each square metre weighs 3 kilograms). Setting this weight as equal to 75 (our upper limit of weight) will allow us to calculate the upper limit of length: 2

3l = 75 We need to get the l2 on its own so we can take the square root of both sides (which will ‘unsquare’ the square), so we divide each side by 3: 2

l = 25 If two sides of an equation are equal, the square roots of both sides will also be equal to each other. Hopefully this seems 44

obvious. If it doesn’t, think about having two actual squares. For the squares to have the same area as each other (which they would if they were equal to each other), their edges must have the same length. It is impossible for two squares to have the same area if they don’t have sides of the same length. We can write our equation of the square roots like this: + 2

+

√ l = √ 25 By the way, going from something to its square root is often called ‘taking’ its square root. So, we have ‘taken’ the square root of each side. On the left hand side we now have the square root of the square of l, which is just l itself. The square root of the square of some item is just the item by itself. Like multiplying and dividing, squaring and taking the square root are the opposites of each other. If you do both, you are back to where you started. 45

So, we can write the equation as:

l=5 Therefore the upper limit of the length is 5 metres. In this case, only the positive square root makes any physical sense. Here is another example. Supposing I am designing a wheel for a train. The wheel is just a very plain solid circular disk made out of steel. The design brief is that the total weight of the wheel as well as the attachment bolt that connects it to the axle must be no greater than 120 kilograms. The attachment bolt has a weight of 15 kilograms and the steel material has a weight of 0.1 kilograms per square centimetre. What is the maximum radius of the wheel, in centimetres? For this, we need to know the equation for the area of a circle:

area = π ×r

2

where π (the Greek letter pi) is a fixed mathematical constant (that is equal to 46

about 3.142) and r is the radius of circle (the distance from the centre to edge). It is usual to write this without × sign (which was just put in to make equation a bit clearer), like so:

area = π r

the the the the

2

As the weight of this particular steel material is 0.1 kilograms per square centimeter (it is very thick and heavy steel), we can now write an equation for the weight:

weight of wheel = 0.1π r

2

The total weight is the weight of the wheel plus the 15 kilo attachment bolt. If we set this equal to 120 as our upper limit, we get: 2

120 = 0.1 π r + 15 We need to get only r 2 on the right hand side. Lets take 15 away from each side:

105 = 0.1 π r

2

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Lets multiply both sides by 10 (which makes the 0.1 into 1:

1050 = π r

2

Lets divide both sides by π:

1050 2 = r π 1050 divided by pi is 334.23 (to 2 decimal places):

334.23 = r

2

If we take the square root of both sides:

√ 334.23 = r In this case, I didn’t bother to put the + sign on the radical symbol, as a negative radius doesn’t make any sense (even though there is a negative square root of 334.23). I also didn’t bother to show that we were taking the square root of r 2, I just went straight to r. As the square root of 48

334.23 is 18.28 (to 2 decimal places), we can now write our final radius:

18.28 = r So, we need a radius of 18.28 cm or less if the total weight is going to be no higher than 120 kilograms. Now, do the following practice questions and see how you are doing so far.

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Practice questions 3: a) If a circle has an area of 33 square centimetres, what is the radius of the circle in centimetres? b) Work out the positive value of x from the following equation: 2

10 x = 1000 c) Work out the negative value of p from the following equation: 2

2 p −20 = 108 d) Get the positive value of m from the following equation: 2

5m + 20 = 65

50

e) From the following equation, work out the positive value of q: 2

q 9= + 5 4 f) Get the negative value of t from the following equation: 2

t − 9 =−5 25

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Answers 3: a) Write the equation for a circle’s area:

area = π r

2

set the area equal to 33:

33 = π r

2

Divide both sides by pi:

10.50 = r

2

Take the square root of both sides:

3.24 = r

b) Divide both sides by 10: 2

x = 100 Take the square root of both sides:

x = 10 52

c) Add 20 to both sides: 2

2 p = 128 Divide both sides by 2: 2

p = 64 Take the square root of both sides (we want the negative square root of 64, as the question asks for the negative value of p):

p =−8 d) Take 20 from each side: 2

5m = 45 Divide each side by 5: 2

m =9 Take the square root of each side:

m =3 53

e) Take 5 from each side: 2

q 4= 4 Multiply each side by 4:

16 = q

2

Take the square root of each side:

4=q f) Add 9 to both sides: 2

t =4 25 Multiply both sides by 25: 2

t = 100 Take the square root of both sides (the question asks for the negative value):

t =−10 54

4: Brackets (and Terms). We mentioned brackets in passing earlier. These symbols () are can also be called parentheses, particularly in the US. Brackets have two inter-related purposes. They specify the order in which we do operations (like addition, subtraction, multiplication, division etc.) and they also group things together). Here is a quick example of where they can make a difference. Supposing I write the following:

5 + 1×3 As you know, multiplication has a higher ‘priority’ than addition, so we would first do the 1 × 3 to get 3, and then add that to 5, to get 8. If we did the addition first, we would get 5 + 1, which is 6 and then if we multiplied that by 3 we would get 18. So, the order makes a difference. How would we write it down if we wanted to do the 5 + 1 first? We could use brackets:

(5 + 1) × 3 55

You always do the operation (or operations) in the brackets first, then take the result of that and operate with what is outside the brackets. In this case, it means do the 5 + 1 first, then multiply the result of that operation by 3. In algebra, we don’t usually put the multiplication sign in, so:

(5 + 1)3 is the same as:

(5 + 1) × 3 We won’t often see this with only numbers, but we will see it a lot with mixtures of numbers and symbols. Here is an example:

5( x + 1) This means add one to x, then multiply that by 5. Of course, if you don’t know what x is, you can’t actually do that calculation at the moment.

