Deciding the Quadratic Method

Exam code: 4024

Written by: Ashika|Reviewed by: Caroline Carroll|Updated 2 July 2026

Quadratic Equation Methods

Deciding the quadratic method

When should I solve by factorisation?

Use factorisation when the question asks to solve by factorisation
- For example
  - part (a) Factorise 6x² + 7x – 3
  - part (b) Solve 6x² + 7x – 3 = 0
Use factorisation when solving two-term quadratic equations
- For example, solve x² – 4x = 0
  - Take out a common factor of x to get x(x – 4) = 0
  - So x = 0 and x = 4
- For example, solve x² – 9 = 0
  - Use the difference of two squares to factorise it as (x + 3)(x – 3) = 0
  - So x = -3 and x = 3
  - (Or rearrange to x² = 9 and use ±√ to get x = ±3)
Factorising can often be the quickest way to solve a quadratic equation

When should I use the quadratic formula?

Use the quadratic formula when the question says to leave solutions correct to a given accuracy (2 decimal places, 3 significant figures etc)
- This is a hint that the equation will not factorise
Use the quadratic formula when it may be faster than factorising
- It's quicker to solve 36x² + 33x – 20 = 0 using the quadratic formula than by factorisation
Use the quadratic formula if in doubt, as it always works

When should I solve by completing the square?

Use 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
Use completing the square when making x the subject of harder formulae containing both 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: $x = - 3 \pm \sqrt{y + 9}$
Completing the square always works
- But it's not always quick or easy to do

Worked example

(a) Solve $x^{2} - 7 x + 2 = 0$ , giving your answers correct to 2 decimal places.

Answer:

“Correct to 2 decimal places” suggests using the quadratic formula
Substitute a = 1, b = -7 and c = 2 into the formula
Put 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 (2 d.p.)

(b) Solve $16 x^{2} - 82 x + 45 = 0$ .

Answer:

Method 1
If you cannot spot the factorisation, use the quadratic formula
Substitute a = 16, b = -82 and c = 45 into the formula
Put 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 = $\frac{5}{8}$

Answer:

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

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

Now solve it
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

$x = - 3 \pm 2$

Find each solution separately using + first, then - second

x = - 1 or x = - 5

Quadratic Equation Methods

Deciding the quadratic method

When should I solve by factorisation?

When should I use the quadratic formula?

When should I solve by completing the square?

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