Quadratic Equation Methods

If you have to solve a quadratic equation but are not told which method to use, here is a guide as to what to do

When should I solve by factorisation?

When the question asks to solve by factorisation
- For example, part (a) Factorise 6x² + 7x – 3, part (b) Solve 6x² + 7x – 3 = 0
When solving two-term quadratic equations
- For example, solve x² – 4x = 0
  - …by taking out a common factor of x to get x(x – 4) = 0
  - ...giving x = 0 and x = 4
- For example, solve x² – 9 = 0
  - …using the difference of two squares to factorise it as (x + 3)(x – 3) = 0
  - ...giving x = -3 and x = 3
  - (Or by rearranging to x² = 9 and using ±√ to get x = = ±3)

When should I use the quadratic formula?

When the question says to leave solutions correct to a given accuracy (2 decimal places, 3 significant figures etc)
When the quadratic formula may be faster than factorising
- It's quicker to solve 36x² + 33x – 20 = 0 using the quadratic formula then by factorisation
If in doubt, use the quadratic formula - it always works

When should I solve by completing the square?

When part (a) of a question says to complete the square and part (b) says to use part (a) to solve the equation
When making x the subject of harder formulae containing x² and x terms
- For example, make x the subject of the formula x² + 6x = y
  - Complete the square: (x + 3)² – 9 = y
  - Add 9 to both sides: (x + 3)² = y + 9
  - Take square roots and use ±: $x + 3 = \pm \sqrt{y + 9}$
  - Subtract 3: $x = - 3 \pm \sqrt{y + 9}$

Examiner Tip

Calculators can solve quadratic equations so use them to check your solutions
If the solutions on your calculator are whole numbers or fractions (with no square roots), this means the quadratic equation does factorise

Worked example

(a)

Solve

x^{2} - 7 x + 2 = 0

, giving your answers correct to 2 decimal places

“Correct to 2 decimal places” suggests using the quadratic formula
Substitute a = 1, b = -7 and c = 2 into the formula, putting brackets around any negative numbers

$x = \frac{- (- 7) \pm \sqrt{{(- 7)}^{2} - 4 \times 1 \times 2}}{2 \times 1}$

Use a calculator to find each solution

x = 6.70156… or 0.2984...

Round your final answers to 2 decimal places

x = 6.70 or x = 0.30

(b)

Solve

16 x^{2} - 82 x + 45 = 0

Method 1
If you cannot spot the factorisation, use the quadratic formula
Substitute a = 16, b = -82 and c = 45 into the formula, putting brackets around any negative numbers

$x = \frac{- (- 82) \pm \sqrt{{(- 82)}^{2} - 4 \times 16 \times 45}}{2 \times 16}$

Use a calculator to find each solution

x = $\frac{9}{2}$ or x = $\frac{5}{8}$

Method 2
If you do spot the factorisation, (2x – 9)(8x – 5), then use that method instead

$(2 x - 9) (8 x - 5) = 0$

Set the first bracket equal to zero

$2 x - 9 = 0$

Add 9 to both sides then divide by 2

$\begin{array}{rcl} 2 x & = & 9 \\ x & = & \frac{9}{2} \end{array}$

Set the second bracket equal to zero

$8 x - 5 = 0$

Add 5 to both sides then divide by 8

$\begin{array}{rcl} 8 x & = & 5 \\ x & = & \frac{5}{8} \end{array}$

x = $\frac{9}{2}$ or x = $\frac{5}{8}$

(c)

By writing

x^{2} + 6 x + 5

in the form

{(x + p)}^{2} + q

, solve

x^{2} + 6 x + 5 = 0

This question wants you to complete the square first
Find p (by halving the middle number)

$p = \frac{6}{2} = 3$

Write x² + 6x as (x + p)² - p²

$\begin{array}{rcl} x^{2} + 6 x & = & {(x + 3)}^{2} - 3^{2} \\ = & {(x + 3)}^{2} - 9 \end{array}$

Replace x² + 6x with (x + 3)² – 9 in the equation

$\begin{array}{rcl} {(x + 3)}^{2} - 9 + 5 & = & 0 \\ {(x + 3)}^{2} - 4 & = & 0 \end{array}$

Make x the subject of the equation (start by adding 4 to both sides)

${(x + 3)}^{2} = 4$

Take square roots of both sides (include a ± sign to get both solutions)

$x + 3 = \pm \sqrt{4} = \pm 2$

Subtract 3 from both sides

$x = \pm 2 - 3$

Find each solution separately using + first, then - second

x = - 5, x = - 1

Even though the quadratic factorises to (x + 5)(x + 1), this is not the method asked for in the question

Quadratic Equation Methods (AQA GCSE Maths)

Revision Note