56

We can make equations with brackets, like:

5( x + 1) = 6 To work this out, and find the value of x, we need to know how to deal with the brackets in this case. As the 5 is written next to the bracket, we know that we need to multiply the things inside the bracket by 5. Instead of adding these together first, we can multiply each of them by the 5, then add the two results together (because they were connected by an addition sign in the bracket): Multiply the first thing in the bracket by 5 first: 5 times x

5(x ⏞ + 1) gives 5x Then multiply the next thing in the bracket by 5: 5 times 1

5( x + 1 ) gives 5 ⏞ 57

Then add these two results together to get:

5x + 5 So, we can write the overall process out as an equation:

5( x + 1) = 5x + 5 Now, if we go back to the equation we had earlier:

5( x + 1) = 6 We can replace the 5(x + 1) with the 5x + 5 to get:

5x + 5 = 6 Taking 5 from each side:

5x = 1 Dividing each side by 5: 58

1 x= 5 or:

x = 0.2 as a decimal. Hopefully, the way you multiply things through the brackets seems like common sense and intuitive. However, to make it clearer, lets thing about it in practical terms. Imagine we have 5 watermelons in a box and we sell them for £2 each. Its obvious that we will get £10 for them in total. We can write this as an equation:

2 × 5 = 10 Another way of writing this is:

2(5)= 10 If we take 3 melons out of the box first, and sell them for £2 each we will get £6. If we then take the other 2 out and sell them 59

for £2 each we will get another £4. In total, we will still have £10. We can write this in the form of an equation:

2(3 + 2)= 10 It doesn’t matter if we add the 3 and 2 together first, then multiply by 2, or if we multiply the 3 by 2, then multiply the 2 by 2, and add the two results together, we will still get 10. Multiply the 2 by each of the numbers in the bracket individually: 2 times 3

2(3 ⏞ + 2)

2 × 3 gives 6

2 times 2

2(3 + 2 ) ⏞

2 × 2 gives 4

Add these two results together:

6 + 4 = 10 60

So:

2(3 + 2)= 10 With factors that are just numbers, we are very used to the fact that it doesn’t matter what order we write them in. For example, 2 × 3 is the same as 3 × 2. The same thing goes for factors written in brackets. If we write the factors in the previous equation, we get:

(3 + 2)2 = 10 And the result is the same (check this yourself by multiplying). So far, we have referred to the things inside the bracket as ‘things’. This is fine, because they actually are ‘things’, but the + symbol and the = symbol could also be called ‘things’ (as they are also actually ‘things’). For numbers, and variables (like 5, x, y, 2), we can call them ‘terms’. This also applies to combinations of them like x2, 4y2, 3p etc.

61

In the following equation, there are three terms on the left hand side (these are x2, 22 and -x) and there is one term on the right hand side (the 4): 2

x + 22 − x = 4 These terms are connected using the +, and = symbols. That’s about all you need to know about terms, it is pretty straightforward and we have been using them without even thinking about it. As we go forward, I will probably use the word ‘term’ rather than ‘thing’. If we think about brackets again, we can say that the term outside the bracket is multiplied by each term inside the bracket in turn, and then all the results are combined together. Here is another example: 2

3(2 + x + x )

62

There is one term, 3, that is outside the bracket. Inside the bracket we have three terms: 2, x and x2. So, we multiply the 3 by each term inside the bracket, in turn, and then we add together all these results: 3 times 2 2

3(2 ⏞ +x+x )

3 × 2 gives 6

3 times x 2 3(2 + x + x ) ⏞ 3 times x

3 × x gives 3x

2

2 ⏞ 3(2 + x + x )

2

3 × x gives 3x

2

We then just add these together, so: 2

3(2 + x + x ) = 6 + 3x + 3x

2

If we have minus symbols in front of the terms inside the brackets, we just multiply and combine in the same way but now the number being added will be a negative number. For instance, if we make a small change to the previous situation: 63

2

3(2 − x + x )= 6 + − 3x + 3x

2

But, adding –3x is just the same as subtracting 3x, so we would actually write it out as: 2

3(2 − x + x )= 6 − 3x + 3x

2

Here is another example:

x(−2 + y − z + x) =−2x + xy − xz + x

2

As you can see, we just multiply the x by every term inside the bracket and combine these. Where x is multiplied by a negative term, we add the negative result (which is just the same as subtracting the result). The process of multiplying the term outside the bracket by the terms inside the bracket is sometimes called ‘multiplying out’ the brackets. Time for you to practice what you have learned in this section.

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Practice questions 4: Multiply out the following brackets: a) x(x + 2) b) x(x − 3) c) −x(x + 5) d) −x(−x − 5) e) Suppose a carpenter works for 5 days a week making cabinets. They get paid £30 each day and also get an extra £20 for each cabinet they make in that day. If the carpenter wants to make the same number of cabinets each day, how many must they make in order to earn £350 in one week?

65

Answers 4: 2

a) x(x + 2) = x + 2x 2

b) x(x − 3)= x − 3x 2

c) −x(x + 5) =−x − 5x 2

d) −x(−x − 5)= x + 5x e) In one day, the carpenter earns £30 + £20 times the number of cabinets:

daily pay = 30 + 20n (where n is the number of cabinets). To get the total amount for the 5-day work week, multiply this by 5:

total pay = 5(30 + 20n) Set this equal to 350:

350 = 5(30 + 20n) 66

Multiply out the bracket:

350 = 150 + 100n Take 150 off both sides:

200 = 100n Divide both sides by 100:

2= n

67

5. Brackets multiplied by brackets To give you some motivation for the following section, later on in the book we will be calculating the distance that a cannonball will fly before it hits the ground. However, in order to do that, we need to think a little bit more about multiplying brackets so that we have the required skills. Previously, we have had one set of brackets (with multiple terms inside) and only one term outside the bracket. If we have another bracket outside the bracket (also with multiple terms inside it), there is more for us to do. To begin with, lets stick to simple numbers:

5 × 7 = 35 So far, pretty straightforward. We can also write this with both the 5 and the 7 in brackets:

(5)(7) = 35 We can then write the 7 as 3 + 4: 68

(5)(3 + 4)= 35 To multiply this out we would multiply 5 by 3 to get 15, then multiply 5 by 4 to get 20, and add these two together to get 35 (which is the same as just multiplying 5 by 7). We can also write the 5 as 3 + 2:

(3 + 2)(3 + 4)= 35 To multiply this out, we now need to multiply each term in the first bracket by each term in the second bracket. This will give us four results which we then need to add together. In this case: 3 times 3

(3⏞ + 2)(3 + 4) gives 9 3 times 4

(3 + 2)(3 + 4 ) gives 12 ⏞ 2 times 3

(3 + 2)(3 ⏞ + 4) gives 6 2 times 4

(3 + ⏞ 2)(3 + 4 ) gives 8 69

When we add them all together:

9 + 12 + 6 + 8 = 35 Which is the same as multiplying 5 by 7. This isn’t really very useful when we’re working just with numbers, but is very useful indeed when we also have variables. Here’s an example of two brackets that each contain a variable:

(x + 3)( x + 2) We do exactly the same thing to multiply this out: x times x

(x⏞ + 3)(x + 2) gives x

2

x times 2

(x + 3)(x + 2 ) gives 2x ⏞ 3 times x

(x + 3)(x ⏞ + 2) gives 3x 3 times 2

(x + ⏞ 3)(x + 2 ) gives 6 70

Then we add them all together: 2

(x + 3)( x + 2)= x + 3x + 2x + 6 As 3x + 2x is the same as 5x, we can rewrite this as: 2

(x + 3)( x + 2)= x + 5x + 6 We will see shortly that the reverse process (finding the terms that go into the two brackets from the multiplied-out form) is actually very useful for a number of practical applications. Before we get to that stage though, its time for you to practice multiplying brackets out.

