chapter 6 exponents and surds

Exponents and surds

6Contents:

A Exponent or index notation [1.4, 1.9]

B Exponent or index laws [1.9, 2.4]

C Zero and negative indices [1.9, 2.4]

D Standard form [1.9]E Surds [1.10]F Properties of surds [1.10]

G Multiplication of surds [1.10]

H Division by surds [1.10]

Opening problem#endboxedheading

Amadeo Avogadro (1776-1856) established that one gram of

hydrogen contains 6:02£ 1023 atoms.

Things to think about:

² How can we write this number as an ordinary number?

² How many atoms would be in one tonne of hydrogen gas?

² Can you find the mass of 1030 atoms of hydrogen?

We have seen previously that 2£ 2£ 2£ 2£ 2, can be written as 25.

25 reads “two to the power of five” or “two with index five”.

In this case 2 is the base and 5 is the exponent, power or index.

We say that 25 is written in exponent or index notation.

EXPONENT OR INDEX NOTATION [1.4, 1.9]A

25exponent,

orpowerindex

base

The use of also called or , allows us to write products of factors and also to

write very large or very small numbers quickly.

exponents powers indices,

IGCSE01magentacyan yellow black

0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IGCSE01\IG01_06\123IGCSE01_06.CDR Friday, 14 November 2008 10:52:42 AM PETER

124 Exponents and surds (Chapter 6)

Example 1 Self Tutor

Find the integer equal to: a 34 b 24 £ 32 £ 7

a 34

= 3£ 3£ 3£ 3

= 9£ 9

= 81

b 24 £ 32 £ 7

= 2£ 2£ 2£ 2£ 3£ 3£ 7

= 16£ 9£ 7

= 1008


Write as a product of prime factors in index form:

a 144 b 4312

a 2 1442 72

2 36

2 18

3 9

3 3

1

) 144 = 24 £ 32

b 2 43122 2156

2 1078

7 539

7 77

11 11

1

) 4312 = 23 £ 72 £ 11

EXERCISE 6A.1

1 Find the integer equal to:

a 23 b 33 c 25 d 53

e 22 £ 33 £ 5 f 23 £ 3£ 72 g 32 £ 52 £ 13 h 24 £ 52 £ 11

2 By dividing continuously by the primes 2, 3, 5, 7, ..... , write as a product of prime factors in index

form:

a 50 b 98 c 108 d 360

e 1128 f 784 g 952 h 6500

3 The following numbers can be written in the form 2n. Find n.

a 32 b 256 c 4096

4 The following numbers can be written in the form 3n. Find n.

a 27 b 729 c 59 049

5 By considering 31, 32, 33, 34, 35 .... and looking for a pattern, find the last digit of 333.

6 What is the last digit of 777?

7 Find n if:

a 54 = n b n3 = 343 c 11n = 161051 d (0:6)n = 0:046 656


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IGCSE01\IG01_06\124IGCSE01_06.CDR Wednesday, 17 September 2008 9:33:15 AM PETER

Exponents and surds (Chapter 6) 125

Historical note#endboxedheading

1 = 13

3 + 5 = 8 = 23

7 + 9 + 11 = 27 = 33 etc.

Nicomachus of Gerasa lived around 100 AD. He discovered

an interesting number pattern involving cubes and sums of odd

numbers:

NEGATIVE BASES

So far we have only considered positive bases raised to a power.

We will now briefly look at negative bases. Consider the statements below:

(¡1)1 = ¡1

(¡1)2 = ¡1£¡1 = 1

(¡1)3 = ¡1£¡1£¡1 = ¡1

(¡1)4 = ¡1£¡1£¡1£¡1 = 1

(¡2)1 = ¡2

(¡2)2 = ¡2£¡2 = 4

(¡2)3 = ¡2£¡2£¡2 = ¡8

(¡2)4 = ¡2£¡2£¡2£¡2 = 16

From the pattern above it can be seen that:

² a negative base raised to an odd power is negative

² a negative base raised to an even power is positive.


Evaluate:

a (¡2)4 b ¡24 c (¡2)5 d ¡(¡2)5

a (¡2)4

= 16

b ¡24

= ¡1£ 24

= ¡16

c (¡2)5

= ¡32

d ¡(¡2)5

= ¡1£ (¡2)5

= ¡1£¡32

= 32

CALCULATOR USE

Power keys

x2 squares the number in the display.

^ raises the number in the display to whatever

power is required. On some calculators this

key is yx , ax or xy .

Notice the effect ofthe brackets inthese examples.

Not all calculators willuse these key sequences.

