Cambridge Additional Mathematics

RHS is short for

Right Hand Side.

Quadratics (Chapter 3) 67

I Consider (2x¡7)^2 =(x+1)^2.

Correct solution

(2x¡7)^2 =(x+1)^2

) (2x¡7)^2 ¡(x+1)^2 =0

) (2x¡7+x+ 1)(2x¡ 7 ¡x¡1) = 0

) (3x¡6)(x¡8) = 0

) x=2or 8

Incorrect solution

(2x¡7)^2 =(x+1)^2

) 2 x¡7=x+1

) x=8

EXERCISE 3A.1

1 Solve the following by factorisation:

a 4 x^2 +7x=0 b 6 x^2 +2x=0 c 3 x^2 ¡ 7 x=0

d 2 x^2 ¡ 11 x=0 e 3 x^2 =8x f 9 x=6x^2

g x^2 ¡ 5 x+6=0 h x^2 =2x+8 i x^2 +21=10x

j 9+x^2 =6x k x^2 +x=12 l x^2 +8x=33

2 Solve the following by factorisation:

a 9 x^2 ¡ 12 x+4=0 b 2 x^2 ¡ 13 x¡7=0 c 3 x^2 =16x+12

d 3 x^2 +5x=2 e 2 x^2 +3=5x f 3 x^2 +8x+4=0

g 3 x^2 =10x+8 h 4 x^2 +4x=3 i 4 x^2 =11x+3

j 12 x^2 =11x+15 k 7 x^2 +6x=1 l 15 x^2 +2x=56

Example 3 Self Tutor

Solve forx: 3 x+^2

x

=¡ 7

3 x+

2

x

=¡ 7

) x

³

3 x+

2

x

́

=¡ 7 x fmultiplying both sides byxg

) 3 x^2 +2=¡ 7 x fexpanding the bracketsg

) 3 x^2 +7x+2=0 fmaking the RHS 0 g

) (x+ 2)(3x+1)=0 ffactorisingg

) x=¡ 2 or¡^13

3 Solve forx:

a (x+1)^2 =2x^2 ¡ 5 x+11 b (x+ 2)(1¡x)=¡ 4

c 5 ¡ 4 x^2 = 3(2x+1)+2 d x+

2

x

=3

e 2 x¡

1

x

=¡ 1 f

x+3

1 ¡x

=¡

9

x

SOLVING BY ‘COMPLETING THE SQUARE’

As you would be aware by now, not all quadratics factorise easily. For example, x^2 +4x+1cannot be

factorised by simple factorisation. In other words, we cannot writex^2 +4x+1in the form (x¡a)(x¡b)

wherea,bare rational.

4037 Cambridge

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Y:\HAESE\CAM4037\CamAdd_03\067CamAdd_03.cdr Thursday, 3 April 2014 4:18:26 PM BRIAN