Rationalising Denominators

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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?

To rationalise the denominator if the denominator is a surd
- STEP 1: Multiply the top and bottom by the surd on the denominator:
  - $\frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}}$
  - This ensures we are multiplying by 1; so not affecting the overall value
- 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
To rationalise the denominator if the denominator is an expression containing a surd:
For example $\frac{2}{1 + \sqrt{3}}$
- STEP 1: Multiply the top and bottom by the expression on the denominator, but with the sign changed
  $\frac{2}{1 + \sqrt{3}} \times \frac{1 - \sqrt{3}}{1 - \sqrt{3}}$
  - This ensures we are multiplying by 1; so not affecting the overall value
- 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
- STEP 3: Simplify your answer if needed
  $\frac{2 (1 - \sqrt{3})}{(1 + \sqrt{3}) (1 - \sqrt{3})} = \frac{2 (1 - \sqrt{3})}{1 - 3} = - \frac{2 (1 - \sqrt{3})}{2} = - (1 - \sqrt{3}) = \sqrt{3} - 1$

Examiner Tip

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

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 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.

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

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

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

By expanding the denominator, you will 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 4 on the numerator and the 2 on the denominator.

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

Expand and write in the form given in the question.

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

$4 + 2 \sqrt{6} p = 4 q = 2 r = 6$

Rationalising Denominators (Cambridge O Level Maths)

Revision Note

Rationalising Denominators

Rationalising Denominators

What does it mean to rationalise a denominator?

How do I rationalise the denominator of a surd?

Examiner Tip

Worked example

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Author: Amber