If you have problems,refer to the calculator

instructions on page .12

Different calculators have different keys for entering powers, but ingeneral they perform raising to powers in a similar manner.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IGCSE01\IG01_06\125IGCSE01_06.CDR Monday, 15 September 2008 2:30:09 PM PETER



Find, using your calculator: a 65 b (¡5)4 c ¡74

Answer

a Press: 6 ^ 5 ENTER 7776

b Press: ( (¡) 5 ) ^ 4 ENTER 625

c Press: (¡) 7 ^ 4 ENTER ¡2401

EXERCISE 6A.2

1 Simplify:

a (¡1)4 b (¡1)5 c (¡1)10 d (¡1)15 e (¡1)8 f ¡18

g ¡(¡1)8 h (¡3)3 i ¡33 j ¡(¡3)3 k ¡(¡6)2 l ¡(¡4)3

2 Simplify:

a 23 £ 32 £ (¡1)5 b (¡1)4 £ 33 £ 22 c (¡2)3 £ (¡3)4

3 Use your calculator to find the value of the following, recording the entire display:

a 28 b (¡5)4 c ¡35 d 74 e 83 f (¡7)6

g ¡76 h 1:0512 i ¡0:62311 j (¡2:11)17

Notice that: ² 23 £ 24 = 2£ 2£ 2 £ 2£ 2£ 2£ 2 = 27

² 25

22=

2£ 2£ 2£ 2£ 2

2£ 2= 23

² (23)2 = 2£ 2£ 2 £ 2£ 2£ 2 = 26

² (3£ 5)2 = 3£ 5£ 3£ 5 = 3£ 3£ 5£ 5 = 3252

²µ2

5

¶3

=2

5£ 2

5£ 2

5=

2£ 2£ 2

5£ 5£ 5=

23

53

These examples can be generalised to the exponent or index laws:

² am £ an = am+n To multiply numbers with the same base, keep the base

and add the indices.

² am

an= am¡n , a 6= 0 To divide numbers with the same base, keep the base

and subtract the indices.

² (am )n = am£n When raising a power to a power, keep the base and

multiply the indices.

² (ab)n = anbn The power of a product is the product of the powers.

²³ab

´n=

an

bn, b 6= 0 The power of a quotient is the quotient of the powers.

EXPONENT OR INDEX LAWS [1.9, 2.4]B

1

1


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IGCSE01\IG01_06\126IGCSE01_06.CDR Thursday, 18 September 2008 12:14:38 PM PETER



Simplify using the laws of indices:

a 23 £ 22 b x4 £ x5

a 23 £ 22 = 23+2

= 25

= 32

b x4 £ x5 = x4+5

= x9


Simplify using the index laws: a35

33b

p7

p3

a35

33= 35¡3

= 32

= 9

bp7

p3= p7¡3

= p4


Simplify using the index laws:

a (23)2 b (x4)5

a (23)2

= 23£2

= 26

= 64

b (x4)5

= x4£5

= x20


Remove the brackets of:

a (3a)2 b

µ2x

y

¶3

a (3a)2

= 32 £ a2

= 9a2

b

µ2x

y

¶3

=23 £ x3

y3

=8x3

y3

To multiply, keepthe base and add

the indices.

To divide, keep thebase and subtract

the indices.

To raise a power toa power, keep thebase and multiply

the indices.

Each factor within thebrackets has to beraised to the power

outside them.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100




Express the following in simplest form, without brackets:

a (3a3b)4 b

µx2

2y

¶3

a (3a3b)4

= 34 £ (a3)4 £ b4

= 81£ a3£4 £ b4

= 81a12b4

b

µx2

2y

¶3

=(x2)3

23 £ y3

=x2£3

8£ y3

=x6

8y3

EXERCISE 6B

1 Simplify using the index laws:

a 23 £ 21 b 22 £ 22 c 35 £ 34 d 52 £ 53

e x2 £ x4 f a3 £ a g n4 £ n6 h b3 £ b5


a24

23b

35

32c

57

53d

49

45

ex6

x3f

y7

y4g a8 ¥ a7 h b9 ¥ b5


a (22)3 b (34)3 c (23)6 d (102)5

e (x3)2 f (x5)3 g (a5)4 h (b6)4


a a5 £ a2 b n3 £ n5 c a7 ¥ a3 d a5 £ a

e b9 ¥ b4 f (a3)6 g an £ a5 h (b2)4

i b6 ¥ b3 j m4 £m3 £m7 k (a3)3 £ a l (g2)4 £ g3

5 Remove the brackets of:

a (ab)3 b (ac)4 c (bc)5 d (abc)3

e (2a)4 f (5b)2 g (3n)4 h (2bc)3

i

µ2

p

¶3

j

³ab

´3k

³mn

´4l

µ2c

d

¶5


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



6 Express the following in simplest form, without brackets:

a (2b4)3 b

µ3

x2y

¶2

c (5a4b)2 d

µm3

2n2

¶4

e

µ3a3

b5

¶3

f (2m3n2)5 g

µ4a4

b2

¶2

h (5x2y3)3

Consider23

23which is obviously 1.

Using the exponent law for division,23

23= 23¡3 = 20

We therefore conclude that 20 = 1.

In general, we can state the zero index law: a0 = 1 for all a 6= 0.

Now consider24

27which is

2£ 2£ 2£ 2

2£ 2£ 2£ 2£ 2£ 2£ 2=

1

23

Using the exponent law of division,24

27= 24¡7 = 2¡3

Consequently, 2¡3 =1

23, which means that 2¡3 and 23 are reciprocals of each other.

