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Radicals (Surds)
4Chapter
Contents: A Radicals on a number lineB Operations with radicals
C Expansions with radicals
D Division by radicals
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Y:\HAESE\IB_MYP4\IB_MYP4_04\087IB_MYP4_04.CDR Friday, 29 February 2008 4:01:39 PM PETERDELL
RESEARCH
88 RADICALS (SURDS) (Chapter 4)
INTRODUCTION
In previous years we used the Theorem of Pythagoras
to find the length of the third side of a triangle.
Our answers often involved radicals such asp
2,p
3,p5, and so on.
A radical is a number that is written using the radical signp
.
Radicals such asp
4 andp
9 are rational sincep
4 = 2 andp
9 = 3.
Radicals such asp
2,p
3 andp
5 are irrational. They have decimal expansions which neither
terminate nor recur. Irrational radicals are also known as surds.
² Where did the names radical and surd come from?
² Why do we use the word irrational to describe some numbers?
² Before we had calculators and computers, finding
decimal representations for numbers like 1p2
to four
or five decimal places was quite difficult and time
consuming.
Imagine having to find 11:414 21 correct to five
decimal places using long division!
A method was devised to do this calculation quickly.
What was the process?
SQUARE ROOTS
The square root of a orpa is the positive number which obeys the rule
pa£p
a = a.
Forpa to have meaning we require a to be non-negative, i.e., a > 0.
For example,p
5 £p5 = 5 or (
p5)2 = 5.
Note thatp
4 = 2, not §2, since the square root of a number cannot be negative.
If we convert a radical such asp
5 to a decimal we can find its approximate position on a
number line.p
5 ¼ 2:236067, sop
5 is close to 214 .
RADICALS ON A NUMBER LINEA
~`5
2
1
~`5
0 1 2 3
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Y:\HAESE\IB_MYP4\IB_MYP4_04\088IB_MYP4_04.CDR Wednesday, 5 March 2008 9:14:46 AM PETERDELL
RADICALS (SURDS) (Chapter 4) 89
We can also construct the position ofp
5 on a number line using a ruler and compass. Since
12 + 22 = (p
5)2, we can use a right-angled triangle with sides of length 1, 2 andp
5.
Step 1: Draw a number line and mark the numbers 0,
Step 2: With compass point on 1, draw an arc above
2. Do the same with compass point on 3 using
the same radius. Draw the perpendicular at 2through the intersection of these arcs, and mark
off 1 cm. Call this point A.
Step 3: Complete the right angled triangle. Its sides are 2, 1 andp
5 cm.
Step 4: With centre O and radius OA, draw an arc through A to meet the
number line. It meets the number line atp
5.
EXERCISE 4A
1 Notice that 12 + 42 = 17 = (p
17)2.
Locatep
17 on a number line using an accurate construction.
2 a The sum of the squares of which two positive integers is 13?
b Accurately construct the position ofp
13 on a number line.
3 Can we construct the exact position ofp
6 on a number line using the method above?
4 7 cannot be written as the sum of two squares
so the above method cannot be used for locatingp7 on the number line.
However, 42¡32 = 7, so 42 = 32+(p
7)2.
We can thus construct a right angled triangle with sides of length 4, 3 andp
7.
Use such a triangle to accurately locatep
7 on a number line.
ADDING AND SUBTRACTING RADICALS
We can add and subtract ‘like radicals’ in the same way as we do ‘like terms’ in algebra.
For example: ² just as 3a + 2a = 5a, 3p
2 + 2p
2 = 5p
2
² just as 6b¡ 4b = 2b, 6p
3 ¡ 4p
3 = 2p
3.
Simplify: a 3p
2 ¡ 4p
2 bp
7 ¡ 2(1 ¡p7)
a 3p
2 ¡ 4p
2 = ¡1p
2
= ¡p
2
bp
7 ¡ 2(1 ¡p
7) =p
7 ¡ 2 + 2p
7
= 3p
7 ¡ 2
OPERATIONS WITH RADICALSB
Example 1 Self Tutor
0 1 2 3
1
~`5
DEMO
1 2 3 1, , and on it, cm apart.
