radicals (surds) · radicals (surds) chapter 4 contents: a radicals on a number line b operations...

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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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



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



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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



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Writep

8 in simplest

form.

p8

=p

4 £ 2

=p

4 £p

2

= 2p

2



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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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


3 where k is an integer:

ap

12 bp

27 cp

48 dp

300


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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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)


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)


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




With practiceyou should

not need themiddle steps.


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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)


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


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


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


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)




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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


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


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:


a (2 +p

3)(2 ¡p3) b (

p3 ¡ 1)(

p3 + 1)

2 What do you notice about your results in 1?


2

1

5

1


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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 .



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


EXERCISE 4D.2


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 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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REVIEW SET 4B


4 a Writep

8 in simplest radical form.

b Hence, simplify 5p

2 ¡p8 .

5 Writep

98 in simplest radical form.


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


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


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)


a24p

3b

1 +p

2

2 ¡p2

c4 ¡p

5

3 +p

5


3 where a, b 2 Q :

a18

5 ¡p3

b¡p

3

3 +p

3


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\IB_MYP4\IB_MYP4_04\100IB_MYP4_04.CDR Monday, 3 March 2008 9:31:13 AM PETERDELL

radicals (surds) · radicals (surds) chapter 4 contents: a radicals on a number line b operations...

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