ZERO AND NEGATIVE INDICES [1.9, 2.4]C

1

1

In general, we can state the negative index law:

If a is any non-zero number and n is an integer, then a¡n =1

an.

This means that an and a¡n are reciprocals of one another.

In particular notice that a¡1 =1

a.

Using the negative index law,¡23

¢¡4=

1¡23

¢4= 1¥ 24

34

= 1£ 34

24

=¡32

¢4So, in general we can see that:

³ab

´¡n

=

µb

a

¶n

provided a 6= 0, b 6= 0.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IGCSE01\IG01_06\129IGCSE01_06.CDR Thursday, 18 September 2008 12:18:04 PM PETER



Simplify, giving answers in simplest rational form:

a 70 b 3¡2 c 30 ¡ 3¡1 d¡53

¢¡2

a 70 = 1 b 3¡2 =1

32= 1

9

c 30 ¡ 3¡1 = 1¡ 13 = 2

3 d¡53

¢¡2=¡35

¢2= 9

25

EXERCISE 6C

1 Simplify, giving answers in simplest rational form:

a 30 b 6¡1 c 4¡1 d 50

e 32 f 3¡2 g 53 h 5¡3

i 72 j 7¡2 k 103 l 10¡3

2 Simplify, giving answers in simplest rational form:

a (12 )0 b

54

54c 2t0 d (2t)0

e 70 f 3£ 40 g53

55h

26

210

i (14 )¡1 j (38 )

¡1 k (23 )¡1 l (15 )

¡1

m 20 + 21 n 50 ¡ 5¡1 o 30 + 31 ¡ 3¡1 p (13 )¡2

q (23 )¡3 r (11

2 )¡3 s (45 )

¡2 t (212)

¡2

3 Write the following without brackets or negative indices:

a (3b)¡1 b 3b¡1 c 7a¡1 d (7a)¡1

e

µ1

t

¶¡2

f (4t)¡2 g (5t)¡2 h (5t¡2)¡1

i xy¡1 j (xy)¡1 k xy¡3 l (xy)¡3

m (3pq)¡1 n 3(pq)¡1 o 3pq¡1 p(xy)3

y¡2

4 Write as powers of 2, 3 or 5:

a 25 b 125 c 27 d 1

27

e 16 f 116 g 2

3 h 35

i 9125 j 32

81 k 225 l 93

8

5 Write as powers of 10:

a 1000 b 0:01 c 100 000 d 0:000 001


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



10 000 = 104

1000 = 103

100 = 102

10 = 101

1 = 100

110 = 10¡1

1100 = 10¡2

11000 = 10¡3

Consider the pattern alongside. Notice that each time we divide by

10, the exponent or power of 10 decreases by one.

We can use this pattern to simplify the writing of very large and very small numbers.

For example, 5 000 000

= 5£ 1 000 000

= 5£ 106

and 0:000 003

=3

1 000 000

=3

1£ 1

1 000 000

= 3£ 10¡6

STANDARD FORM

Standard form (or scientific notation) involves writing any given number as a number between 1and 10, multiplied by an integer power of 10,

i.e., a£ 10n where 1 6 a < 10 and n 2 Z :


Write in standard form: a 37 600 b 0:000 86

a 37 600 = 3:76£ 10 000

= 3:76£ 104fshift decimal point 4 places to the left and £10 000g

b 0:000 86 = 8:6¥ 104

= 8:6£ 10¡4

fshift decimal point 4 places to the right and ¥10 000g


Write as an ordinary number:

a 3:2£ 102 b 5:76£ 10¡5

a 3:2£ 102

= 3:20£ 100

= 320

b 5:76£ 10¡5

= 000005:76¥ 105

= 0:000 057 6

STANDARD FORM [1.9]D

÷10

÷10

÷10

�1

�1

�1

÷10

÷10

÷10

÷10

�1

�1

�1

�1


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

y:\HAESE\IGCSE01\IG01_06\131IGCSE01_06.CDR Friday, 10 October 2008 9:08:28 AM PETER



Simplify the following, giving your answer in standard form:

a (5£ 104)£ (4£ 105) b (8£ 105)¥ (2£ 103)

a (5£ 104)£ (4£ 105)

= 5£ 4£ 104 £ 105

= 20£ 104+5

= 2£ 101 £ 109

= 2£ 1010

b (8£ 105)¥ (2£ 103)

=8£ 105

2£ 103

= 82 £ 105¡3

= 4£ 102

To help write numbers in standard form:

² If the original number is > 10, the power of 10 is positive (+).

² If the original number is < 1, the power of 10 is negative (¡).

² If the original number is between 1 and 10, leave it as it is and multiply it by 100.