4~`7
3
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Y:\HAESE\IB_MYP4\IB_MYP4_04\089IB_MYP4_04.CDR Wednesday, 5 March 2008 9:19:04 AM PETERDELL
90 RADICALS (SURDS) (Chapter 4)
Simplifying: a (3p
2)2 b 3p
3 £ (¡2p
3)
a (3p
2)2
= 3p
2 £ 3p
2
= 9 £ 2
= 18
b 3p
3 £ (¡2p
3)
= 3 £¡2 £p
3 £p
3
= ¡6 £ 3
= ¡18
Write in simplest form:
ap
2 £p5 b 3
p2 £ 4
p11
ap
2 £p
5
=p
2 £ 5
=p
10
b 3p
2 £ 4p
11
= 3 £ 4 £p
2 £p
11
= 12 £p2 £ 11
= 12p
22
Example 4 Self Tutor
Example 3 Self Tutor
With practiceyou should
not need themiddle steps.
SIMPLIFYING PRODUCTS
We have established in previous years that:papa = (
pa)2 = a
papb =
pab
papb
=
ra
b
Simplify: a (p
2)2 b (p
2)3 c
µ4p2
¶2
a (p
2)2
=p
2 £p
2
= 2
b (p
2)3
=p
2 £p
2 £p
2
= 2p
2
c
µ4p2
¶2
=42
(p
2)2
= 162
= 8
Example 2 Self Tutor
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Y:\HAESE\IB_MYP4\IB_MYP4_04\090IB_MYP4_04.CDR Wednesday, 5 March 2008 9:24:17 AM PETERDELL
RADICALS (SURDS) (Chapter 4) 91
Simplify: a
p75p3
b
p32
2p
2
a
p75p3
=q
753
=p
25
= 5
b
p32
2p
2
= 12
q322
= 12
p16
= 12 £ 4
= 2
EXERCISE 4B.1
1 Simplify:
a 3p
2 + 7p
2 b 11p
3 ¡ 8p
3
c 6p
5 ¡ 7p
5 d ¡p2 + 2
p2
ep
3 ¡ (2 ¡p3) f ¡p
2 ¡ (3 +p
2)
g 5p
2 ¡p3 +
p2 ¡p
3 hp
7 ¡ 2p
2 +p
7 ¡p2
i 3p
3 ¡p2 ¡ (1 ¡p
2) j 2(p
3 + 1) + 3(1 ¡p3)
k 3(p
3 ¡p2) ¡ (
p2 ¡p
3) l 3(p
3 ¡ 1) ¡ 2(2 ¡p3)
2 Simplify:
a (p
3)2 b (p
3)3 c (p
3)5 d
µ1p3
¶2
e (p
7)2 f (p
7)3 g
µ1p7
¶2
h
µ3p7
¶2
i (p
5)2 j (p
5)4 k
µ5p5
¶2
l
µ10p
5
¶2
3 Simplify:
a (2p
2)2 b (4p
2)2 c (2p
3)2
d (3p
3)2 e (2p
5)2 f (3p
5)2
g (2p
7)2 h (2p
10)2 i (7p
10)2
j 3p
2 £ 4p
2 k 5p
3 £ 2p
3 l 7p
2 £ 5p
2
m (¡4p
2)2 n (¡7p
3)2 op
2 £ (¡3p
2)
p (¡2p
3)(¡5p
3) q (¡2p
7) £ 3p
7 rp
11 £ (¡2p
11)
4 Simplify:
a
q614 b
q179 c
q21425 d
q719
Example 5 Self Tutor
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Y:\HAESE\IB_MYP4\IB_MYP4_04\091IB_MYP4_04.CDR Thursday, 13 March 2008 11:42:19 AM PETERDELL
Writep
8 in simplest
form.
p8
=p
4 £ 2
=p
4 £p
2
= 2p
2
Example 6 Self Tutor
92 RADICALS (SURDS) (Chapter 4)
5 Simplify:
ap
2 £p3 b
p2 £p
7 cp
2 £p17
dp
7 £p3 e 2
p2 £ 5
p3 f (3
p2)2
g 5p
2 £p7 h 2
p6 £ 3
p5 i ¡5
p2 £ 2
p7
j (¡p7) £ (¡2
p3) k (2
p3)2 £ 2
p5 l (2
p2)3 £ 5
p3
6 Simplify:
a
p8p2
b
p3p27
c
p18p3
d
p2p50
e
p75p5
f
p5p75
g
p18p2
h
p3p60
i3p
6p2
j4p
12p3
k4p
6p24
l3p
98
2p
2
7 a Isp
9 +p
16 =p
9 + 16 ? Isp
25 ¡p16 =
p25 ¡ 16 ?
b Arepa +
pb =
pa + b and
pa ¡ p
b =pa¡ b possible laws for radical
numbers?