71

Practice questions 5. Multiply out the following brackets. As you do this, see if you notice any patterns emerging between the numbers in the brackets and the numbers in the answers. a) (x + 2)(x + 2) b) (x + 10)( x + 5) c) (x + 10)( x + 10) d) (x + 10)( x − 10) e) (x − 3)(x + 3) f) (x + 1)(x + 1) g) (2x + 1)(x + 1) h) (−x + 1)(x + 1) i) (−x + 1)(−x + 1)

72

Answers 5: 2

a) x + 4x + 4 2

b) x + 15x + 50 2

c) x + 20x + 100 2

d) x − 100 2

e) x − 9 2

f) x + 2x + 1 2

g) 2x + 3x + 1 2

h) −x + 1 2

i) x −2x + 1 For questions a)-i), try swapping the order of the brackets and check that they give the same answers. 73

6. Factorisation (and throwing balls in the air). Imagine I have a ball in my hand and I throw it straight up into the air. We all know what will happen. It will go up, but its speed will start to decrease and eventually it will stop going up. After this, it will start to come back down again. Its speed will increase until it hits the ground (if I don’t catch it). We might be interested in working out how long it will take to come back down. There is an equation that we can use to work this out:

gt h = h 0 + vt − 2

2

Here, the letter ‘h’ represents the height at a particular time, ‘h0’ is the height at the beginning (if we were using a stopwatch, this would be when we started the stopwatch, at time = zero). This is just a number. The ‘v’ is the velocity of the object at the beginning (when time = 74

zero). The ‘t’ represents the time (e.g. the time on the stopwatch). The ‘g’ represents the acceleration due to gravity. On earth, this force has a value of 10 meters per second per second, so this is just a number (it is actually about 9.806 to be precise, but we will use the value of 10 to make the calculations simpler). So, if h0, v and g are just numbers that we can put into the equation, we only have two variables (h and t), and the equation tells us how they are related (in other words how the height varies with time). I should point out that this equation is only correct for heights that are not too high. If we are talking about leaving earth’s atmosphere, then we would need to use another equation (because the effect of gravity obviously changes as you start to leave earth’s atmosphere, when you get high enough it becomes practically zero!). With the example mentioned above, lets suppose that when I let go of the ball, it is at a height of zero (I could possibly do this if I lay on the ground). We would put a 75

value of zero for h0. If I threw it with a velocity of 2 metres per second, we could put 2 for v. By the way, a positive value for the velocity means it was thrown in an upward direction. We can also put a value of 10 in for g. This gives us:

10t h = 0 + 2t − 2

2

We can get rid of the zero and, because 10 divided by 2 is 5, write the equation as:

h = 2t − 5t

2

This is the equation that tells us what the height will be at any particular time. If we want to know when the ball will come back down and hit the ground again, we want to know when its height will be zero. Lets put a value of zero in for h:

0 = 2t − 5t

2

So, we need to find out what values of t make the right hand side zero. This is where our knowledge of brackets comes 76

in. If we look at each of the terms on the right hand side, we can see that 2t is the result of 2 times t, and 5t2 is the result of 5t times t. In other words, two things can be multiplied by t and the combined results give 2t – 5t2. This sounds like brackets. If we write:

t(2 − 5t ) and multiply out the brackets, we can see that:

t(2 − 5t )= 2t − 5t

2

What we have done is sometimes called ‘taking t outside the bracket’. So, we can now replace the 2t – 5t2 in our equation for the height:

0 = t(2 − 5t ) We are now very close to our answer. Before we get there, however, we will take a slight diversion and talk about factors, and also about zero. 77

Factors are just things that are multiplied together to give something else. For instance, if I multiply 5 and 7 and get 35:

5 × 7 = 35 then the 5 and the 7 are factors of 35. Likewise, 6 and 10 are factors of 60:

6 × 10 = 60 If I said to you that I was looking for two factors that multiply to give 60, and that one of these factors was 6, you would know that the other one must be 10. I could also tell you that I was looking for two factors of 60 and that one of them was 30. You would then be able to tell me that the other one must be 2:

30 × 2 = 60 If, instead of using numbers, I used variables instead, I could write an equation like: 78

x × y = 60 If I then said that x was 30, you would reply that y must be 2. In other words, whether one number is a factor of 60 or not depends on the other factor. This is the same for almost all numbers. If I give you a number, whether another number is a factor of it or not depends on the other factor. There is one special number where things are different, and this is the number zero. If two numbers are multiplied together to give zero, then one (or both) of them must be zero. Think about the following equation:

x× y=0 Then, if x was zero, the equation would be true regardless of what y was. For instance, if y was 2, the equation would be correct, as zero times 2 equals zero:

0×2=0

79

If y was 7, the equation would still be true, as zero times 7 is zero:

0×7=0 It doesn’t matter what y is, a value of zero for x will always make the right hand side equal zero. The same is true of y. If y was zero, then the right hand side would always be zero, regardless of the value of x. Another thing we can say about factors of zero is that at least one of them must be zero. There is no combination of non-zero numbers that multiply together to give zero. Lets go back to our equation for the height of the ball:

0 = t(2 − 5t ) On one side (the right hand side in this case) we have two factors that are multiplied together and the result is equal to zero. One of the factors is t and the other factor is (2 – 5t). 80

Well, as I have just mentioned, if two factors multiply to give zero, this will be true if either of them are zero. So, either t or (2 – 5t) could be zero and the equation would be correct. This gives us two possibilities. Lets see this for t equal to zero:

0 = 0(2 − 5t ) which is correct. We don’t even need to worry about the right hand bracket (2 – 5t), which will actually be equal to 2 if t equals zero, because the overall right hand side must be zero, because this is being multiplied by zero. Anything multiplied by zero gives zero. Lets see this for the other option of having (2 – 5t) equal to zero:

0=t ×0 which is also correct, whatever t is.