EXERCISE 6D.1

1 Write the following as powers of 10:

a 100 b 1000 c 10 d 100 000

e 0:1 f 0:01 g 0:0001 h 100 000 000

2 Express the following in standard form:

a 387 b 38 700 c 3:87 d 0:0387

e 0:003 87 f 20:5 g 205 h 0:205

i 20 500 j 20 500 000 k 0:000 205

3 Express the following in standard form:

a The circumference of the Earth is approximately 40 075 kilometres.

b The distance from the Earth to the Sun is 149 500 000 000 m.

c Bacteria are single cell organisms, some of which have a diameter of 0:0004 mm.

d There are typically 40 million bacteria in a gram of soil.

e The probability that your six numbers will be selected

for Lotto on Saturday night is 0:000 000 141 62.

f Superfine sheep have wool fibres as low as 0:01 mm in

diameter.

4 Write as an ordinary decimal number:

a 3£ 102 b 2£ 103 c 3:6£ 104 d 9:2£ 105

e 5:6£ 106 f 3:4£ 101 g 7:85£ 106 h 9£ 108


a 3£ 10¡2 b 2£ 10¡3 c 4:7£ 10¡4 d 6:3£ 10¡5

e 1:7£ 100 f 9:5£ 10¡4 g 3:49£ 10¡1 h 7£ 10¡6


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100


Scientific and graphics calculators are able to display very large and very small numbers in standard form.

Likewise, if you perform 0:0024 ¥ 10 000 000 your calculator might display or ,

which actually represent 2:4£ 10¡10.

If you perform your calculator might display or or ,

all of which actually represent .

2300000 400000

9 2 10

� � � � �

� �

££: 11

9.2 11� �E9.211

2.4-10

9.2 +11�E�

2.4 -10� �E



a The wavelength of visible light is 9£ 10¡7 m.

b In 2007, the world population was approximately 6:606£ 109.

c The diameter of our galaxy, the Milky Way, is 1£ 105 light years.

d The smallest viruses are 1£ 10¡5 mm in size.

e 1 atomic mass unit is approximately 1:66£ 10¡27 kg.

7 Write in standard form:

a 18:17£ 106 b 0:934£ 1011 c 0:041£ 10¡2

8 Simplify the following, giving your answer in standard form:

a (8£ 103)£ (2£ 104) b (8£ 103)£ (4£ 105)

c (5£ 104)£ (3£ 105) d (2£ 103)3

e (6£ 103)2 f (7£ 10¡2)2

g (9£ 104)¥ (3£ 103) h (8£ 105)¥ (4£ 106)

STANDARD FORM ON A CALCULATOR

You will find instructions for graphics calculators on page 16.

EXERCISE 6D.2

1 Write each of the following as it would appear on the display of your calculator:

a 4 650 000 b 0:000 051 2 c 5:99£ 10¡4

d 3:761£ 1010 e 49 500 000 f 0:000 008 44

2 Calculate each of the following, giving your answers in standard form. The decimal part should be

correct to 2 decimal places:

a 0:06£ 0:002¥ 4000 b 426£ 760£ 42 000 c 627 000£ 74 000

d 320£ 600£ 51 400 e 0:004 28¥ 120 000 f 0:026£ 0:00 42£ 0:08


Use your calculator to find:

a (1:42£ 104)£ (2:56£ 108) b (4:75£ 10¡4)¥ (2:5£ 107)

Instructions are given for the Casio fx-6890G : Answer:

a 1:42 EXP 4 £ 2:56 EXP 8 EXE 3:6352£ 1012

b 4:75 EXP (¡) 4 ¥ 2:5 EXP 7 EXE 1:9£ 10¡11


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



3 Find, in standard form, with decimal part correct to 2 places:

a (5:31£ 104)£ (4:8£ 103) b (2:75£ 10¡3)2 c8:24£ 10¡6

3£ 104

d (7:2£ 10¡5)¥ (2:4£ 10¡6) e1

4:1£ 104f (3:2£ 103)2

4 For the following give answers in standard form correct to 3 significant figures:

a How many millimetres are there in 479:8 kilometres?

b How many seconds are there in one year?

c How many seconds are there in a millennium?

d How many kilograms are there in 0:5 milligrams?

5 If a missile travels at 3600 km/h, how far will it travel in:

a 1 day b 1 week c 2 years?

6 Light travels at a speed of 3£ 108 metres per second. How far will light travel in:

a 1 minute b 1 day c 1 year?

Give your answers in standard form with decimal part correct to 2 decimal places.

Assume that 1 year = 365 days.

For the remainder of this chapter we consider surds and radicals, which are numbers that are written using

the radical or square root signp

:

Surds and radicals occur frequently in mathematics, often as solutions to equations involving squared terms.

We will see a typical example of this in Chapter 8 when we study Pythagoras’ theorem.

RATIONAL AND IRRATIONAL RADICALS

Some radicals are rational, but most are irrational.

For example, some rational radicals include:p1 =

p12 = 1 or 1

1p4 =

p22 = 2 or 2

1q14 =

q¡12

¢2= 1

2

Two examples of irrational radicals arep2 ¼ 1:414 214

andp3 ¼ 1:732 051:

Strictly speaking, a surd is an irrational radical. However, in this and many other courses, the term surd

is used to describe any radical. It is reasonable to do so because the properties of surds and radicals are the

same.

