Solving Quadratics by Factorising (Cambridge (CIE) IGCSE Additional Maths): Revision Note

• Where permitted, and if you calculator has a quadratic solving feature, you can use it to check your final solutions!

  • Such calculators also help you to factorise (if you're struggling with that step)

  • e.g.  A calculator gives solutions to $6x^2+x-2=0$ as x = $-\frac{2}{3}$  and x = $\frac{1}{2}$

    • "Reverse" the method above to factorise!

    • $6x^2+x-2=\left(3x+2\right)\left(2x-1\right)$

  • Warning: a calculator (correctly) gives solutions to 12x² + 2x – 4 = 0 as x = $-\frac{2}{3}$ and x = $\frac{1}{2}$

    • But 12x² + 2x – 4 ≠ $\left(3x+2\right)\left(2x-1\right)$ as these brackets expand to 6x² + ... not 12x² + ...

    • Multiply by 2 to correct this

    • 12x² + 2x – 4 = $2\left(3x+2\right)\left(2x-1\right)$

(a) Solve $\left(x-2\right)\left(x+5\right)=0$
 

Set the first bracket equal to zero

x – 2 = 0

Add 2 to both sides

x = 2

Set the second bracket equal to zero

x + 5 = 0

Subtract 5 from both sides

x = -5

Write both solutions together using “or”

x = 2 or x = -5

(b) Solve $\left(8x+7\right)\left(2x-3\right)=0$
 

Set the first bracket equal to zero

8x + 7 = 0

Subtract 7 from both sides

8x = -7

Divide both sides by 8

x = $-\frac{7}{8}$

Set the second bracket equal to zero

2x - 3 = 0

Add 3 to both sides

2x = 3

Divide both sides by 2

x = $\frac{3}{2}$

Write both solutions together using “or”

x = $\mathbf{-}\frac{\mathbf{7}}{\mathbf{8}}$ or x = $\frac{\mathbf{3}}{\mathbf{2}}$

(c) Solve $x\left(5x-1\right)=0$
 

Do not divide both sides by x (this will lose a solution at the end)
Set the first “bracket” equal to zero

(x) = 0

Solve this equation to find x

x = 0

Set the second bracket equal to zero

5x - 1 = 0

Add 1 to both sides

5x = 1

Divide both sides by 5

x = $\frac{1}{5}$

Write both solutions together using “or”

x = 0 or x = $\frac{\mathbf{1}}{\mathbf{5}}$