81

As (2 – 5t) equals zero, we can write an equation to work out what t is itself in this case:

2 − 5t = 0

Add 5t to both sides:

2 = 5t Divide both sides by 5:

or:

2 =t 5

0.4 = t in decimals. So, we now have two possible values for t: 0 and 0.4. As the velocity and acceleration in the original equation were in terms of seconds, these correspond to 0 seconds or 0.4 seconds. These are the two times when the ball will be at ground level.

82

0 seconds clearly corresponds to the very start of the process, when I threw the ball and started my stopwatch (time = zero). The second value of t, 0.4 seconds, is when the ball will land again. Before we move onto another example, lets think about a really good way to check our answers. Well, if these actually are values of t that fit the original equation, then if we put them into that equation in place of t, the equation should be correct. Lets do this with our values. First, with t = 0:

0 = 2t − 5t

2

becomes: 2

0 = 2(0)− 5(0) which is:

0=0 so this value of t works.

83

With t = becomes:

0.4,

the

original

equation

2

0 = 2(0.4)− 5(0.4) which gives:

0 = 0.8 − 5(0.16) which gives:

0 = 0.8 − 0.8 which is correct. So, this value of t is also correct when put into the original equation. This is a very good way of checking your answers. Okay, lets have another example. Suppose that I throw the ball into the air with an upward velocity of 5 metres per second, but instead of throwing it from exactly ground level, I throw it from a height of 30 metres. When will it hit the ground in this case? Well, we just write the equation for the height, but put 30 for h0 (instead of zero), and 5 for v (instead of 2): 84

10t h = 30 + 5t − 2

2

As we want to know what time the ball will come back to ground level, we set h equal to zero again:

10t 0 = 30 + 5t − 2

2

As 10 divided by 2 is 5, we can rewrite this as:

0 = 30 + 5t − 5t

2

Now, this is where all of our earlier work on factoring with two sets of brackets comes in. To make the numbers easier to work with, let’s multiply both sides by -1 (which turns all positive values into negative values, but obviously the zero is still a zero as zero times -1 is zero):

0 =−30 − 5t + 5t

2

Let’s also reorder the terms on the right hand side: 85

2

0 = 5t − 5t − 30 We can also make this easier by dividing both sides by 5: 2

0=t −t −6 Knowing what we know about factors that multiply to make zero, if we could turn the right hand side into two factors (in other words two sets of brackets) then we could get two possible values for t (as we did in the last example). This one is a little more difficult, but if you noticed a pattern as you were doing the last set of practice questions, then you might be able to do this. We need two brackets, that both contain t and another number. The two numbers need to add together to make -1, and when multiplied together they need to give -6 (we’ll go through this in more detail shortly, if you didn’t see the pattern). What are the sets of two numbers that could give -6? (lets assume they are whole 86

numbers for now): We have 6 and -1, -6 and 1, -3 and 2, -2 and 3. Which of these pairs add up to -1? Well, hopefully you can see that -3 and 2 add up to -1. So, we can now write the right hand side in the form of two brackets:

0 = (t − 3)(t + 2) Check this yourself: 2

(t − 3)(t + 2)= t − t − 6 Also, check that swapping the order of the brackets makes no difference whatsoever: 2

(t + 2)(t − 3)= t − t − 6 So, with the following equation:

0 = (t − 3)(t + 2) We have two factors of (t – 3) and (t + 2). We know that if either of them was zero, the equation would be correct. Lets work out the values of t that this would give us. For (t - 3) equal to zero: 87

t −3=0 Add 3 to both sides:

t =3 For (t + 2) equal to zero:

t + 2= 0 Take 2 from both sides:

t =−2 So, we have two values of t: 3 seconds and -2 seconds. Obviously, only one of these values, 3, makes any sense, so the ball will hit the ground in 3 seconds. What about the -2? If we were to play time backwards, from when the ball is in the air to when we threw it, and instead of catching the ball as it came back to us (approaching time = zero) we just let it fall, then it would hit the ground at -2 seconds. Another way to think of it is that the trajectory of the ball after I threw it up from 30 metres is 88

exactly the same as if I threw it from the ground level with a certain velocity so that it was travelling with an upwards velocity of 5 metres per second when it reached 30 metres. Anyway, we now know that the ball will hit the ground when my stopwatch reads 3 seconds. Lets have a more detailed look at factoring again (if you didn’t do the practice questions from the last chapter, you should do them before going any further!). Let’s suppose we have two brackets that both contain the variable x, and they also have two other numbers. Instead of putting actual numbers in them, I’ll call them a and b, to keep it general:

(x + a)( x + b) Lets multiply out the brackets: 2

(x + a)( x + b)= x + ax + bx + ab Rather than writing ax + bx, lets combine these two x terms as (a + b)x: 89

2

(x + a)( x + b)= x + (a + b)x + ab So, we will get three terms when we multiply out the brackets: an x2 term, an x term (which is x multiplied by the sum of a and b), and a number (without x) that is the product of a and b. Here’s an example: 2

(x + 2)(x + 7)= x + 9x + 14 Here’s another one: 2

(x + 11)(x − 9)= x + 2x − 99 Knowing this pattern, it should be possible to try to go backwards, from the ‘multiplied-out’ form to the two sets of brackets (which are two factors that multiply to make the ‘multiplied-out’ form). Let’s try with this one: 2

(x + ?)(x + ?) = x + 11x + 24 Well, we know that the two missing numbers multiply to give 24, and they also add up to 11. Could it be 12 and 2 (which 90

multiply to give 24)? No, because these add up to give 14. How about 6 and 4? No, because these add up to 10. How about 8 and 3? Yes, these add up to 11, so: 2

(x + 8)(x + 3) = x + 11x + 24 Check this yourself by multiplying out the brackets. Here’s another one, this time with some negative numbers in the x and number terms: 2

(x + ?)(x + ?) = x − 4 x − 21 As the last number is negative (-21), we know that one of the two numbers in the brackets must be negative and one must be positive (this is the only way to get a negative number by multiplying two numbers together). As the x term has a negative number in front of it, we know that the negative number in the brackets must be bigger than the positive number (as the sum of the two numbers will then 91

be negative). 7 and 3 are factors of 21, so we could either have -7 and 3, or 7 and -3 (each pair gives -21 when they are multiplied together), but it must be -7 and 3 because these add together to give -4: 2

(x − 7)( x + 3)= x − 4 x − 21 Again, multiply out the brackets to check this. Here’s an interesting one: 2

(x + ?)(x + ?) = x − 9 In this case, there is no x term. That means that the two numbers in the brackets need to add up to zero. The only way they can do this is if they are the opposite of each other. They must also multiply to give -9. So, it must be 3 and -3. These add up to zero and multiply together to give -9: 2

(x + 3)( x − 3) = x − 9 (check this by multiplying out.)