SURDS [1.10]E

Give your answers in standard form with decimal partcorrect to places. Assume that year days.2 1 = 365


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



Historical note#endboxedheading

The name surd and the radical signp

both had a rather absurd past. Many

centuries after Pythagoras, when the Golden Age of the Greeks was past,

the writings of the Greeks were preserved, translated, and extended by Arab

mathematicians.

The Arabs thought of a square number as growing out of its roots. The roots

had to be extracted. The Latin word for “root” is radix, from which we get

the words radical and radish! The printed symbol for radix was first R, then r,

which was copied by hand asp

.

The word surd actually came about because of an error of translation by the

Arab mathematician Al-Khwarizmi in the 9th century AD. The Greek word

a-logos means “irrational” but also means “deaf”. So, the Greek a-logos was

interpreted as “deaf” which in Latin is surdus. Hence to this day, irrational

radicals likep2 are called surds.

BASIC OPERATIONS WITH SURDS

We have seen square roots and cube roots in previous courses. We can use their properties to help with

some simplifications.


Simplify:

a (p5)2 b

µ1p5

¶2

a (p5)2

=p5£

p5

= 5

b

µ1p5

¶2

=1p5£ 1p

5

= 15


Simplify:

a (2p5)3 b ¡2

p5£ 3

p5

a (2p5)3

= 2p5£ 2

p5£ 2

p5

= 2£ 2£ 2£p5£

p5£

p5

= 8£ 5£p5

= 40p5

b ¡2p5£ 3

p5

= ¡2£ 3£p5£

p5

= ¡6£ 5

= ¡30


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



ADDING AND SUBTRACTING SURDS

‘Like surds’ can be added and subtracted in the same way as ‘like terms’ in algebra.

Consider 2p3 + 4

p3, which has the same form as 2x+ 4x.

If we interpret this as 2 ‘lots’ ofp3 plus 4 ‘lots’ of

p3, we have 6 ‘lots’ of

p3.

So, 2p3 + 4

p3 = 6

p3, and we can compare this with 2x+ 4x = 6x.


Simplify:

a 3p2 + 4

p2 b 5

p3¡ 6

p3

a 3p2 + 4

p2

= 7p2

fCompare: 3x+ 4x = 7xg

b 5p3¡ 6

p3

= ¡1p3

= ¡p3

fCompare: 5x¡ 6x = ¡xg

EXERCISE 6E

1 Simplify:

a (p7)2 b (

p13)2 c (

p15)2 d (

p24)2

e

µ1p3

¶2

f

µ1p11

¶2

g

µ1p17

¶2

h

µ1p23

¶2

2 Simplify:

a ( 3p2)3 b ( 3

p¡5)3 c

³13p5

´33 Simplify:

a 3p2£ 4

p2 b ¡2

p3£ 5

p3 c 3

p5£ (¡2

p5)

d ¡2p2£ (¡3

p2) e (3

p2)2 f (3

p2)3

g (2p3)2 h (2

p3)3 i (2

p2)4

4 Simplify:

ap2 +

p2 b

p2¡p

2 c 3p2¡ 2

p2

d 2p3¡p

3 e 5p7 + 2

p7 f 3

p5¡ 6

p5

g 3p2 + 4

p2¡p

2 h 6p2¡ 9

p2 i

p5 + 7

p5

j 3p2¡ 5

p2¡p

2 k 3p3¡p

3 + 2p3 l 3

p5 + 7

p5¡ 10

5 Simplify:

a 3p2 + 2

p3¡p

2 + 5p3 b 7

p2¡ 4

p3¡ 2

p2 + 3

p3

c ¡6p2¡ 2

p3¡p

2 + 6p3 d 2

p5 + 4

p2 + 9

p5¡ 9

p2

e 3p2¡ 5

p7¡p

2¡ 5p7 f 3

p2 + 4

p11 + 6¡p

2¡p11¡ 3

g 6p6¡ 2

p2¡p

2¡ 5p6 + 4 h 5

p3¡ 6

p7¡ 5 + 4

p3 +

p7¡ 8


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



Discovery Properties of surds#endboxedheading

Notice thatp4£ 9 =

p36 = 6 and

p4£p

9 = 2£ 3 = 6, which suggests thatp4£p

9 =p4£ 9:

Also,

r36

4=

p9 = 3 and

p36p4

=6

2= 3, which suggests that

p36p4

=

r36

4.

What to do:

Test the following possible properties or rules for surds by substituting different values of a and b. Use

your calculator to evaluate the results.

1pa£p

b =pab for all a > 0, b > 0.

2

qa

b=

papb

for all a > 0, b > 0.

3pa+ b =

pa+

pb for all a > 0, b > 0.

4pa¡ b =

pa¡p

b for all a > 0, b > 0.

