Solving Quadratics by Factorising (Cambridge International IGCSE Further Maths)

Solving quadratics by factorising

How do I solve a quadratic equation using factorisation?

• Rearrange it into the form ax² + bx + c = 0

  • zero must be on one side

    • it is easier to use the side where a is positive

• Factorise the quadratic and solve each bracket equal to zero

  • If (x + 4)(x - 1) = 0, then either x + 4 = 0 or x - 1 = 0

    • Because if A × B = 0, then either A = 0 or B = 0

• To solve (x−3)(x+7)=0

  • …solve “first bracket = 0”:

    • x – 3 = 0
    • add 3 to both sides: x = 3

  • …and solve “second bracket = 0”

    • x + 7 = 0
    • subtract 7 from both sides: x = -7

  • The two solutions are x = 3 or x = -7

    • The solutions have the opposite signs to the numbers in the brackets

• To solve (2x−3)(3x+5)=0

  • …solve “first bracket = 0”

    • 2x – 3 = 0
    • add 3 to both sides: 2x = 3
    • divide both sides by 2: x = $\frac{3}{2}$

  • …solve “second bracket = 0”

    • 3x + 5 = 0
    • subtract 5 from both sides: 3x = -5
    • divide both sides by 3: x = $-\frac{5}{3}$

  • The two solutions are x = $\frac{3}{2}$ or x = $-\frac{5}{3}$

• To solve x (x−4)=0

  • it may help to think of x as (x – 0) or (x)

  • …solve “first bracket = 0”

    • (x) = 0, so x = 0

  • …solve “second bracket = 0”

    • x – 4 = 0
    • add 4 to both sides: x = 4

  • The two solutions are x = 0 or x = 4

    • It is a common mistake to divide both sides by x at the beginning - you will lose a solution (the x = 0 solution)

Exam Hint

• 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!
    • 6x²+x−2=(3x+2)(2x−1)

  • 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 ≠ (3x+2)(2x−1) as these brackets expand to 6x² + ... not 12x² + ...
    • Multiply by 2 to correct this
    • 12x² + 2x – 4 = 2 (3x+2)(2x−1)

Worked Example (a) Solve (x−2)(x+5)=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”

Final answer: x = 2 or x = -5

(b) Solve (8x+7)(2x−3)=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”

Final answer: x = $-\frac{7}{8}$or x =$\frac{3}{2}$

(c) Solve x (5x−1)=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”

Final answer: x = 0 or x = $\frac{1}{5}$