92

This is a general thing, if you have x2 minus a number (and no x term), then one of the brackets has the positive square root of the number and the other bracket has the negative square root of the number: 2

x − a =(x + √ a)(x − √ a) Now its time for you to try some more practice questions.

93

Practice questions 6: Find factors (in two brackets) for the following (remember that it doesn’t matter which order the two brackets are written – so the brackets in your answer might not be written in the same order as the brackets in the answers given below): 2

a) x + 2x − 8 2

b) x + 11x + 30 2

c) x − 36 2

d) 2x + 8x + 6 In this one you will need 2x in one of the brackets. You will have to think about things a bit because instead of the two numbers adding up to give 8, it will be 2 times one of them plus the other one adding up to 8.

94

2

e) x + 4 x In this one, only one bracket needs a number in it (in other words, one of the numbers is zero, which is the same as not having a number in one of the brackets. For the following question, here is a reminder of the formula for heights of objects under gravity (h is height, h0 is the initial height, v is the initial upward velocity, t = time and g is the force of gravity (10):

gt h = h 0 + vt − 2

2

If we put 10 for g, we can divide the 10 by 2 to simplify the equation:

h = h 0 + vt − 5t

2

f) A person throws a ball straight up in the air from a height of 110 metres. If the initial upward velocity was 45 meters per second, when will the ball reach ground level (zero metres)? 95

Anwers 6: 2

a) x + 2x − 8 =( x − 2)( x + 4) 2

b) x + 11x + 30 =( x + 6)(x + 5) 2

c) x − 36 =(x − 6)(x + 6) 2

d) 2x + 8x + 6 = (2x + 2)( x + 3) 2

e) x + 4 x = x( x + 4) f) h = 110 + 45t − 5t

2

put zero for h (as we want to know when the ball will reach ground level again):

0 = 110 + 45t − 5t

2

Divide both sides by 5:

0 = 22 + 9t − t

2

Multiply both sides by -1:

0 =−22 − 9t + t

2

96

reorder for clarity: 2

0 = t − 9t − 22 find the factors: 2

t − 9t − 22 =(t + 2)(t − 11) So:

0 = (t + 2)(t − 11) Which means:

t + 2= 0 or:

t − 11 = 0 If t + 2 = 0:

t + 2= 0 Subtract 2 from both sides: 97

t =−2 If t – 11 = 0:

t − 11 = 0 add 11 to both sides:

t = 11 So, the two values for t are -2 or 11. Only 11 makes sense. So, the ball will reach ground level (height = zero) at 11 seconds (see the discussion in the text above about the meaning of the negative value).

98

7: Quadratic functions and factorisation with non-whole numbers. So far, I have carefully chosen all the values in the factorisation questions and examples so that all the numbers in the factor brackets were whole numbers. This definitely won’t always be the case and this aspect of algebra wouldn’t be much use if it only worked with such whole numbers. As you can imagine, using our current technique for factorisation probably won’t work for these examples (not easily anyway). It is possible to use a special formula to get the numbers directly from the equation, and we will certainly get to that point. However, there is another method that is pretty easy and is actually the method you can use to work out the special formula for yourself, so that is what we will do. Before we do that, there are some words we need to explain. Firstly, there is the word ‘function’. We have actuallyalready been using functions without explaining what the word ‘function’ means in mathematics. Anything (or more or less anything) that tells you 99

what to do with a number can be called a ‘function’. For instance, x2 is a function. It tells you to take a number (which is represented by x) and multiply it by itself. Also, x2 + x + 2 is a function. It says ‘take the number, multiply it by itself, add the same number and then add 2’. The functions that we have seen in the last couple of chapters contain x2 terms, x terms and number terms (like x2 + x + 2), and these are called ‘quadratic’ functions. I have no idea where that name actually comes from (look it up on the internet if you want to find out) and I have been doing mathematics with quadratic functions for many years without knowing, so it probably isn’t so important. Anyway, now you know what a ‘function’ is, and what a ‘quadratic function’ is (by the way, functions that only contain x terms and number terms (like 4x + 3 for instance) are called ‘linear’ functions). Now, lets get to our new method of factorisation. Suppose I multiplied out the following brackets: 100

(x + 3)( x + 3) We would get: 2

(x + 3)( x + 3)= x + 6x + 9 Now, if we look at the two brackets, we notice that they are the same. So, we can actually say that the left hand side is the square of (x + 3). In other words: 2

2

(x + 3) = x + 6x + 9 So, if we following:

had

an

equation

like

the

2

x + 6x + 9 = 16 we could rewrite this as: 2

(x + 3) = 16 This tells us what (x + 3)2 is. If we want to get what (x + 3) is, we need to take the square root of each side: ±

x + 3 = √ 16 101

We have used the ± symbol on the radical on the right hand side because there are two square roots of 16: 4 and -4 (both these numbers give 16 when squared. If we take 3 from both sides we get: ±

x =−3 + √ 16 So:

x =−3 + 4 which means:

x =1 or:

x =−3 − 4 which means:

x =−7 (check these values by putting them into the original quadratic equation!). 102

