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CHAPTER 2
C1: Surds
2
Learning objectives
After studying this chapter, you should be able to:
■ distinguish between rational and irrational numbers
■ understand what is meant by a surd
■ simplify expressions involving surds
■ perform arithmetic involving surds.
2.1 Special sets of numbers
When you learned to count, you started to use the numbers 1, 2,
3, 4, … which are known as natural numbers and denoted by
the symbol .
As the need arose to use zero and negative numbers the set of
integers, denoted by = {… 3, 2, 1, 0, 1, 2, 3, 4, …}, was
constructed.
Some numbers such as fractions cannot be written as a
a
whole number. Those which can be written in the form ,
b
where a and b are integers and b is not equal to zero, are
called rational numbers and are denoted by , since they
can be written as the quotient of two integers.
When rational numbers are written as decimals they either
71
3
terminate such as 0.75 or 0.071 or they have a
1000 5
4
sequence of recurring digits such as 0.45454545… or
11
5
0.714285714285714….
7
However, numbers that cannot be written as a fraction in
a
the form , where a and b are integers, are called
b
irrational.
The decimal representation of an irrational number neither
terminates nor has a recurring pattern of digits, no matter
how many decimal places we write down.
Are these sensible or reasonable
numbers?
The term rational is used
because these numbers can be
written as the ratio of two
integers.
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3
Examples include 3
1.73205… and 5
1.709975947… and
3.14159… which each has a non-repeating pattern of digits
in its decimal representation.
The set of rationals combined with the set of irrationals gives us
the set of real numbers and is denoted by .
2.2 Surds
In mathematics, we often arrive at answers that contain root
signs (they may be square roots, cube roots, etc.).
We will find that some of these numbers with a root sign are easy
to deal with since they have an exact decimal representation.
1
3
6
4, 8
2, 1
1
.5
6
3.4, 5 0.5.
For instance 1
32
This is because each of these numbers is rational.
Remember that although the
equation x2 9 has solutions
x 3, the symbol means
the positive square root so that
9
3.
Expressions with root signs involving irrational numbers
3
2 or 5
are called surds.
such as 7
EXERCISE 2A
State whether each of the following is a rational or irrational
number.
.
1 7
6
27 .
8
3
2 9
.
3
3
7 2 6
.
5
.
16
1
5
8 32 .
4
50
.
72
9 7 4
.
3
5 6
4
.
10 2.
Sometimes, you will be required to give answers to problems in
surd form since these answers are exact. If you use your calculator
to get a decimal form, it can only give the answer to a certain
number of decimal places and so can only be an approximation.
is exact, whereas 1.732050808 is the full
For example, 3
.
calculator display, but is still only an approximation to 3
Order of a surd
The order of a surd is determined by the root symbol. For
example:
3
is a surd of order 2 (sometimes called a quadratic surd).
3
5
is a surd of order 3.
n
x is a surd of order n.
A rational quantity may be expressed in the form of a root of
3
4
2
7
8
1
, etc.
any required order. For example, 3 9
In general, you will be handling
surds of order 2.
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15
Worked example 2.1
4
3
6
Write the following surds in ascending size: 9
, 5
, 2
6
.
Solution
2
It is difficult to see which is the biggest at a glance since they
are all of a different order.
Since we have the fourth root, the third root and the sixth root
of various numbers, we can raise each of the terms to the power
12, because the lowest common multiple of 4, 3 and 6 is 12.
4
3
6
, b 5
and c 26
.
Let a 9
Hence, a 4 9 ⇒ a12 93 729
b3 5 ⇒ b12 54 625
and
c 6 26 ⇒ c12 262 676.
We can now see that b c a and hence rearrange the surds in
3
6
4
, 26
, 9
.
ascending size as 5
2.3 Simplest form of surds
When simplifying surds, we try to make the number under the
root sign as small as possible.
Worked example 2.2
Simplify the following as far as possible.
8
.
1 1
3
8
.
