Solving Quadratic Equations (Edexcel IGCSE Further Pure Maths)

Revision Note

Reviewed by: Dan Finlay

Solving Quadratic Equations

You should be familiar with solving quadratic equations from your IGCSE Mathematics course.
This is a quick revision guide about the different methods and when to use them.

When should I solve by factorisation?

When the question asks to solve by factorisation
- For example, part (a) Factorise $6 x^{2} + 7 x - 3$ , part (b) Solve $6 x^{2} + 7 x - 3 = 0$
  - Factorises as $(3 x - 1) (2 x + 3) = 0$
  - Solutions are $x = \frac{1}{3}$ and $x = - \frac{3}{2}$
When solving two-term quadratic equations
- For example, solve $x^{2} - 4 x = 0$
  - Take out a common factor of $x$ to get $x (x - 4) = 0$
  - Solutions are $x = 0$ and $x = 4$
- For example, solve $x^{2} - 9 = 0$
  - Use difference of two squares to factorise it as $(x + 3) (x - 3) = 0$
  - Solutions are $x = - 3$ and $x = 3$
  - (Could also rearrange to $x^{2} = 9$ and use ±√ to get $x = \pm 3$ )

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 $36 x^{2} + 33 x - 20 = 0$ using the quadratic formula than by factorisation
If in doubt, use the quadratic formula - it always works
You must remember the formula however - it isn't on the exam formula sheet
If $a x^{2} + b x + c = 0$ , the solutions are
- $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

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 $x^{2}$ and $x$ terms
- For example, make $x$ the subject of the formula $x^{2} + 6 x = y$
  - 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: $x = - 3 \pm \sqrt{y + 9}$
Like the quadratic formula, completing the square will always work
- But it is not always quick or easy to use the method

Examiner Tips and Tricks

Some calculators can solve quadratic equations
- Even if you need to show working you can use a calculator to check your solutions
- If the calculator solutions are whole numbers or fractions (with no square roots), this means the quadratic can be factorised

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

$x = 6.7015 . . . or x = 0.2984 . . .$

Round your final answers to 2 decimal places

$x = 6.70 or x = 0.30$ (2 d.p.)

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

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

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

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

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

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

Write $x^{2} + 6 x$ as ${(x + p)}^{2} - p^{2}$

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

Replace $x^{2} + 6 x$ 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

$x = - 3 \pm 2$

Find each solution separately using + first, then - second

$x = - 5 or x = - 1$

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

Last updated: 19 October 2024

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