8 a Prove thatpapb =
pab for all positive numbers a and b.
Hint: Consider (papb)2 and (
pab)2.
b Prove that
papb
=
ra
bfor a > 0 and b > 0.
SIMPLEST RADICAL FORM
A radical is in simplest form when the number under the radical sign is the smallest
possible integer.
p32 =
p4 £ 8 = 2
p8 is not in simplest form as
p8 can be further simplified into 2
p2.
In simplest form,p
32 = 4p
2.
We look for the largestperfect square that canbe taken out as a factor
of this number.
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Y:\HAESE\IB_MYP4\IB_MYP4_04\092IB_MYP4_04.CDR Wednesday, 5 March 2008 9:25:31 AM PETERDELL
RADICALS (SURDS) (Chapter 4) 93
Writep
432 in simplest
radical form.
p432
=p
24 £ 33
=p
24 £p
33
= 4 £ 3p
3
= 12p
3
EXERCISE 4B.2
1 Write in the form kp
2 where k is an integer:
ap
18 bp
50 cp
72 dp
98
ep
162 fp
200 gp
20 000 hp
2 000 000
2 Write in the form kp
3 where k is an integer:
ap
12 bp
27 cp
48 dp
300
3 Write in the form kp
5 where k 2 Z :
ap
20 bp
80 cp
320 dp
500
4 Write in simplest radical form:
ap
99 bp
52 cp
40 dp
63
ep
48 fp
125 gp
147 hp
175
ip
176 jp
150 kp
275 lp
2000
5 Write in simplest radical form a + bpn where a, b 2 Q , n 2 Z :
a4 +
p8
2b
6 ¡p12
2c
4 +p
18
4d
8 ¡p32
4
e12 +
p72
6f
18 +p
27
6g
14 ¡p50
8h
5 ¡p200
10
The rules for expanding radical expressions containing brackets are identical to those for
ordinary algebra.
EXPANSIONS WITH RADICALSC
Example 7 Self Tutor It may be usefulto do a prime
factorisation of thenumber under the
radical sign.
a(b + c) = ab+ ac
(a+ b)(c + d) = ac+ ad+ bc+ bd
(a+ b)2 = a2 + 2ab+ b2
(a+ b)(a¡ b) = a2 ¡ b2
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Y:\HAESE\IB_MYP4\IB_MYP4_04\093IB_MYP4_04.CDR Wednesday, 5 March 2008 9:26:42 AM PETERDELL
94 RADICALS (SURDS) (Chapter 4)
Simplify: a 2(2 +p
3) bp
2(5 ¡ 2p
2)
a 2(2 +p
3)
= 2 £ 2 + 2 £p
3
= 4 + 2p
3
bp
2(5 ¡ 2p
2)
=p
2 £ 5 +p
2 £¡2p
2
= 5p
2 ¡ 4
Expand and simplify:
a ¡p3(2 +
p3) b ¡p
2(p
2 ¡p3)
a ¡p
3(2 +p
3)
= ¡p
3 £ 2 + ¡p
3 £p
3
= ¡2p
3 ¡ 3
b ¡p
2(p
2 ¡p
3)
= ¡p
2 £p
2 + ¡p
2 £¡p
3
= ¡2 +p
6
EXERCISE 4C
1 Expand and simplify:
a 4(3 +p
2) b 3(p
2 +p
3) c 5(4 ¡p7)
d 6(p
11 ¡ 4) ep
2(1 +p
2) fp
2(p
2 ¡ 5)
gp
3(2 + 2p
3) hp