2 4
Hint. Look for a square number
that divides exactly into the
number under the square root
sign (or a perfect cube if the surd
is of order 3, etc.).
3 2
1
6
.
Solution
9 is a square number that
is also a factor of 18.
1 1
8
⇒
1
8
9
2 9
2
32
1
8
32
8 is a perfect cube which
is also a factor of 48.
3
2 4
8
3
⇒
3
3
3
3
4
8
8
6 8
6
26
3
3
4
8
26
3 2
1
6
⇒
2
1
6
4
4
5 25
4
25
4
29
6 29
6
66
2
1
6
66
but 5
4
can itself be simplified.
4 is a square number that is a
factor of 216.
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2.4 Manipulating square roots
There are some simple rules which apply to all positive numbers
and they will help us when working with square roots. The first
rule generalises the idea demonstrated in the last worked
example.
Rules
1 ab
a b
2
a a
b b
Worked example 2.3
Simplify the following.
6
3
2 .
3
1 1
1
2
3
9.
25
4 1
2
2
1
.
Solution
1 1
1
2
1
6
7 1
6
7
47
6
3
9
7 3 7
2 7
3
3
3
3
2
5
9 9 3
25
5
4 1
2
2
1
3
4
3
7
4
(3
3
) 7
2 3 7
67
EXERCISE 2B
Simplify each of the surd expressions 1–18.
.
1 8
2 1
2
.
3 2
0
.
5
.
4 7
5 5
2
.
6 1
2
0
.
4
5
.
7 2
8 2
5
2
.
9 1
9
2
.
1
2
5
13 .
5
2
7
11 .
3
4
4
8
14 .
4
5
2
7
.
16 7
17 2
0
1
5
6
.
10 1
0
0
0
.
3
2
12 .
4
15 3
5
7
.
4
8
1
4
18 .
6
1
8
5
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19 Express each of the following in the form kp where k is an
integer and p is a prime number.
(a) 7
2
8
,
(b) 52
8
6
3
,
2
(c) 21
4
7
54
8
7
5
.
3
4
6
, 5
, 10
.
20 Arrange the following surds in ascending size: 3
2.5 Use in geometry
One of the main areas in which you will need to work with
surds is in the use of Pythagoras’ theorem in geometry. Often
you need the exact value of a particular length.
Worked example 2.4
The hypotenuse of a right-angled triangle has length 18 cm and
one of the other sides has length 6 cm. Find the length of the
remaining side.
18 cm
6 cm
x cm
Solution
Let the remaining side have length x cm.
By Pythagoras’ theorem
x 2 62 182 so that x 2 324 36 288
⇒
x 1
4
4
2
122
So the remaining side has length 122
cm.
2.6 Like and unlike surds
‘Like’ surds have the same irrational factor.
, 65
, 135
are ‘like’ surds since they have
For example 35
the same number under the root sign.
‘Unlike’ surds have different irrational factors.
, 26
, 41
1
are ‘unlike’ surds since they have
For instance, 73
different numbers under the root sign.
288 144 2
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Student health warning!
It is well worth noting that:
a
b a b
and
a
b a b
For example, consider
9
6
1 2
5
5
However,
9 + 16 = 3 + 4 = 7.
This, of course, is not the same and demonstrates that we cannot split up
9
6
1 as 9 1
6
. This is a common mistake and must be guarded
against.
2.7 Adding and subtracting surds
We can add or subtract surds as long as they are ‘like’ surds.
‘Like’ surds can be collected.
‘Unlike’ surds cannot be collected.
Worked example 2.5
Simplify each of the following as much as possible.
65
,
(a) 145
(b) 7
5
22
7
51
0
8
,
96
,
(c) 43
(d) 55
43
25
73
.
Solution
(a) 145
65
85
;
5
22
7
51
0
8
(b) 7
We first need to simplify these surds as far as possible.
(2
5
3
) (2 9
3
) (5 3
6
3
)
53
2 33
5 63
53
63
303
These are all ‘like’ surds and, therefore, can be collected.
