Date post: | 10-Jan-2017 |
Category: |
Education |
Upload: | sarah-sue-calbio |
View: | 561 times |
Download: | 10 times |
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
Example 2 Self Tutor
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
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\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.
Example 3 Self Tutor
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.
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\125IGCSE01_06.CDR Monday, 15 September 2008 2:30:09 PM PETER
126 Exponents and surds (Chapter 6)
Example 4 Self Tutor
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
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\126IGCSE01_06.CDR Thursday, 18 September 2008 12:14:38 PM PETER
Exponents and surds (Chapter 6) 127
Example 5 Self Tutor
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
Example 6 Self Tutor
Simplify using the index laws: a35
33b
p7
p3
a35
33= 35¡3
= 32
= 9
bp7
p3= p7¡3
= p4
Example 7 Self Tutor
Simplify using the index laws:
a (23)2 b (x4)5
a (23)2
= 23£2
= 26
= 64
b (x4)5
= x4£5
= x20
Example 8 Self Tutor
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.
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\127IGCSE01_06.CDR Monday, 15 September 2008 2:34:48 PM PETER
128 Exponents and surds (Chapter 6)
Example 9 Self Tutor
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
2 Simplify using the index laws:
a24
23b
35
32c
57
53d
49
45
ex6
x3f
y7
y4g a8 ¥ a7 h b9 ¥ b5
3 Simplify using the index laws:
a (22)3 b (34)3 c (23)6 d (102)5
e (x3)2 f (x5)3 g (a5)4 h (b6)4
4 Simplify using the index laws:
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
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\128IGCSE01_06.CDR Monday, 15 September 2008 3:29:53 PM PETER
Exponents and surds (Chapter 6) 129
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.
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\129IGCSE01_06.CDR Thursday, 18 September 2008 12:18:04 PM PETER
130 Exponents and surds (Chapter 6)
Example 10 Self Tutor
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
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\130IGCSE01_06.CDR Monday, 15 September 2008 2:38:27 PM PETER
Exponents and surds (Chapter 6) 131
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 :
Example 11 Self Tutor
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
Example 12 Self Tutor
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
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\131IGCSE01_06.CDR Friday, 10 October 2008 9:08:28 AM PETER
132 Exponents and surds (Chapter 6)
Example 13 Self Tutor
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
5 Write as an ordinary decimal number:
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
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\132IGCSE01_06.CDR Wednesday, 17 September 2008 9:43:26 AM PETER
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
Exponents and surds (Chapter 6) 133
6 Write as an ordinary decimal number:
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
Example 14 Self Tutor
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
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\133IGCSE01_06.CDR Wednesday, 17 September 2008 9:49:45 AM PETER
134 Exponents and surds (Chapter 6)
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
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\134IGCSE01_06.CDR Monday, 15 September 2008 2:59:38 PM PETER
Exponents and surds (Chapter 6) 135
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.
Example 15 Self Tutor
Simplify:
a (p5)2 b
µ1p5
¶2
a (p5)2
=p5£
p5
= 5
b
µ1p5
¶2
=1p5£ 1p
5
= 15
Example 16 Self Tutor
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
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\135IGCSE01_06.CDR Monday, 15 September 2008 3:01:23 PM PETER
136 Exponents and surds (Chapter 6)
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.
Example 17 Self Tutor
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
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\136IGCSE01_06.CDR Friday, 10 October 2008 9:08:51 AM PETER
Exponents and surds (Chapter 6) 137
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:
Example 18 Self Tutor
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
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\137IGCSE01_06.CDR Monday, 15 September 2008 3:02:50 PM PETER
138 Exponents and surds (Chapter 6)
Example 19 Self Tutor
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
Example 20 Self Tutor
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.
Example 21 Self Tutor
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.
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\138IGCSE01_06.CDR Monday, 15 September 2008 3:04:28 PM PETER
Exponents and surds (Chapter 6) 139
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
5 Write the following in the form kp3:
ap12 b
p27 c
p75 d
q13
6 Write the following in the form kp5:
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
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\139IGCSE01_06.CDR Monday, 15 September 2008 3:06:19 PM PETER
140 Exponents and surds (Chapter 6)
Example 22 Self Tutor
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
Example 23 Self Tutor
Expand and simplify:
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
Example 24 Self Tutor
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
Example 25 Self Tutor
Expand and simplify:
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
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\140IGCSE01_06.CDR Monday, 15 September 2008 3:21:04 PM PETER
Exponents and surds (Chapter 6) 141
Example 26 Self Tutor
Expand and simplify:
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)
2 Expand and simplify:
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)
3 Expand and simplify:
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)
4 Expand and simplify:
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
5 Expand and simplify:
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)
6 Expand and simplify:
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
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\141IGCSE01_06.CDR Monday, 15 September 2008 3:11:33 PM PETER
142 Exponents and surds (Chapter 6)
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.
Example 27 Self Tutor
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
Example 28 Self Tutor
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.
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\142IGCSE01_06.CDR Friday, 10 October 2008 9:10:01 AM PETER
Exponents and surds (Chapter 6) 143
Example 29 Self Tutor
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
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\143IGCSE01_06.CDR Monday, 15 September 2008 3:24:16 PM PETER
144 Exponents and surds (Chapter 6)
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
9 Write as an ordinary decimal number:
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.
13 Expand and simplify:
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.
16 Write in the form a+ bp3 where a, b 2 Q :
a
p32 + 1
1¡p32
b2 +
p3
2¡p3¡ 2
p3
2 +p3
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\144IGCSE01_06.CDR Monday, 15 September 2008 3:25:03 PM PETER
Exponents and surds (Chapter 6) 145
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
9 Write as an ordinary decimal number:
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
14 Expand and simplify:
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
16 Write in the form a+ bp5 where a, b 2 Q :
a1 +
p5
2¡p5
b3¡p
5
3 +p5¡ 4
3¡p5
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\145IGCSE01_06.CDR Wednesday, 5 November 2008 2:15:20 PM PETER
Challenge#endboxedheading
146 Exponents and surds (Chapter 6)
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:
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\146IGCSE01_06.cdr Friday, 31 October 2008 9:51:20 AM PETER