Cambridge International Mathematics

b 2 ¡ 3 x^2 =8

) ¡ 3 x^2 =6 fsubtracting 2 from both sidesg

) x^2 =¡ 2 fdividing both sides by¡ 3 g

which has no solutions asx^2 cannot be< 0 :

Example 2 Self Tutor

Solve forx:

a (x¡3)^2 =16 b (x+2)^2 =11

For equations of

the form

(x§a)^2 =k

we do not

a (x¡3)^2 =16 expand the LHS.

) x¡3=§

p

16

) x¡3=§ 4

) x=3§ 4

) x=7or¡ 1

b (x+2)^2 =11

) x+2=§

p

11

) x=¡ 2 §

p

11

EXERCISE 21A

1 Solve forx:

a x^2 = 100 b 2 x^2 =50 c 5 x^2 =20

d 6 x^2 =54 e 5 x^2 =¡ 45 f 7 x^2 =0

g 3 x^2 ¡2=25 h 4 ¡ 2 x^2 =12 i 4 x^2 +2=10

2 Solve forx:

a (x¡1)^2 =9 b (x+4)^2 =16 c (x+2)^2 =¡ 1

d (x¡4)^2 =5 e (x¡6)^2 =¡ 4 f (x+2)^2 =0

g (2x¡5)^2 =0 h (3x+2)^2 =4 i^13 (2x+3)^2 =2

For quadratic equations which are not of the formx^2 =k, we need an alternative method of solution. One

method is tofactorisethe quadratic into the product of linear factors and then apply theNull Factor law:

When the product of two (or more) numbers is zero, thenat least oneof them must be zero.

If ab=0then a=0or b=0.

Example 3 Self Tutor

Solve forxusing the Null Factor law:

a 3 x(x¡5) = 0 b (x¡4)(3x+7)=0

a 3 x(x¡5) = 0

) 3 x=0or x¡5=0

) x=0or 5

b (x¡4)(3x+7)=0

) x¡4=0or 3 x+7=0

) x=4 or 3 x=¡ 7

) x=4or ¡^73

B THE NULL FACTOR LAW [2.10]

Quadratic equations and functions (Chapter 21) 423

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Y:\HAESE\IGCSE01\IG01_21\423IGCSE01_21.CDR Monday, 27 October 2008 2:08:55 PM PETER