193
(c) 43
96
;
(d) 55
43
25
73
We can collect all the terms with 5
and then separately
collect the terms with 3
55
25
43
73
35
113
There are 14 lots of 5
minus 6
lots of 5
giving 8 lots of 5
.
3
and 6
are ‘unlike’ surds and,
therefore, cannot be collected so
this expression cannot be
simplified any further.
Note that we cannot combine
these any further since 5
and
3
are ‘unlike’ surds.
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19
EXERCISE 2C
Simplify each of the expressions 1–10 as far as possible.
37
.
1 7
2 1
2
3
.
0
35
.
3 2
4 7
2
1
2
.
1
8
.
5 8
6 1
2
1
0
8
2
7
.
0
2
4
5
35
.
7 2
8 46
32
4
1
5
0
.
5
32
0
1
8
0
.
9 154
2
10 31
7
5
61
8
32
8
47
2
.
11 The two shorter sides of a right-angled triangle have lengths
3 m and 6 m. Find the length of the hypotenuse in the form
ab metres, where a and b are prime numbers.
12 The hypotenuse of a right-angled triangle has length 24 cm
and another one of the sides has length 18 cm. Find the
exact length of the third side.
13 Find the length of the remaining side in each of the
right-angled triangles below, giving your answers as simply
as possible in surd form.
(a)
5
2 cm
(b)
3 cm
7 cm
3
5
3 cm
2.8 Multiplying surds
We multiply the rational factors then the irrational factors and
then simplify (if possible).
Worked example 2.6
Simplify each of the following:
53
.
1 32
2 23
33
.
3 56
33
.
Solution
1 32
53
3 5 2
3
156
2 23
33
69
6 3 18
33
151
8
3 56
15 9
2
452
You should be able to recognise
that 3
3
3 and in general
n n n.
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C1: Surds
An expression such as 3
5
is called a compound surd.
To multiply compound surds, we use the same idea as
multiplying out brackets in algebra.
Worked example 2.7
Simplify each of the following.
5
)(2
3
).
1 (3
2 (3 2
)(5 7
).
35
)(32
25
).
3 (42
4 (3 5
)2.
Solution
1 (3
5
)(2
3
) 6
1
0
9
1
5
6
1
0
3 1
5
2 (3 2
)(5 7
) 15 37
52
1
4
3 (42
35
)(32
25
) 124
81
0
91
0
62
5
24 81
0
91
0
30
54 171
0
4 (3 5
)2 (3 5
)(3 5
)
9 35
35
5
14 65
EXERCISE 2D
Multiply out the following and simplify the answers as far as
possible.
26
.
1 53
0
52
.
2 61
(6
2).
3 5
(2
8
).
4 8
(3 65
).
5 25
)(7 3
).
6 (3 2
3
)(5
3
).
7 (5
7
)(6
27
).
8 (36
0
2)(1
0
2).
9 (1
1
23
)(31
1
2).
10 (41
Recall that
(a b)(c d) ac bc bd ad.
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C1: Surds
2.9 Rationalising the denominator
Whenever we have a fraction in which the denominator is a
surd, we can rewrite the fraction so that the denominator no
longer contains a surd.
This is done by multiplying the top and bottom of the fraction
by the same number so that the final answer has a rational
number in the denominator.
)(2 3
) 22 (3
)2 1, we could simplify
Since (2 3
Since 3
3
3, we could
2
simplify by writing
3
2
2
3
23
.
3 3 3
3
Since (p q)(p q) p2 q2.
4
4
2 3
by writing 4(2 3
).
2 3
2 3
2 3
This process is called ‘rationalising the denominator’.
The method is as follows.
If the denominator is of the form a then multiply the top
and bottom by a.
If the denominator is of the form a b then multiply top
and bottom by a b.
If the denominator is of the form a b then multiply top
and bottom by a b.
Worked example 2.8
Rationalise the denominators of the following:
3
2 .