You should have discovered the following properties of surds:

² pa£

pb =

pa£ b for a > 0, b > 0

²papb

=

ra

bfor a > 0, b > 0

However, in general it is not true thatpa+ b =

pa+

pb or that

pa¡ b =

pa¡p

b:


Write in simplest form:

ap3£p

2 b 2p5£ 3

p2

ap3£

p2

=p3£ 2

=p6

b 2p5£ 3

p2

= 2£ 3£p5£

p2

= 6£p5£ 2

= 6p10

PROPERTIES OF SURDS [1.10]F


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100




Simplify: a

p32p2

b

p12

2p3

a

p32p2

=q

322

=p16

= 4

b

p12

2p3

= 12

q123 fusing

papb=

ra

bg

= 12

p4

= 12 £ 2

= 1


Writep32 in the form k

p2.

p32

=p16£ 2

=p16£

p2 fusing

pab =

pa£

pbg

= 4p2

SIMPLEST SURD FORM

A surd is in simplest form when the number under the radical sign is the smallest

integer possible.


Writep28 in simplest surd form.

p28

=p4£ 7 f4 is the largest perfect square factor of 28g

=p4£

p7

= 2p7

EXERCISE 6F

1 Simplify:

ap2£p

5 bp3£p

7 cp3£p

11

dp7£p

7 ep3£ 2

p3 f 2

p2£p

5

Look for thelargest perfectsquare factor.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



2 Simplify:

a 3p3£ 2

p2 b 2

p3£ 3

p5 c

p2£p

3£p5

dp3£p

2£ 2p2 e ¡3

p2£ (

p2)3 f (3

p2)3 £ (

p3)3

3 Simplify:

a

p8p2

b

p2p8

c

p18p2

d

p2p18

e

p20p5

f

p5p20

g

p27p3

h

p18p3

i

p3p30

j

p50p2

k2p6p24

l5p75p3

4 Write the following in the form kp2:

ap8 b

p18 c

p50 d

p98

ep200 f

p288 g

p20 000 h

q12


ap12 b

p27 c

p75 d

q13


ap20 b

p45 c

p125 d

q15

7 a Find:

ip16 +

p9 ii

p16 + 9 iii

p25¡p

9 ivp25¡ 9

b Copy and complete: In general,pa+ b 6= :::::: and

pa¡ b 6= ::::::

8 Write the following in simplest surd form:

ap24 b

p50 c

p54 d

p40 e

p56 f

p63

gp52 h

p44 i

p60 j

p90 k

p96 l

p68

mp175 n

p162 o

p128 p

p700

9 Write the following in simplest surd form:

a

q59 b

q184 c

q1216 d

q7536

The rules for expanding brackets involving surds are identical to those for ordinary algebra.

We can thus use:a(b+ c) = ab+ ac

(a+ b)(c+ d) = ac+ ad+ bc+ bd

(a+ b)2 = a2 + 2ab+ b2

(a¡ b)2 = a2 ¡ 2ab+ b2

(a+ b)(a¡ b) = a2 ¡ b2

MULTIPLICATION OF SURDS [1.10]G


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100




Expand and simplify:

ap2(p2 +

p3) b

p3(6¡ 2

p3)

ap2(p2 +

p3)

=p2£

p2 +

p2£

p3

= 2 +p6

bp3(6¡ 2

p3)

= (p3)(6 +¡2

p3)

= (p3)(6) + (

p3)(¡2

p3)

= 6p3 +¡6

= 6p3¡ 6



a ¡p2(p2 + 3) b ¡p

3(7¡ 2p3)

a ¡p2(p2 + 3)

= ¡p2£

p2 + ¡

p2£ 3

= ¡2¡ 3p2

b ¡p3(7¡ 2

p3)

= (¡p3)(7¡ 2

p3)

= (¡p3)(7) + (¡

p3)(¡2

p3)

= ¡7p3 + 6


Expand and simplify: (3¡p2)(4 + 2

p2)

(3¡p2)(4 + 2

p2)

= (3¡p2)(4) + (3¡

p2)(2

p2)

= 12¡ 4p2 + 6

p2¡ 4

= 8 + 2p2



a (p3 + 2)2 b (

p3¡p

7)2

a (p3 + 2)2

= (p3)2 + 2£

p3£ 2 + 22

= 3 + 4p3 + 4

= 7 + 4p3

b (p3¡

p7)2

= (p3)2 ¡ 2£

p3£

p7 + (

p7)2

= 3¡ 2p21 + 7

= 10¡ 2p21


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100





a (3 +p2)(3¡p

2) b (2p3¡ 5)(2

p3 + 5)

a (3 +p2)(3¡

p2)

= 32 ¡ (p2)2

= 9¡ 2

= 7

b (2p3¡ 5)(2

p3 + 5)