So, if we can get a single bracket squared on one side, and a single number on the other side, then we can just take the square root of each side. In this particular case, we already knew that x2 + 6x + 9 was equal to (x + 3)2 but what if we have another quadratic function, or if it cannot be factored like this? Well, there is a way of getting everything to work if we make some modifications. Lets go straight into an example. Supposing we have the following equation: 2

x + 2x − 1 = 0 Factoring this in our usual way won’t be easy (the numbers in the brackets won’t be whole numbers). Lets think about what the factors might look like though. We need the two numbers in the brackets to add up to 2. Bearing in mind we are trying to get to some point where the two factors are the same (so we can take a square root of 103

something), we would like the numbers in the two brackets to be the same. In other words, we would like both factors to be (x + 1), which will at least provide x2 and 2x terms when we multiply out, as well as a number term. Lets think about what we would actually get if we had (x + 1)(x + 1), which we can write as (x + 1)2 2

(x + 1)(x + 1)= x + 2x + 1 which isn’t the same as the x2 + 2x – 1 which is in the equation. But, if we take 2 away from both sides, we get to: 2

(x + 1)(x + 1)− 2 = x + 2x − 1 Which does now match the x2 + 2x – 1. So, we can put the left hand side of this equation into our earlier equation:

(x + 1)(x + 1)− 2 = 0 Let’s rewrite the left hand side: 2

(x + 1) − 2 = 0 104

If we add 2 to both sides, we get: 2

(x + 1) = 2 This method of solving quadratic equations is usually called ‘completing the square’. Hopefully this makes sense, as we have now, through some manipulations, got a square on the left hand side, on its own. Because of this, we can now take the square root of each side: ±

x + 1 = √2 If we now take 1 from both sides, we will have x on the left hand side: ±

x =−1 + √ 2 By the way, we can also write this as:

x =−1 ± √ 2 This is actually the more common way of writing it. The two forms mean the same thing, which in words is ‘x equals minus 1, plus or minus the square root of 2’. 105

The square root of 2 is approximately 1.414. So:

x =−1 + 1.414 = 0.414 or:

x =−1 − 1.414 =−2.414 Okay, here’s another example: 2

x + 6x = 4 We could take 4 from both sides to get zero on the right hand side, but this method of solving the equation doesn’t need that so we won’t bother. Let’s focus on the x2 + 6x part. For two identical factors to multiply to give an x2 term and a 6x term, we need them to be (x + 3), but: 2

(x + 3)( x + 3)= x + 6x + 9 Taking 9 away from both sides we get: 106

2

(x + 3)( x + 3)− 9 = x + 6 x This means we can replace the x2 + 6x in the original equation with the left hand side of this one:

(x + 3)( x + 3)− 9 = 4 Adding 9 to both sides:

(x + 3)( x + 3)= 13 or:

2

(x + 3) = 13 Now we can take the square root of each side: ±

x + 3 = √ 13

Taking 3 from both sides:

x =−3 ± √ 13 As the square root of 13 is about 3.606, we either have:

x =−3 + 3.606

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which gives:

x = 0.606 or we have:

x =−3 − 3.606 which gives:

x =−6.606 Before you do some practice questions, lets make this a little more general. Let’s suppose we have x2 + ax (where we are using ‘a’ as a general number), we want our two identical factors to be:

a a (x + )(x + ) 2 2 If we multiply out the brackets, we get: 2

a a a 2 (x + )(x + ) = x + ax + 2 2 4

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So, to get the right hand side equal to x2 + ax, we need to take a2/4 from each side: 2

a a a 2 (x + )(x + ) − = x + ax 2 2 4 Now, you can use this approach in the following practice questions. By the way, a good way to work out new ways to do things in algebra is to try some examples, see if you can see a pattern forming and then try to work out a more general approach.

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Practice questions 7: Use the ‘completing the square’ method to solve the following equations: 2

a) x + 2x = 3 2

b) x + 5x = 7 2

c) x + 8x = 11 2

d) x + 12x = 7

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Answers 7: a)

(x + 1)(x + 1)− 1 = 3 rewrite: 2

(x + 1) − 1 = 3 add 1 to both sides: 2

(x + 1) = 4 take the square root of both sides:

x + 1 =± 2 take 1 from both sides:

x =−1 ± 2 So x = 1 or -3.

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b)

2

5 25 (x + ) − = 7 2 4 add 25/4 to both sides: 2

5 25 (x + ) = 7 + 2 4 take the square root of both sides:

5 ± 25 x+ = 7+ 2 4

√

take 5/2 away from both sides:

5 25 x + =− ± 7 + 2 4

√

Which gives x = 1.14 or -6.14

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c)

2

(x + 4) − 16 = 11 add 16 to both sides: 2

(x + 4) = 27 take the square root of both sides: ±

x + 4 = √ 27 take 4 from both sides:

x =−4 ± √ 27 Which gives x = 1.196 or -9.196

d)

2

(x + 6) − 36 = 7 add 36 to both sides: 2

(x + 6) = 43 113

take the square root of both sides: ±

x + 6 = √ 43 take 6 from both sides:

x =−6 ± √ 43 Which gives x = 0.557 or -12.557

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8. Deriving quadratics.

a

formula

for

solving

Okay, hopefully you have now got the idea behind ‘completing the square’. We will now use our skills to come up with a general formula that works for all quadratic equations that are written in the following form: 2

x + bx + c = 0 where b and c are any numbers (they can be whole numbers or fractions). To do this, we’ll just use the exact same method of completing the squares that we have been using, but now we will have the symbols ‘b’ and ‘c’ to represent any general numbers. Later, when we do have actual numbers, we can just put them in the formula, in place of the ‘b’ and ‘c’.

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As we can rewrite x2 + bx as (x + b/2)2 – b2/4, the equation can be rewritten as: 2

2

b b (x + ) − + c = 0 2 4 Just like we did with specific numbers. We then add b2/4 to both sides: 2

2

b b (x + ) + c = 2 4

We then take c from both sides: 2

2

b b (x + ) = − c 2 4 We then take the square root of both sides:

b ± b2 x+ = −c 2 4

√

Then we take b/2 away from both sides:

b b2 x =− ± −c 2 4

√

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We now have x on its own on the left hand side. You will shortly practice using this formula, but before that, we will briefly discuss some ways of simplifying equations involving fractions (as you might need to do this to answer some of the practice questions, and its also a good skill to have for some of the things coming up later). Suppose I have the following equation:

9 x = −2 4 2

We could get the value of x just by taking the square root of both sides. In this case, this is quite easy once we have simplified the right hand side. At the moment, the 9 is over 4 (this is 9 divided by 4, or nine quarters). If the two was represented as some number over 4 (or some number of quarters), we could just combine them. How many quarters is 2? Well, if we remember that multiplication and division by the same number just gives us back the same number, we can multiply the 2 by 4 117

and also divide it by 4. This will still be equal to 2, but it will be written differently: 2

x =

9 8 − 4 4

So, instead of writing the 2 as ‘2’, it is now written as ‘8 divided by 4’ (or 8 quarters). Obviously, 8 divided by 4 is still 2, so the equation is still okay. As we now have 9 quarters minus 8 quarters, this is just (9 minus 8) quarters, or 1 quarter: 2

x =

9−8 4

1 x = 4 2

Now we take square roots of both sides:

1 x= 4

√

±

so: 118

1 x =± 2 x = ½ or x = -½ Now its time to practice what you have learned in this section.

