# Rationalising the denominator

> Edexcel International GCSE Further Pure Mathematics · 4PM1 (2016 spec)
> Source: https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s1-rationalising-the-denominator/

This guide covers the two core rationalisation techniques required for Edexcel IGCSE Further Pure Math (4PM1) S1, including single surd denominators and conjugate pairs for binomial surd denominators, aligned strictly to the 2016 specification.

**Prerequisites:** [Basic surd simplification](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s1-basic-surd-simplification/); [Index and log laws](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s1-index-log-laws/)

## Learning objectives

- Rationalise denominators of the form $1/\sqrt{a}$ by multiplying by $\sqrt{a}/\sqrt{a}$
- Rationalise denominators of the form $1/(a \pm \sqrt{b})$ using the conjugate pair
- Leave final answers in fully simplified, exact rationalised surd form as required for 4PM1 exams

## Rationalising Single Surd Denominators

When a denominator contains only a single surd of the form $\sqrt{a}$, you can rationalise it by multiplying both the numerator and denominator by $\sqrt{a}$. This works because $\sqrt{a} \times \sqrt{a} = a$, a rational number, and you are effectively multiplying by 1 so the value of the expression does not change.

**Single surd denominator** — A denominator consisting only of a radical (square root) term with no additional constants or added/subtracted terms

**Worked example:** Rationalise $\frac{10}{\sqrt{2}}$, leaving your answer in simplest form.

1. Multiply numerator and denominator by $\sqrt{2}$ to eliminate the surd in the denominator

   $$\frac{10}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$$
2. Simplify the denominator using $\sqrt{2} \times \sqrt{2} = 2$

   $$\frac{10\sqrt{2}}{2}$$
3. Divide numerator and denominator by their highest common factor of 2 to simplify fully

   $$5\sqrt{2}$$

> **Exam tip:** Always simplify any resulting fraction after rationalising to avoid losing marks for incomplete working.

*Calculator:* allowed

## Rationalising Binomial Surd Denominators

When a denominator is a binomial containing a surd of the form $a \pm \sqrt{b}$, you use the conjugate of the denominator to rationalise. Multiplying a binomial by its conjugate eliminates the surd term, as the cross terms cancel out when you expand the product using the difference of squares identity.

**Conjugate of a binomial surd** — For a binomial of the form $a + \sqrt{b}$, the conjugate is $a - \sqrt{b}$; for $a - \sqrt{b}$, the conjugate is $a + \sqrt{b}$. The product of a binomial and its conjugate is always rational.

**Worked example:** Rationalise $\frac{1}{2 - \sqrt{3}}$, giving your answer in exact form.

1. Identify the conjugate of the denominator $2 - \sqrt{3}$: this is $2 + \sqrt{3}$
2. Multiply numerator and denominator by the conjugate

   $$\frac{1}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}}$$
3. Expand the denominator using the difference of squares identity $(x - y)(x + y) = x^2 - y^2$

   $$\frac{2 + \sqrt{3}}{2^2 - (\sqrt{3})^2} = \frac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3}$$

> **Exam tip:** Use the difference of squares identity instead of FOIL to expand the denominator, to avoid common sign errors.

*Calculator:* allowed

## Combined Rationalisation Problems

Exam questions often require you to simplify surds first before rationalising, or combine rationalised terms with other expressions to get the final answer in the required form. Always check that your final answer has no common factors between numerator and denominator, and all surds are fully simplified.

