Rationalising Denominators

What does it mean to rationalise a denominator?

If a fraction has a surd in the denominator, it is often useful to rationalise it
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 if the denominator is a surd?

STEP 1
Multiply the top and bottom by the surd in the denominator:

$\frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}}$

- We are multiplying by 1, so the overall value does not change
STEP 2
Multiply the numerators and denominators

$\sqrt{b} \times \sqrt{b} = b$ so the denominator is no longer a surd

STEP 3
Simplify your answer if needed

How do I rationalise the denominator if the denominator is a linear expression containing a surd?

For example $\frac{2}{1 + \sqrt{3}}$

STEP 1
Multiply the top and bottom by the expression in the denominator, but with the sign in the middle changed

$\frac{2}{1 + \sqrt{3}} \times \frac{1 - \sqrt{3}}{1 - \sqrt{3}}$

- We are multiplying by 1, so the overall value does not change
STEP 2
Multiply out the expressions in the numerators and denominators

- $(a + \sqrt{b}) (a - \sqrt{b}) = a^{2} - a \sqrt{b} + a \sqrt{b} - b = a^{2} - b$ so the denominator no longer contains a surd
- Note that this is an example of 'difference of two squares'
STEP 3
Simplify your answer if needed

$\frac{2 (1 - \sqrt{3})}{(1 + \sqrt{3}) (1 - \sqrt{3})} = \frac{2 (1 - \sqrt{3})}{1 - 3} = \frac{2 (1 - \sqrt{3})}{- 2} = - (1 - \sqrt{3}) = - 1 + \sqrt{3}$

Exam Tip

Remember that the aim is to remove the surd from the denominator
- If this doesn't happen, check your working or rethink the expression you used in your calculation
Your calculator can rationalise denominators
- You can use this to check your answer
- But on a 'show that' question you must show your working to get full marks

Worked example

Write $\frac{4}{\sqrt{6} - 2}$ in the form $p + q \sqrt{r}$ where $p, q$ and $r$ are integers and $r$ has no square factors.

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

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

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

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

When expanding the denominator, notice that it is a difference of two squares problem

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

Simplify by cancelling out the 2 in the denominator against the 4 in the numerator

$2 (\sqrt{6} + 2)$

Expand and write in the form given in the question

$2 \sqrt{6} + 4 = 4 + 2 \sqrt{6}$

This is now in the required form, with $p = 4$ , $q = 2$ and $r = 6$

$4 + 2 \sqrt{6}$

Rationalising Denominators (Edexcel IGCSE Further Maths)

Revision Note