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Practice Questions 8: Use the formula from the previous section to solve the following quadratic equations: 2

a) x + 3x + 2 = 0 2

b) x + 4 x + 3 = 0 2

c) x − 2x − 2 = 0 2

d) x + 5x + 6 = 0

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Answers 8: a) b = 3 and c = 2 so:

3 32 x =− ± −2 2 4

√

simplifying:

3 9 8 x =− ± − 2 4 4

√

which gives:

3 1 x =− ± 2 4

√

taking the root:

3 1 x =− ± 2 2 combining the fractions (as they are both ‘over 2’ or halves):

−3 ± 1 x= 2 121

So:

4 x =− 2 which is -2 or:

2 x =− 2 which is -1 b) b = 4 and c = 3 so:

4 42 x =− ± −3 2 4 simplifying:

√

x =− 2 ±

√

16 12 − 4 4

which gives:

x =− 2 ±

√

4 4 122

or:

x =− 2 ± √ 1 taking the root:

x =− 2 ± 1 so x = -3 or x = -1 c) b = -2 and c = -2 so:

2 22 x= ± +2 2 4

√

simplifying:

x = 1 ± √1 + 2 which gives:

x = 1 ± √3 This can’t be simplified any further. 123

Using a calculator, we get x = 2.732 or x = -0.732. d) b = 5 and c = 6 so:

5 52 x =− ± −6 2 4

√

simplifying:

5 25 24 x =− ± − 2 4 4

√

combining the fractions in the radical:

5 1 x =− ± 2 4

√

Taking the root:

5 1 x =− ± 2 2 combining the fractions: 124

x

−5±1 2

So x = -6/2 (which is -3) or x = -4/2 (which is -2).

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9. Extending quadratics.

the

formula

for

In the last chapter, the formula we derived to solve quadratics was for equations that involved x2, like: 2

x + bx + c = 0 But what if we have something that contains 2x2, or 7x2 (for instance). We need to derive a slightly different formula for that. Let’s use the symbol ‘a’ to refer to this number: 2

ax + bx + c = 0 To get it to x2 instead of ax2 (so it looks more like the equations we derived the formula for), lets divide both sides by a:

b c 2 x + x + =0 a a Now we can just carry on using the ‘completing the square’ method. As usual, convert the x2 term and the x term into a 126

squared bracket and a number term (that doesn’t contain x): 2

2

b b b x 2 + x =( x+ a) − 2 a 2 4a If you are not yet 100% convinced that this is correct, re-read the previous couple of chapters, and try multiplying out the bracket on the right hand side and see what you get. Now we can use the right hand side to replace the x2 and x terms of the previous equation, giving us: 2

2

b b c (x+ ) − 2 + = 0 2a 4a a adding b2/4a2 to both sides, we get: 2

2

b c b (x+ ) + = 2 2a a 4a taking c/a from both sides, we get: 2

2

b b c (x + ) = 2 − 2a a 4a 127

now we can take the square root of both sides: 2

b ± b c x+ = − 2 2a a 4a

√

if we take b/2a from both sides, we get: 2

b b c x =− ± − 2 2a a 4a

√

We could leave it like this, as we now have x on its own on the left hand side, but it is normally written in a slightly different form in most textbooks, so I will show you how to rearrange the equation to match that form. Lets look at the terms inside the radical: 2

b c − 2 a 4a If we multiply the c/a term by 4 and also divide it by 4 we won’t have actually changed anything (because multiplying by 128

4 and dividing by 4 cancel each other out), but it will be written differently: 2

2

b c b 4c − = − 2 2 a 4a 4a 4a If we now multiply the 4c/4a term we have got by a, and also divide it by a, we will still won’t have changed anything (it will just be written differently): 2

2

b 4c b 4ac − = − 2 4a 4a2 4a2 4a As we now have two fractions which are both things divided by 4a2, we can combine them: 2

2

b 4ac b − 4ac − = 2 2 2 4a 4a 4a Now, lets put this back into our formula: 2

b b − 4ac x =− ± 2 2a 4a

√

129

Using the following fact about square roots of fractions:

√

u √u = l √l

(where ‘u’ and ‘l’ are the ‘upper’ and ‘lower’ parts of the fraction), we can rewrite the previous equation as: 2

b b − 4ac √ x =− ± 2a √ 4a2 Now, as the square root of 4a2 is 2a, we can rewrite this as: 2

b √ b − 4ac x =− ± 2a 2a and after combining the fractions (which are both things divided by 2a), we have: 2

− b ± √ b − 4ac x= 2a

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Depending on the algebra you have done previously, you may have seen this formula before but perhaps didn’t know where it came from. Well, now you do and (far more importantly) you also know how to derive it from scratch. Learning formulas is fine, but knowing how to derive them yourself is infinitely more powerful, and usually means you can remember them more easily.