**Worked example:** Simplify fully: $\frac{12}{\sqrt{8}} + \frac{1}{3 + \sqrt{2}}$

1. Simplify $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$, then simplify the first term

   $$\frac{12}{2\sqrt{2}} = \frac{6}{\sqrt{2}}$$
2. Rationalise the first term and simplify

   $$\frac{6}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$$
3. Rationalise the second term using its conjugate $3 - \sqrt{2}$

   $$\frac{1}{3 + \sqrt{2}} \times \frac{3 - \sqrt{2}}{3 - \sqrt{2}} = \frac{3 - \sqrt{2}}{9 - 2} = \frac{3 - \sqrt{2}}{7}$$
4. Combine the terms over a common denominator and simplify

   $$3\sqrt{2} + \frac{3 - \sqrt{2}}{7} = \frac{21\sqrt{2} + 3 - \sqrt{2}}{7} = \frac{3 + 20\sqrt{2}}{7}$$

**Check your understanding**

1. What is the conjugate of $5 + \sqrt{7}$?

   - $5 - \sqrt{7}$
   - $-5 + \sqrt{7}$
   - $\sqrt{7} + 5$
   - $\sqrt{7} - 5$

   *Answer:* $5 - \sqrt{7}$

   *Why:* Correct: you only swap the sign of the surd term to find the conjugate.

2. Rationalise $\frac{4}{\sqrt{3}}$: what is the simplified answer?

   - $4\sqrt{3}$
   - $\frac{4\sqrt{3}}{3}$
   - $\frac{\sqrt{3}}{4}$
   - $12\sqrt{3}$

   *Answer:* $\frac{4\sqrt{3}}{3}$

   *Why:* Correct: multiply numerator and denominator by $\sqrt{3}$ to get $\frac{4\sqrt{3}}{3}$.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Only multiplying the denominator by $\sqrt{a}$ or the conjugate, not the numerator
  - Why it fails: This changes the value of the entire expression, leading to an incorrect answer
  - Correct: Always multiply both numerator and denominator by the same term (equivalent to multiplying by 1) to preserve the expression's value
- **Wrong:** Changing the sign of the constant term when finding the conjugate, e.g., using $-2 + \sqrt{3}$ as the conjugate of $2 - \sqrt{3}$
  - Why it fails: The product will not be rational if you change the constant sign, so the surd in the denominator will not be eliminated
  - Correct: Only change the sign of the surd term when finding the conjugate: the conjugate of $2 - \sqrt{3}$ is $2 + \sqrt{3}$
- **Wrong:** Failing to simplify the final fraction after rationalising, e.g., leaving $\frac{10\sqrt{2}}{2}$ as the final answer
  - Why it fails: Examiners require answers to be in simplest form, so you will lose marks for incomplete simplification
  - Correct: Always divide numerator and denominator by their highest common factor after rationalising to get the simplest form
- **Wrong:** Expanding the denominator of a binomial surd using FOIL instead of the difference of squares formula
  - Why it fails: Manual expansion often leads to mistakes with negative signs when combining cross terms
  - Correct: Use the difference of squares identity $(x - y)(x + y) = x^2 - y^2$ to expand the denominator quickly and accurately

## Cheatsheet

| Denominator Type | Rationalisation Method | Example |
| --- | --- | --- |
| Single surd ($\frac{1}{\sqrt{a}}$) | Multiply numerator & denominator by $\sqrt{a}$ | $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ |
| Binomial ($\frac{1}{a + \sqrt{b}}$) | Multiply numerator & denominator by $a - \sqrt{b}$ | $\frac{1}{3 + \sqrt{5}} = \frac{3 - \sqrt{5}}{4}$ |
| Binomial ($\frac{1}{a - \sqrt{b}}$) | Multiply numerator & denominator by $a + \sqrt{b}$ | $\frac{1}{2 - \sqrt{3}} = 2 + \sqrt{3}$ |

## What's next

Now that you have mastered rationalising denominators, you can apply this skill to a wide range of topics in Edexcel IGCSE Further Pure Math, including coordinate geometry problems involving surd distances, trigonometry exact value calculations, and solving quadratic equations with surd roots. This skill is also a prerequisite for more advanced surd manipulation you will encounter if you progress to A-level Mathematics. Make sure you practice past paper questions to familiarise yourself with how rationalisation is tested in combination with other S1 topics.

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