Answers − Solving quadratics of the form x² + bx + c = 0 by factorising first

1.   Solve x² + 6x + 8 = 0

i.		From answer 1 in Factorising Quadratics − 1 the factors are: (x + 2)(x + 4)
ii.		So (x + 2)(x + 4) = 0

iii.		x + 2 = 0		=>		x = −2		Or
		x + 4 = 0		=>		x = −4

2.   Find the values of x where x² + 9x + 18 = 0

i.		From answer 2 in Factorising Quadratics − 1 the factors are: (x + 3)(x + 6)
ii.		So (x + 3)(x + 6) = 0

iii.		x + 3 = 0		=>		x = −3		Or
		x + 6 = 0		=>		x = −6

3.   Solve x² + x − 56 = 0

i.		From answer 3 in Factorising Quadratics − 1 the factors are: (x + 8)(x − 7)
ii.		So (x + 8)(x − 7) = 0

iii.		x + 8 = 0		=>		x = −8		Or
		x − 7 = 0		=>		x = 7

4.   Find the values of x where x² + 10x − 24 = 0

i.		From answer 4 in Factorising Quadratics − 1 the factors are: (x + 12)(x − 2)
ii.		So (x + 12)(x − 2) = 0

iii.		x + 12 = 0		=>		x = −12		Or
		x − 2 = 0		=>		x = 2

5.   Solve x² − 16x + 60 = 0

i.		From answer 5 in Factorising Quadratics − 1 the factors are: (x − 6)(x − 10)
ii.		So (x − 6)(x − 10) = 0

iii.		x − 6 = 0		=>		x = 6		Or
		x − 10 = 0		=>		x = 10

6.   Find the values of x where x² − 15x + 36 = 0

i.		From answer 6 in Factorising Quadratics − 1 the factors are: (x − 3)(x − 12)
ii.		So (x − 3)(x − 12) = 0

iii.		x − 3 = 0		=>		x = 3		Or
		x − 12 = 0		=>		x = 12

7.   Solve x² − 8x − 48 = 0

i.		From answer 7 in Factorising Quadratics − 1 the factors are: (x + 4)(x − 12)
ii.		So (x + 4)(x − 12) = 0

iii.		x + 4 = 0		=>		x = −4		Or
		x − 12 = 0		=>		x = 12

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