3(p
3 ¡p2) i
p5(6 ¡p
5)
jp
5(2p
5 ¡ 1) kp
5(2p
5 +p
3) lp
7(2 +p
7 +p
2)
2 Expand and simplify:
a ¡p2(4 +
p2) b
p2(3 ¡p
2) c ¡p2(p
2 ¡p7)
d ¡p3(3 +
p3) e ¡p
3(5 ¡p3) f ¡p
3(2p
3 +p
5)
g ¡p5(2
p2 ¡p
3) h ¡2p
2(p
2 +p
3) i ¡2p
3(1 ¡ 2p
2)
j ¡p7(2
p7 + 4) k ¡p
11(2 ¡p11) l ¡(
p2)3(4 ¡ 2
p2)
Expand and simplify:
a (2 +p
2)(3 +p
2) b (3 +p
5)(1 ¡p5)
a (2 +p
2)(3 +p
2)
= (2 +p
2)3 + (2 +p
2)p
2
= 6 + 3p
2 + 2p
2 + 2
= 8 + 5p
2
b (3 +p
5)(1 ¡p
5)
= (3 +p
5)(1 + ¡p
5)
= (3 +p
5)1 + (3 +p
5)(¡p
5)
= 3 +p
5 ¡ 3p
5 ¡ 5
= ¡2 ¡ 2p
5
Example 10 Self Tutor
Example 9 Self Tutor
Example 8 Self Tutor
With practiceyou should
not need themiddle steps.
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Y:\HAESE\IB_MYP4\IB_MYP4_04\094IB_MYP4_04.CDR Wednesday, 5 March 2008 10:23:22 AM PETERDELL
RADICALS (SURDS) (Chapter 4) 95
3 Expand and simplify:
a (2 +p
2)(3 +p
2) b (3 +p
2)(3 +p
2)
c (p
2 + 2)(p
2 ¡ 1) d (4 ¡p3)(3 +
p3)
e (2 +p
3)(2 ¡p3) f (2 ¡p
6)(5 +p
6)
g (p
7 + 2)(p
7 ¡ 3) h (p
11 +p
2)(p
11 ¡p2)
i (3p
2 + 1)(3p
2 + 3) j (6 ¡ 2p
2)(2 +p
2)
Expand and simplify:
a (p
2 + 3)2 b (p
5 ¡p3)2
a (p
2 + 3)2
= (p
2)2 + 2p
2(3) + 32
= 2 + 6p
2 + 9
= 11 + 6p
2
b (p
5 ¡p
3)2
= (p
5 + ¡p
3)2
= (p
5)2 + 2p
5(¡p
3) + (¡p
3)2
= 5 ¡ 2p
15 + 3
= 8 ¡ 2p
15
4 Expand and simplify:
a (1 +p
3)2 b (p
2 + 5)2 c (3 ¡p2)2
d (1 +p
7)2 e (p
3 ¡p2)2 f (4 ¡p
5)2
g (p
3 +p
5)2 h (3 ¡p6)2 i (
p6 ¡p
3)2
j (2p
2 + 3)2 k (3 ¡ 2p
2)2 l (3 ¡ 5p
2)2
Expand and simplify:
a (4 +p
2)(4 ¡p2) b (2
p2 + 3)(2
p2 ¡ 3)
a (4 +p
2)(4 ¡p
2)
= 42 ¡ (p
2)2
= 16 ¡ 2
= 14
b (2p
2 + 3)(2p
2 ¡ 3)
= (2p
2)2 ¡ 32
= 8 ¡ 9
= ¡1
5 Expand and simplify:
a (3 +p
2)(3 ¡p2) b (
p3 ¡ 1)(
p3 + 1)
c (5 +p
3)(5 ¡p3) d (
p3 ¡ 4)(
p3 + 4)
e (p
7 ¡ 3)(p
7 + 3) f (2 + 5p
2)(2 ¡ 5p
2)
g (p
7 ¡p11)(
p7 +
p11) h (2
p5 + 6)(2
p5 ¡ 6)
Example 12 Self Tutor
Example 11 Self Tutor
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Y:\HAESE\IB_MYP4\IB_MYP4_04\095IB_MYP4_04.CDR Friday, 29 February 2008 4:51:00 PM PETERDELL
INVESTIGATION 1 DIVISION BYpa
96 RADICALS (SURDS) (Chapter 4)
i (3p
2 + 2)(3p
2 ¡ 2) j (p
3 ¡p2)(
p3 +
p2)
k (p
3 ¡p7)(
p3 +
p7) l (2
p2 + 1)(2
p2 ¡ 1)
In numbers like6p2
and9p
5 ¡p2
we have divided by a radical.