= (2p3)2 ¡ 52

= (4£ 3)¡ 25

= 12¡ 25

= ¡13

EXERCISE 6G

1 Expand and simplify:

ap2(p5 +

p2) b

p2(3¡p

2) cp3(p3 + 1) d

p3(1¡p

3)

ep7(7¡p

7) fp5(2¡p

5) gp11(2

p11¡ 1) h

p6(1¡ 2

p6)

ip3(p3 +

p2¡ 1) j 2

p3(p3¡p

5) k 2p5(3¡p

5) l 3p5(2

p5 +

p2)


a ¡p2(3¡p

2) b ¡p2(p2 +

p3) c ¡p

2(4¡p2) d ¡p

3(1 +p3)

e ¡p3(p3 + 2) f ¡p

5(2 +p5) g ¡(

p2 + 3) h ¡p

5(p5¡ 4)

i ¡(3¡p7) j ¡p

11(2¡p11) k ¡(

p3¡p

7) l ¡2p2(1¡p

2)

m ¡3p3(5¡p

3) n ¡7p2(p2 +

p6) o (¡p

2)3(3¡p2)


a (1 +p2)(2 +

p2) b (2 +

p3)(2 +

p3) c (

p3 + 2)(

p3¡ 1)

d (4¡p2)(3 +

p2) e (1 +

p3)(1¡p

3) f (5 +p7)(2¡p

7)

g (p5 + 2)(

p5¡ 3) h (

p7¡p

3)(p7 +

p3) i (2

p2 +

p3)(2

p2¡p

3)

j (4¡p2)(3¡p

2)


a (1 +p2)2 b (2¡p

3)2 c (p3 + 2)2 d (1 +

p5)2

e (p2¡p

3)2 f (5¡p2)2 g (

p2 +

p7)2 h (4¡p

6)2

i (p6¡p

2)2 j (p5 + 2

p2)2 k (

p5¡ 2

p2)2 l (6 +

p8)2

m (5p2¡ 1)2 n (3¡ 2

p2)2 o (1 + 3

p2)2


a (4 +p3)(4¡p

3) b (5¡p2)(5 +

p2) c (

p5¡ 2)(

p5 + 2)

d (p7 + 4)(

p7¡ 4) e (3

p2 + 2)(3

p2¡ 2) f (2

p5¡ 1)(2

p5 + 1)

g (5¡ 3p3)(5 + 3

p3) h (2¡ 4

p2)(2 + 4

p2) i (1 + 5

p7)(1¡ 5

p7)


a (p3 +

p2)(

p3¡p

2) b (p7 +

p11)(

p7¡p

11) c (px¡p

y)(py +

px)

Did you notice thatthese answers are

?integers


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



When an expression involves division by a surd, we can write the expression with an integer denominator

which does not contain surds.

If the denominator contains a simple surd such aspa then we use the rule

pa£p

a = a:

For example:6p3

can be written as6p3£

p3p3

since we are really just multiplying

the original fraction by 1.

6p3£

p3p3

then simplifies to6p3

3or 2

p3.


Express with integer denominator: a7p3

b10p5

c10

2p2

a7p3

=7p3£

p3p3

=7p3

3

b10p5

=10p5£

p5p5

= 105

p5

= 2p5

c10

2p2

=10

2p2£

p2p2

=10p2

4

=5p2

2


Express1

3 +p2

with integer denominator.

1

3 +p2=

µ1

3 +p2

¶µ3¡p

2

3¡p2

¶=

3¡p2

32 ¡ (p2)2

fusing (a+ b)(a¡ b) = a2 ¡ b2g

=3¡p

2

7

DIVISION BY SURDS [1.10]H

We are reallymultiplying by one,

which does not changethe value of the

original expression.

If the denominator has the form a+pb then we can remove the surd from the denominator by multiplying

both the numerator and the denominator by its radical conjugate a ¡ pb. This produces a rational

denominator, so the process is called rationalisation of the denominator.


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100




Write1¡ 2

p3

1 +p3

in simplest form.

1¡ 2p3

1 +p3

=

µ1¡ 2

p3

1 +p3

¶µ1¡p

3

1¡p3

¶

=1¡p

3¡ 2p3 + 6

1¡ 3

=7¡ 3

p3

¡2

=3p3¡ 7

2

EXERCISE 6H

1 Express with integer denominator:

a1p2

b2p2

c4p2

d10p2

e

p7p2

f1p3

g3p3

h4p3

i18p3

j

p11p3

k1p5

l3p5

m

p3p5

n15p5

o125p5

p

p10p2

q1

2p3

r2p2p3

s15

2p5

t1

(p2)3

2 Rationalise the denominator:

a1

3¡p5

b1

2 +p3

c1

4¡p11

d

p2

5 +p2

e

p3

3 +p3

f5

2¡ 3p2

g¡p

5

3 + 2p5

h3¡p

7

2 +p7

3 Write in the form a+ bp2 where a, b 2 Q :

a4

2¡p2

b¡5

1 +p2

c1¡p

2

1 +p2

d

p2¡ 2

3¡p2

e

1p2

1¡ 1p2

f1 + 1p

2

1¡ 1p2

g1

1¡p23

h

p22 + 1

1¡p24

Review set 6A#endboxedheading

1 Find the integer equal to: a 34 b 5£ 23

2 Write as a product of primes in index form: a 36 b 242


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



3 Simplify:

a (¡2)3 b ¡(¡1)8 c (¡1)3 £ (¡2)2 £¡33

4 Simplify, giving your answers in simplest rational form:

a 3¡3 b¡43

¢¡2c 30 ¡ 31

5 Write 116 as a power of 2.

6 Simplify, using the exponent laws:

a 56 £ 5 b b7 ¥ b2 c (x4)3

7 Express in simplest form, without brackets:

a (4c3)2 b (2a2b)3 c

³ s

3t2

´48 Write in standard form:

a 9 b 34 900 c 0:0075


a 2:81£ 106 b 2:81£ 100 c 2:81£ 10¡3

10 Simplify, giving your answer in standard form:

a (6£ 103)£ (7:1£ 104) b (2:4£ 106)¥ (4£ 102)

11 The Earth orbits around the Sun at a speed of approximately 1:07£ 105 km/h. How far does the

Earth move, relative to the Sun, in:

a 1 day b 1 week c 1 year?

