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Examples, videos, and solutions to help GCSE Maths students learn about surds and rationalising denominators by working through some examination questions.
Rationalizing the Denominator
It’s common practice in mathematics to express fractions without surds in the denominator. The goal is to rewrite the expression so that the denominator contains only rational numbers. This process is called rationalizing the denominator.
The following diagram shows examples of how to rationalize the denominator. Scroll down the page for more examples and solutions on how to rationalize the denominator.
Algebra Worksheets
Practice your skills with the following Algebra worksheets:
Printable & Online Algebra Worksheets
Case 1: Single Surd in the Denominator
Multiply both the numerator and the denominator by the surd in the denominator.
\(\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{a}}{\sqrt{b}}\times \frac{\sqrt{b}}{\sqrt{b}}=\frac{\sqrt{ab}}{b}\)
Example: Rationalize \(\frac{2}{\sqrt{3}}\)
\(\frac{2}{\sqrt{3}}=\frac{2}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\)
Example: Rationalize \(\frac{5}{3\sqrt{7}}\)
\(\frac{5}{3\sqrt{7}}=\frac{5}{3\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}}=\frac{5\sqrt{7}}{3\times 7}=\frac{5\sqrt{7}}{21}\)
Case 2: Binomial Surd in the Denominator (Using the Conjugate)
If the denominator is in the form \((a+ \sqrt{b})\) or \((a- \sqrt{b})\), multiply both the numerator and the denominator by its conjugate.
The conjugate is formed by changing the sign between the two terms.
The conjugate of \((a+ \sqrt{b})\) is \((a- \sqrt{b})\)).
The conjugate of \((a- \sqrt{b})\) is \((a + \sqrt{b})\)).
The conjugate of \((\sqrt{a} + \sqrt{b})\) is \((\sqrt{a} - \sqrt{b})\)).
The conjugate of \((\sqrt{a} - \sqrt{b})\) is \((\sqrt{a} + \sqrt{b})\)).
This uses the difference of squares identity:
(x + y)(x - y) = x2 - y2
\((a+ \sqrt{b})\)\((a- \sqrt{b})= a^2 - (\sqrt{b})^2 = a^2 - b\)
Example: Rationalize \(\frac{1}{2+\sqrt{3}}\)
The conjugate of \((2 + \sqrt{3})\) is \((2 - \sqrt{3})\)
\(\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{2^{2}-(\sqrt{3})^2}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}\)
Videos
Rationalising Denominators - Part 1
A single surd in the denominator
Rationalising Denominators - Part 2
More complex surds in the denominator
Rationalise the Denominator: Practice Questions
Rationalising (or rationalizing) the denominator
How to do Surds GCSE A/A* Maths revision Higher level exam questions (simplifying, rationalising)?
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