3 6
4
1 .
7
1
3 .
1
1
7
6
32
4 .
23
5
Solution
4
1 .
7
.
Multiply the top and bottom by 7
7
4
7
47
4
7
7
4
9
7
⇒
7
4
4
7
7
We now have a rational number
in the denominator.
3
2 .
3 6
3
6) 3(3 6
(3 + 6
) 3(3 )
3 6
(3 6
) (3 + 6
)
96
3
⇒
3
3 6
3 6
21
Multiply the top and bottom by
(3 6
).
2
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1
3 .
1
1
7
Multiply the top and bottom by (1
1
7
).
1
1
(1
1
7
)
1
7
1
1
7
(1
1
7
) (1
1
7
) 1
2
1
4
9
4
1
1
1
7
⇒ 1
1
7
4
Note that the number that we
multiply the top and bottom by
in order to rationalise the
denominator is sometimes called
the conjugate.
The conjugate of 1
1
7
is 1
1
7
.
The conjugate of 8 35
is 8 35
.
6
32
4 .
2 3
5
0
21
8
31
0
66
(6
32
) (5
23
) 3
2
5
49
23
) (5
23
)
(5
Multiply top and bottom by the
conjugate (5
23
).
3
0
62
31
0
66
7
⇒
3
0
62
31
0
66
6
32
7
23
5
66
3
0
31
0
62
7
EXERCISE 2E
Rationalise the denominators of the following:
1
1 .
1
0
3
2 .
2
7
3 .
25
1
4 .
2
1
3
5 .
2
1
3
2
6 .
5
2
5
7 .
1
4
2
6
8 .
6
3
7
5
9 .
7
3
23
7
10 .
7
53
1
3
5
11 .
3
5
1
5
10
12 .
21
5 5
42
5
13 .
5
42
22
47
14 .
37
52
Although this answer is perfectly
correct, it is customary to try to
leave a positive denominator,
where possible.
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23
2.10 Equations and inequalities
involving surds
An equation or inequality may look more complicated because it
involves surds. The same techniques are used as in chapter 1.
However, it may be necessary to rationalise the denominator if
this is in surd form.
2
Worked example 2.9
Solve the equation 5
3x 25
x 7.
Solution
⇒
⇒
⇒
⇒
5
3x 25
x 7
5
7 25
x 3x (25
3)x
5
7
x
25
3
5
7
3
25
x 25
3
25
3
10 145
35
21
31 175
x 20 9
11
Although this answer is now
correct it is better if we
rationalise the denominator.
Worked example 2.10
Solve the inequality 3
2x 43
x 5
Solution
⇒
⇒
⇒
⇒
⇒
⇒
3
2x 43
x 5
3
5 43
x 2x
3
5 (43
2)x
3
5
x
43
2
3
5
2
43
x 43
2
43
2
203
10
12 23
x 48 4
1 93
2 183
x or x 44
22
EXERCISE 2F
Solve each of the following:
1 3x 5 73
x 11.
2 25
x 5 5
x.
3 73
3x 53
x 6.
4 67
2x 5 37
x.
Although this answer is again
correct, it is better if you
rationalise the denominator. In
fact, in an examination, you may
well be required to do so.
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5 22
3x 2
x 4.
6 43
2x 7 53
x.
x 6.
7 3x 4 53
8 35
x 1 5
2x.
9 37
5x 8 27
x.
10 32
5x 42
x 7.
Worked examination question 2.11
4 53
Express in the form p + q3
where p and q are rational
2 73
numbers.
Solution
4 53
2 73
2 73
2 73
8 283
103
35(3
)2
4 49(3
)2
We need to rationalise the
denominator and do this by
multiplying top and bottom by
2 73
.
8 105 183
4 147
A common mistake is to think
that (3
)2 is 9.
97 183
143
97
3
18
143
143
97
18
which is in the given form p = and q = .