Give your answers in standard form with decimal part correct to 2 decimal places. Assume that

1 year = 365 days.

12 a Simplify (3p2)2: b Simplify ¡2

p3£ 4

p3:

c Simplify 3p2¡p

8: d Writep48 in simplest surd form.


a 2p3(4¡p

3) b (3¡p7)2 c (2¡p

3)(2 +p3)

d (3 + 2p5)(2¡p

5) e (4¡p2)(3 + 2

p2)

14 Rationalise the denominator:

a8p2

b15p3

c

p3p2

d5

6¡p3

15 Write

q17 in the form k

p7.


a

p32 + 1

1¡p32

b2 +

p3

2¡p3¡ 2

p3

2 +p3


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100



Review set 6B#endboxedheading

1 Find the integer equal to: a 73 b 32 £ 52

2 Write as a product of primes in index form: a 42 b 144

3 Simplify:

a ¡(¡1)7 b ¡43 c (¡2)5 £ (¡3)2

4 Simplify, giving your answers in simplest rational form:

a 6¡2 b¡112

¢¡1c¡35

¢¡2

5 Simplify, using the exponent laws:

a 32 £ 36 b a5 ¥ a5 c (y3)5

6 Write as powers of 2, 3 or 5:

a 1625 b 40

81 c 180 d 1119

7 Express in simplest form, without brackets or negative indices:

a (5c)¡1 b 7k¡2 c (4d2)¡3

8 Write in standard form:

a 263:57 b 0:000 511 c 863 400 000


a 2:78£ 100 b 3:99£ 107 c 2:081£ 10¡3

10 Simplify, giving your answer in standard form:

a (8£ 103)2 b (3:6£ 105)¥ (6£ 10¡2)

11 How many kilometres are there in 0:21 millimetres? Give your answer in standard form.

12 Simplify:

a 2p3£ 3

p5 b (2

p5)3 c 5

p2¡ 7

p2

d ¡p2(2¡p

2) e (p3)4 f

p3£p

5£p15

13 Write in simplest surd form: ap75 b

q209


a (5¡p3)(5 +

p3) b ¡(2¡p

5)2

c 2p3(p3¡ 1)¡ 2

p3 d (2

p2¡ 5)(1¡p

2)

15 Express with integer denominator:

a14p2

b

p2p3

c

p2

3 +p2

d¡5

4¡p3


a1 +

p5

2¡p5

b3¡p

5

3 +p5¡ 4

3¡p5


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IGCSE01\IG01_06\145IGCSE01_06.CDR Wednesday, 5 November 2008 2:15:20 PM PETER

Challenge#endboxedheading


Discovery Continued square roots#endboxedheading

X =

s2 +

r2 +

q2 +

p2 +

p2 + :::::: is an example of a

continued square root.

Some continued square roots have actual values which are integers.

1 Use your calculator to show thatp2 ¼ 1:41421p

2 +p2 ¼ 1:84776q

2 +p

2 +p2 ¼ 1:96157 :

2 Find the values, correct to 6 decimal places, of:

a

r2 +

q2 +

p2 +

p2 b

s2 +

r2 +

q2 +

p2 +

p2

3 Continue the process and hence predict the actual value of X.

4 Use algebra to find the exact value of X.

Hint: Find X2 in terms of X, and solve by inspection.

5 Can you find a continued square root whose actual value is 3?

1 Find

s3 + 2

p2

3¡ 2p2

giving your answer in the form a+ bp2 where a, b 2 Q .

2 If x =p5¡p

3, find x2 and x4. Hence find the value of x4 ¡ 16x2.

Copy and complete: x =p5¡p

3 is one of the solutions of the equation x4 ¡ 16x2 = 0

3 a We know that in general,pa+ b 6= p

a+pb

Deduce that ifpa+ b =

pa+

pb then at least one of a or b is 0.

b What can be deduced about a and b ifpa¡ b =

pa¡p

b?

4 a Find the value of

µ1 +

p5

2

¶n

¡µ1¡p

5

2

¶n

for n = 1, 2, 3 and 4.

b What do you suspect about

µ1 +

p5

2

¶n

¡µ1¡p

5

2

¶n

for all n 2 Z +?

What to do:


0 05 5

25

25

75

75

50

50

95

95

100

100 0 05 5

25

25

75

75

50

50

95

95

100

100

Y:\HAESE\IGCSE01\IG01_06\146IGCSE01_06.cdr Friday, 31 October 2008 9:51:20 AM PETER

chapter 6 exponents and surds

Education