GCSEFurther MathsAQARevision Notes1. NumberSurdsRationalising Denominators

Rationalising Denominators (AQA GCSE Further Maths): Revision Note

Exam code: 8365

Written by: Amber

Reviewed by: Dan Finlay

Updated on 18 August 2025

Rationalising denominators

What does it mean to rationalise a denominator?

If a fraction has a surd on the denominator, it is not in its simplest form and must be rationalised
Rationalising a denominator changes a fraction with surds in the denominator into an equivalent fraction
- The denominator will be an integer and any surds are in the numerator

How do I rationalise the denominator of a surd?

The denominator is a surd

STEP 1
Multiply the top and bottom by the surd on the denominator:
- This ensures you are multiplying by 1; so not affecting the overall value
  - $\frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}}$
STEP 2
Multiply the numerator and denominators together
- $\sqrt{b} \times \sqrt{b} = b$ so the denominator is no longer a surd
STEP 3
Simplify your answer if needed

The denominator is an expression containing a surd

STEP 1
Multiply the top and bottom by the expression on the denominator, but with the sign changed
- This ensures you are multiplying by 1; so not affecting the overall value
  - $\frac{2}{1 + \sqrt{3}} \times \frac{1 - \sqrt{3}}{1 - \sqrt{3}}$
STEP 2
Multiply the expressions on the numerator and denominator together
- $(a + \sqrt{b}) (a - \sqrt{b}) = a^{2} + a \sqrt{b} - a \sqrt{b} - b = a^{2} - b$ so the denominator no longer contains a surd
  - $\frac{2 (1 - \sqrt{3})}{(1 + \sqrt{3}) (1 - \sqrt{3})} = \frac{2 (1 - \sqrt{3})}{1 - 3}$
STEP 3
Simplify your answer if needed
- $- \frac{2 (1 - \sqrt{3})}{2} = - (1 - \sqrt{3}) = \sqrt{3} - 1$

The denominator is an expression containing more than one surd

STEP 1
Multiply the top and bottom by the expression on the denominator, but with the sign changed (this is called the conjugate)
- This ensures you are multiplying by 1; so not affecting the overall value
  - $\frac{1}{a \sqrt{b} + c \sqrt{d}} \times \frac{a \sqrt{b} - c \sqrt{d}}{a \sqrt{b} - c \sqrt{d}}$
STEP 2
Multiply the expressions on the numerator and denominator together
- so the denominator no longer contains a surd
  - $\begin{array}{rcl} \frac{1 \times (a \sqrt{b} - c \sqrt{d})}{(a \sqrt{b} + c \sqrt{d}) (a \sqrt{b} - c \sqrt{d})} & = & \frac{a \sqrt{b} - c \sqrt{d}}{a^{2} {(\sqrt{b})}^{2} - a \sqrt{b} c \sqrt{d} + a \sqrt{b} c \sqrt{d} - c^{2} {(\sqrt{d})}^{2}} \end{array}$
STEP 3
Simplify your answer if needed
- $\frac{1}{a \sqrt{b} + c \sqrt{d}} \begin{array}{rcl} = \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \frac{a \sqrt{b} - c \sqrt{d}}{a^{2} b - c^{2} d} \end{array}$

Examiner Tips and Tricks

When you have an expression on the denominator you can use the FOIL technique from multiplying out double brackets
- Remember that the aim is to remove the surd from the denominator, so if this doesn't happen you need to check your working or rethink the expression you are using in your calculation

Worked Example

Rationalise and simplify.

$\frac{4 - \sqrt{6}}{\sqrt{6} - 2}$

Give your answer in the form $p + q \sqrt{r}$ where $p, q$ and $r$ are integers and $r$ has no square factors.

There is an expression on the denominator, so the fraction will need to be multiplied by a fraction with this expression on both the numerator and denominator, but with the sign changed.

$\frac{4 - \sqrt{6}}{\sqrt{6} - 2} \times \frac{\sqrt{6} + 2}{\sqrt{6} + 2}$

Multiply the fractions together by multiplying across the numerator and the denominator.

$\frac{(4 - \sqrt{6}) (\sqrt{6} + 2)}{(\sqrt{6} - 2) (\sqrt{6} + 2)}$

Expand the numerator and the denominator. You can expand the denominator quickly by using the difference of two squares

$\begin{array}{rcl} \frac{4 \sqrt{6} + 8 - {(\sqrt{6})}^{2} - 2 \sqrt{6}}{{(\sqrt{6})}^{2} - 4} \\ = & \frac{4 \sqrt{6} + 8 - 6 - 2 \sqrt{6}}{6 - 4} \end{array}$

Simplify the numerator and denominator.

$\begin{array}{rcl} \frac{2 + 2 \sqrt{6}}{2} \end{array}$

Divide both terms in the numerator by 2.

$\frac{4 - \sqrt{6}}{\sqrt{6} - 2} = 1 + 1 \sqrt{6}$

$1 + \sqrt{6} p = 1, q = 1, r = 6$

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