143
143
MIXED EXERCISE
1 Simplify the following as far as possible:
(a) 75
,
(b) 300
,
(c) 128
,
(d) 2
8
1
7
5
6
3
,
(e) 128
98
3
2
,
(f) 2
4
2
1
6
150
.
2 Expand the following and simplify where possible.
(a) 3
(5 23
),
(b) 5
(3 45
),
(c) 2
(50
18
),
(d) (3
2)(3
7),
(e) (5
7)(35
4),
(f) (3 23
)2.
3 Rationalise the denominator, simplifying where possible:
73
(a) ,
5
5
1
(c) ,
1
5
1
(b) ,
3
1
1
(d) ,
13
11
This would now score full marks,
but it is good to state the actual
values of p and q.
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11
(e) ,
33
7
53
1
(f) ,
3 23
5 32
(g) ,
3
22
7
(h) .
2
(1
4 7
)
4 Express each of the following in the form p q2
where
p and q are rational:
1
)2,
(b) 2 .
[A]
(a) (3 2
(3 2
)
5 (a) Write down:
(i) a rational number which lies between 4 and 5,
(ii) an irrational number which lies between 4 and 5.
(b) A student says, ‘When you multiply two irrational
numbers together the answer is always an irrational
number‘.
Simplify (2 3
)(2 3
) and comment on the
student’s statement.
[A]
where
6 Express each of the following in the form p q7
p and q are rational numbers:
(5 7
)
)(5 27
),
(b) .
[A]
(a) (2 37
)
(3 7
7 (a) Express each of the following in the form k5
:
20
,
(ii) .
(i) 45
5
20
(b) Hence write 45
in the form n5
, where n is an
5
integer.
[A]
8 (a) Express (7
1)2 in the form a b7
, where a and b
are integers.
(7
1)2
, where
(b) Hence express in the form p q7
(7
2)
p and q are rational numbers.
[A]
:
9 Express each of the following in the form p q3
(a) (2 3
)(5 23
),
26
(b) .
4 3
[A]
2
1
10 (a) Express in the form a2
b, where a and b are
2
1
integers.
(b) Solve the inequality 2
(x 2
) x 22
.
[A]
25
2
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Key point summary
1 A rational number is one which can be written in the p13
a
form , where a and b are integers.
b
2 A real number that is not rational is called irrational.
p13
Its decimal representation neither terminates nor has a
recurring pattern of digits.
3 A surd is an irrational number containing a root sign.
p14
a b
4 ab
p16
5
a
b b
a
p16
6 ‘Like’ surds can be collected.
‘Unlike’ surds cannot be collected.
p18
7 To rationalise the denominator of the form a,
multiply top and bottom by a.
p21
8 To rationalise the denominator of the form a b,
multiply top and bottom by a b.
p21
9 To rationalise the denominator of the form a b,
multiply top and bottom by a b.
p21
Test yourself
What to review
1 State whether each of the following is rational or irrational:
Section 2.1
1
2 3
1
2
(b) ,
(c) ,
(d) .
3
5
3
2 Simplify each of the surd expressions below:
Sections 2.3 and 2.4
3
(a) 8
,
9
0
3
1
5
(b) ,
(c) .
5
1
6
3 Simplify each of the following surd expressions as far
as possible.
1
3
7
(a) 2
7
23
,
(b) 2
0
4
5
,
(c) .
2
4 Find the length of the hypotenuse of a right-angled triangle
if the two smaller sides have lengths 33
cm and 35
cm.
(a) 2
7
6
,
5 Rationalise the denominator of the expression:
96
8
.
6
1
Section 2.7
Section 2.5
Section 2.9
2
1 (a) rational;
(b) irrational;
(b) 55
;
;
3 (a) 3
(b) 6
;
;
2 (a) 92
(c) irrational;
(d) rational.
5
(c) .
2
(c) cannot simplify.
cm.
4 62
46 + 6
5 .
5
Test yourself
ANSWERS
C1: Surds
B_Chap02_013-027.qxd
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10:41 am
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