It is customary to ‘simplify’ these numbers by rewriting them without the radical in the
denominator.
In this investigation we consider fractions of the formbpa
where a and
b are real numbers. To remove the radical from the denominator, there are
two methods we could use:
² ‘splitting’ the numerator ² rationalising the denominator
What to do:
1 Consider the fraction6p2
.
a Since 2 is a factor of 6, ‘split’ the 6 into 3 £p2 £p
2.
b Simplify6p2
.
2 Can the method of ‘splitting’ the numerator be used to simplify7p2
?
3 Consider the fraction7p2
.
a If we multiply this fraction by
p2p2
, are we changing its value?
b Simplify7p2
by multiplying both its numerator and denominator byp
2.
4 The method in 3 is called ‘rationalising the denominator’. Will this method work
for all fractions of the formbpa
where a and b are real?
From the Investigation above, you should have found that for any fraction of the formbpa
,
we can remove the radical from the denominator by multiplying by
papa
. Since
papa
= 1,
we do not change the value of the fraction.
DIVISION BY RADICALSD
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Y:\HAESE\IB_MYP4\IB_MYP4_04\096IB_MYP4_04.CDR Monday, 3 March 2008 9:19:26 AM PETERDELL
INVESTIGATION 2 RADICAL CONJUGATES
RADICALS (SURDS) (Chapter 4) 97
Multiplying the original
number by
p3p3
orp7p7
does not change
its value.
Write with an integer denominator:
a6p3
b35p
7
a6p3
=6p3£
p3p3
=6p
3
3
= 2p
3
b35p
7
=35p
7£
p7p7
=35p
7
7
= 5p
7
EXERCISE 4D.1
1 Write with integer denominator:
a1p3
b3p3
c9p3
d11p
3e
p2
3p
3
f2p2
g6p2
h12p
2i
p3p2
j1
4p
2
k5p5
l15p
5m
¡3p5
n200p
5o
1
3p
5
p7p7
q21p
7r
2p11
s26p13
t1
(p
3)3
RADICAL CONJUGATES
Radical expressions such as 3 +p
2 and 3 ¡p2 which are identical except for opposing
signs in the middle, are called radical conjugates.
The radical conjugate of a +pb is a¡p
b.
Fractions of the formc
a +pb
can also be simplified to remove the
radical from the denominator. To do this we use radical conjugates.
What to do:
1 Expand and simplify:
a (2 +p
3)(2 ¡p3) b (
p3 ¡ 1)(
p3 + 1)
2 What do you notice about your results in 1?
Example 13 Self Tutor
2
1
5
1
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Y:\HAESE\IB_MYP4\IB_MYP4_04\097IB_MYP4_04.CDR Friday, 11 April 2008 4:41:01 PM PETERDELL
98 RADICALS (SURDS) (Chapter 4)
3
a (a +pb)(a¡p
b) b (pa¡ b)(
pa + b)
4 a Copy and complete:
To remove the radicals from the denominator of a fraction, we can multiply the
denominator by its ......
b What must we do to the numerator of the fraction to ensure we do not change
its value?
From the Investigation above, we should have found that:
to remove the radicals from the denominator of a fraction, we multiply both the numerator
and the denominator by the radical conjugate of the denominator.
Write14
3 ¡p2
with an integer denominator.
14
3 ¡p2
=
µ14
3 ¡p2
¶Ã3 +
p2
3 +p
2
!
=14
9 ¡ 2£ (3 +
p2)
= 2(3 +p
2)
= 6 + 2p
2
Write1
5 +p
2: a with integer denominator
b in the form a + bp
2 where a, b 2 Q .
a1
5 +p
2
=
µ1
5 +p
2
¶£Ã
5 ¡p2
5 ¡p2
!
