IGCSEMathsEdexcelMaths A (Modular)Higher Unit 1Revision NotesNumberSurdsRationalising Denominators

Rationalising Denominators (Edexcel IGCSE Maths A (Modular))

Revision Note

Amber

Written by: Amber

Reviewed by: Dan Finlay

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Rationalising Denominators

What does rationalising the denominator mean?

If a fraction has a denominator containing a surd then it has an irrational denominator
- E.g. $\frac{4}{\sqrt{5}}$ or $\sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$
The fraction can be rewritten as an equivalent fraction, but with a rational denominator
- E.g. $\frac{4 \sqrt{5}}{5}$ or $\frac{\sqrt{6}}{3}$
The numerator may contain a surd, but the denominator is rationalised

How do I rationalise simple denominators?

If the denominator is a surd:
- Multiply the top and bottom of the fraction by the surd on the denominator
  - $\frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}}$
  - This is equivalent to multiplying by 1, so does not change the value of the fraction
  - $\sqrt{b} \times \sqrt{b} = b$ so the denominator is no longer a surd
- Multiply the fractions as you would usually, and simplify if needed
  - $\frac{a \sqrt{b}}{b}$

How do I rationalise harder denominators?

If the denominator is an expression containing a surd:
- For example $\frac{2}{3 + \sqrt{5}}$
- Multiply the top and bottom of the fraction by the expression on the denominator, but with the sign changed
  - $\frac{2}{3 + \sqrt{5}} = \frac{2}{3 + \sqrt{5}} \times \frac{3 - \sqrt{5}}{3 - \sqrt{5}}$
  - This is equivalent to multiplying by 1, so does not change the value of the fraction
- Multiply the fractions as you would usually (use brackets to help)
  - $\frac{2}{3 + \sqrt{5}} = \frac{2 (3 - \sqrt{5})}{(3 + \sqrt{5}) (3 - \sqrt{5})}$
- Expand any brackets, and simplify
  - $\frac{2}{3 + \sqrt{5}} = \frac{6 - 2 \sqrt{5}}{(3 \times 3) + 3 \sqrt{5} - 3 \sqrt{5} - (\sqrt{5} \sqrt{5})} = \frac{6 - 2 \sqrt{5}}{9 - (5)}$
- You can use the difference of two squares to expand the denominator quickly
  - $(a + \sqrt{b}) (a - \sqrt{b}) = a^{2} - {(\sqrt{b})}^{2} = a^{2} - b$
  - This is what makes the denominator rational
- Simplify
  - $\frac{2}{3 + \sqrt{5}} = \frac{6 - 2 \sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}$

Examiner Tips and Tricks

If your answer still has a surd on the bottom, go back and check your working!

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.

Multiply the top and bottom of the fraction by the expression on the denominator, but with the sign changed

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

Multiply the fractions as you would usually

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

Expand the brackets
The denominator can be expanded using the difference of two squares

$\frac{4}{\sqrt{6} - 2} = \frac{4 \sqrt{6} + 8}{{(\sqrt{6})}^{2} - {(2)}^{2}} = \frac{4 \sqrt{6} + 8}{6 - 4}$

Simplify

$\frac{4}{\sqrt{6} - 2} = \frac{4 \sqrt{6} + 8}{2} = 2 \sqrt{6} + 4$

Write in the form given in the question

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

Last updated: 17 October 2024

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Amber

Author: Amber

Expertise: Maths

Amber gained a first class degree in Mathematics & Meteorology from the University of Reading before training to become a teacher. She is passionate about teaching, having spent 8 years teaching GCSE and A Level Mathematics both in the UK and internationally. Amber loves creating bright and informative resources to help students reach their potential.

Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.