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Get Your Head Around - Basic Algebra I

Year: 2019
Language: english
Pages: 146
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Forging Python

Year: 2019
Language: english
File: PDF, 3.86 MB
1

Get Your Head Around:
Basic Algebra I

by
Austin Hartnell-Jones

© Copyright 2019 Austin Hartnell-Jones.

2

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.

4

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

9

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:

10

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.

11

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

14

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.

15

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.

16

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

17

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.

21

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?

23

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.

26

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

28

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

31

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.

33

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

35

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

38

g) Take 13 from both sides:

35 = 5m
Divide both sides by 5:

7=m

39

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).

42

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

47

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.

49

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

51

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.

64

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

107

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

108

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.

109

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

110

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.

111

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

112

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

114

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’.

115

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

√

116

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.

119

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

120

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).

125

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

130

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.

131

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
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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
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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.

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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
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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.

140

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.

146