=5 ¡p
2
25 ¡ 2
=5 ¡p
2
23
b5 ¡p
2
23= 5
23 ¡ 123
p2
So, a = 523 and b = ¡ 1
23 .
Example 15 Self Tutor
Example 14 Self Tutor
Show that for any integers a and b, the following products are integers:
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Y:\HAESE\IB_MYP4\IB_MYP4_04\098IB_MYP4_04.CDR Thursday, 13 March 2008 2:11:59 PM PETERDELL
REVIEW SET 4A
RADICALS (SURDS) (Chapter 4) 99
EXERCISE 4D.2
1 Write with integer denominator:
a1
3 +p
2b
2
3 ¡p2
c1
2 +p
5d
p2
2 ¡p2
e1 +
p2
1 ¡p2
f
p3
4 ¡p3
g¡2
p2
1 ¡p2
h1 +
p5
2 ¡p5
2 Write in the form a + bp
2 where a, b 2 Q :
a3p
2 ¡ 3b
4
2 +p
2c
p2p
2 ¡ 5d
¡2p
2p2 + 1
3 Write in the form a + bp
3 where a, b 2 Q :
a4
1 ¡p3
b6p
3 + 2c
p3
2 ¡p3
d1 + 2
p3
3 +p
3
4 a If a, b and c are integers, show that (a + bpc)(a¡ b
pc) is an integer.
b Write with an integer denominator: i1
1 + 2p
3ii
p2
3p
2 ¡ 5.
5 a If a and b are integers, show that (pa +
pb)(
pa¡p
b) is also an integer.
b
i1p
2 +p
3ii
p3p
3 ¡p5
1 Simplify:
a (2p
3)2 b
µ4p2
¶2
c 3p
2 £ 2p
5 dq
1214
2 a Copy and complete: 12 + 32 = (::::::)2
b Use a to accurately construct the position ofp
10 on a number line using a ruler
and compass.
3 Simplify:
a
p15p3
b
p35p7
c
p35p5
d
p2p20
LINKSclick here
HOW A CALCULATOR CALCULATES RATIONALNUMBERSAreas of interaction:Human ingenuity
Write with an integer denominator:
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Y:\HAESE\IB_MYP4\IB_MYP4_04\099IB_MYP4_04.CDR Thursday, 13 March 2008 11:45:11 AM PETERDELL
REVIEW SET 4B
100 RADICALS (SURDS) (Chapter 4)
4 a Writep
8 in simplest radical form.
b Hence, simplify 5p
2 ¡p8 .
5 Writep
98 in simplest radical form.
6 Expand and simplify:
a 2(p
3 + 1) bp
2(3 ¡p2) c (1 +
p7)2
d (2 ¡p5)2 e (3 +
p2)(3 ¡p
2) f (3 +p
2)(1 ¡p2)
7 Write with an integer denominator:
a10p
5b
p3 + 2p3 + 1
c1 +
p7
1 ¡p7
8 Write in the form a + bp
5 where a, b 2 Q :
a3
2 ¡p5
b2p
5p5 + 1
1 Simplify:
ap
3p
2 b
p8p2
c (3p
5)2 dq
549
2 Find the exact position ofp
12 on a number line using a ruler and compass
construction. Explain your method.
Hint: Look for two positive integers a and b such that a2 ¡ b2 = 12.
3 Simplify: a
p21p3
b
p3p24
4 Simplify:p
3 ¡p27
5 Write in simplest radical form: ap
12 bp
63
6 Expand and simplify:
a 3(2 ¡p3) b
p7(p
2 ¡ 1) c (3 ¡p2)2
d (p
3 +p
2)2 e (2 ¡p5)(2 +
p5) f (2 +
p3)(3 ¡p
3)
7 Write with integer denominator:
a24p
3b
1 +p
2
2 ¡p2
c4 ¡p
5
3 +p
5
8 Write in the form a + bp
3 where a, b 2 Q :
a18
5 ¡p3
b¡p
3
3 +p
3
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Y:\HAESE\IB_MYP4\IB_MYP4_04\100IB_MYP4_04.CDR Monday, 3 March 2008 9:31:13 AM PETERDELL