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Practice questions 9. Use the quadratic solver formula to get the value of x in the following equations. 2

a) 2x + 7x − 15 = 0 2

b) 2x − 3x − 3 = 0 2

c) 11x + 7x + 1 = 0 2

d) 0.33x + 2.71x + 3.2 = 0

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Answers 9: a) a = 2, b = 7, c = -15

− 7 ± √ 49 − 4 × 2 × (−15) x= 2×2 x=

− 7 ± √ 169 4

− 7 ± 13 x= 4 so x = -5 or x = 1.5 b) a = 2, b = -3, c = -3

3 ± √ 9 − 4(2)(−3) x= 2(2) 3 ± √ 33 x= 4 x = 2.186 or x = -0.686

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c) a = 11, b = 7, c = 1

−7 ± √ 49 − 4(11)(1) x= 2(11) x=

−7 ± √ 5 22

x = - 0.2165 or x = - 0.4198 d) a = 0.33, b = 2.71, c = 3.2

−2.71 ± √ 2.712 − 4(0.33)(3.2) x= 2(0.33) −2.71 ± √ 7.3441 − 4.224 x= 0.66 x=

−2.71 ± √ 3.1201 0.66

−2.71 ± 1.7664 x= 0.66 x = -6.7824 or x = -1.4297 134

10. Cannonballs. After all of our hard work, we can now apply our knowledge to the matter of ballistics. Earlier, we had a look at what happened when a tennis ball (or any other object for that matter) was thrown straight up in the air. In all cases, it will obviously land back in the spot that it was thrown from, and we were just working out the time it would take to do that. Now let’s extend this and look at cases where the object (for instance a cannonball) is projected at an angle, rather than being projected straight up. We don’t just want to know how long it will take to hit the ground again, we want to know how far it will have travelled horizontally. Have a look at the picture below:

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It shows the cannon that the cannonball is being shot from. It is not at ground level, but on top of a building (a castle?). The path of the cannonball is shown with a dotted line. Its important to remember that gravity only acts in a vertical direction (towards the centre of the earth). It doesn’t affect how things move horizontally. When projectiles are in motion, they have a horizontal velocity (which isn’t affected by gravity) and a vertical velocity (which is affected by gravity). The actual velocity is the sum of these two components. When the cannonball is fired, the overall velocity depends on the force and on the weight of the cannonball, but how much of the velocity ends up being vertical and how 136

much ends up being horizontal depends on the angle that the cannon is pointed at. We won’t go into the details of that (because it involves trigonometry, another area of mathematics). However, if we place the cannon at 45 degrees (halfway between horizontal and vertical) this will mean that the horizontal velocity and the vertical velocity are the same (at the start). One of the very useful things about the horizontal and vertical motion is that they can be dealt with independently. In other words, we can use our previous equation for the height to work out when the cannonball will reach ground level, and we can then work out how far it has travelled horizontally in that time. The horizontal velocity won’t change (of course, we are ignoring air resistance, which should be relatively minor, and would make the equation more complicated). If it is 50 metres per second, the overall horizontal distance the cannonball travels in t seconds will be 50t.

137

Okay, suppose the cannonball was fired from a cannon at 45 degrees, from a height of 23 metres, with initial horizontal and vertical velocities of 73 metres per second each (the horizontal and vertical velocities are the same because the angle is 45 degrees, if the angle was different they wouldn’t be the same). Let’s remind ourselves of the equation for height of an object under gravity:

gt h = h 0 + vt − 2

2

As g = 10, h0 = 23, and v = 73, we can rewrite this as:

h = 23 + 73t − 5t

2

Setting h to zero (ground level):

0 = 23 + 73t − 5t

2

I don’t think we will get very far trying to factorise this, so we can either use the 138

completing-the-square method or just go for the formula. Let’s go for the formula, which is: 2

− b ± √ b − 4ac t= 2a I’ll rewrite the quadratic equation in the usual order: 2

−5t + 73t + 23 = 0 So, a = -5, b = 73 and c = 23. Putting these into the quadratic solver equation:

− 73 ± √(73)2 − 4(−5)(23) t= 2(−5) − 73 ± √ 5329 + 460 t= −10 − 73 ± √ 5789 t= −10 139

− 73 ± 76.085 t= −10 So t = -0.3085 or t = 14.9085 (seconds). Only the positive value makes any physical sense in this situation. This is how long it will take to hit the ground. As the horizontal velocity (which doesn’t change over time, if we ignore air resistance) was also 73 metres per second, the horizontal distance travelled in this time will be equal to 73 × 14.9085 which is 1088 metres (rounded to the nearest metre), or just over 1 kilometre. Now, it’s time for you to practice what you have learnt about projectiles.

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Practice questions 10. a) A cannonball is fired from a cannon at an angle of 45 degrees, and from a height of 88 metres. The initial horizontal and vertical velocities are both 51 metres per second. How long will it take the cannonball to hit the ground again and how far will it travel horizontally during this time?

b) A cannonball is fired from a cannon at an angle of 45 degrees, and from a height of 10 metres. The initial horizontal and vertical velocities are both 37 metres per second. How long will it take the cannonball to hit the ground again and how far will it travel horizontally during this time?

c) A cannonball is fired from ground level, at an angle of 45 degrees toward the base of a castle on the top of a hill which is 50 141

metres high. The initial horizontal and vertical velocities are both 47 metres per second. How long will it take the cannonball to hit its target, and how far will it travel horizontally during this time?

142

Answers 10: a)

0 = 88 + 51t − 5t

2

− 51 ± √(51)2 − 4(−5)(88) t= 2(−5) − 51 ± √ 2601 + 1760 t= −10 − 51 ± √ 4361 t= −10 t = 11.704 or t = -1.504. Using the positive value, the horizontal distance = 11.704 × 51= 596.904 metres. b)

0 = 10 + 37t − 5t

2

− 37 ± √(37)2 − 4(−5)(10) t= 2(−5) − 37 ± √ 1569 t= −10 143

t = 7.661 or t = -0.261. Using the positive value, the horizontal distance = 7.661 × 37= 283.457 metres. c)

50 = 0 + 47t − 5t

2

The height is now set to 50, which is the height of the target, we need to take 50 from both sides:

0 =−50 + 47t − 5t

2

− 47 ± √(47)2 − 4(−5)(−50) t= 2(−5) − 47 ± √ 2209 − 1000 t= −10 t=

− 47 ± √ 1209 −10

t = 1.223 or t = 8.177. Notice that there are two positive values. One is for when 144

the cannonball reaches 50 metres height on the way up, and one is when it reaches 50 metres height on the way down (in this example the cannonball starts off lower than the height we are interested in, so it will actually be at this height at two moments in time). Using the value for reaching 50 metres on the way down, the horizontal distance = 8.177 × 47 = 384.319 metres.

145

Conclusion. If you have got this far, congratulations! Give yourself a huge pat on the back. You have made great progress and are well on your way to developing serious mathematical understanding. You now have many of the fundamental building blocks needed to build a solid foundation. Other books in the series will build on this.

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