Cambridge Additional Mathematics

104 Surds, indices, and exponentials (Chapter 4)

EXERCISE 4A.2

1 Simplify:

a 2

p

2+3

p

2 b 2

p

2 ¡ 3

p

2 c 5

p

5 ¡ 3

p

5 d 5

p

5+3

p

5

e 3

p

5 ¡ 5

p

5 f 7

p

3+2

p

3 g 9

p

6 ¡ 12

p

6 h

p

2+

p

2+

p

2

2 Simplify:

a

p

2(3¡

p

2) b

p

5(

p

5+1) c

p

10(3 + 2

p

10) d

p

7(3

p

7 ¡4)

e ¡

p

3(5 +

p

3) f 2

p

6(

p

6 ¡7) g ¡

p

8(

p

8 ¡5) h ¡ 3

p

2(4¡ 6

p

2)

3 Simplify:

a (5 +

p

2)(4 +

p

2) b (7 + 2

p

3)(4 +

p

3) c (9¡

p

7)(4 + 2

p

7)

d (

p

3 + 1)(2¡ 3

p

3) e (

p

8 ¡6)(2

p

8 ¡3) f (2

p

5 ¡7)(1¡ 4

p

5)

Example 5 Self Tutor

Simplify:

a (5¡

p

2)^2 b (7 + 2

p

5)(7¡ 2

p

5)

a (5¡

p

2)^2

=25¡ 10

p

2+2

=27¡ 10

p

2

b (7 + 2

p

5)(7¡ 2

p

5)

=7^2 ¡(2

p

5)^2

=49¡(4£5)

=29

4 Simplify:

a (3 +

p

2)^2 b (6¡

p

3)^2 c (

p

5+1)^2 d (

p

8 ¡3)^2

e (4 + 2

p

3)^2 f (3

p

5+1)^2 g (7¡ 2

p

10)^2 h (5

p

6 ¡4)^2

5 Simplify:

a (3 +

p

7)(3¡

p

7) b (

p

2 + 5)(

p

2 ¡5) c (4¡

p

3)(4 +

p

3)

d (2

p

2 + 1)(2

p

2 ¡1) e (4 + 3

p

8)(4¡ 3

p

8) f (9

p

3 ¡5)(9

p

3+5)

DIVISION BY SURDS

Numbers like

6

p

2

and

9

5+

p

2

involve dividing by a surd.

It is customary to ‘simplify’ these numbers by rewriting them without the surd in the denominator.

For any fraction of the form

b

p

a

, we can remove the surd from the denominator by multiplying by

p

a

p

a

.

Since

p

a

p

a

=1, this does not change the value of the fraction.

=5^2 + 2(5)(¡

p

2) + (

p

2)^2

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Additional Mathematics
Y:\HAESE\CAM4037\CamAdd_04\104CamAdd_04.cdr Friday, 31 January 2014 11:18:18 AM BRIAN