Rationalising Denominators
Learn rationalising the denominator for Queensland Year 11 Mathematical Methods (QCAA). It means rewriting a fraction so no surd is left on the bottom, giving a tidy exact answer with a whole-number denominator.
You will learn to clear a single surd by multiplying top and bottom by that surd, handle a binomial denominator by multiplying by its conjugate, and simplify the result — a key exact-value skill for working with surds.
Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), rationalising the denominator means rewriting a fraction so there is no surd on the bottom. For a single surd, multiply top and bottom by that surd; for a binomial such as \(a+\sqrt{b}\), multiply by its conjugate \(a-\sqrt{b}\). The value is unchanged — only the form becomes an exact answer with a rational denominator.
To rationalise the denominator is to remove the surd from the bottom of a fraction while keeping the fraction equal in value. We do this by multiplying by a cleverly chosen form of \(1\).
For a single surd denominator, multiply the numerator and denominator by that surd, because \(\sqrt{b}\times\sqrt{b}=b\) is rational.
For a binomial surd denominator such as \(a+\sqrt{b}\), multiply by its conjugate \(a-\sqrt{b}\). Using the difference of two squares, \((a+\sqrt{b})(a-\sqrt{b})=a^2-b\), which is rational.
Single-surd denominator — multiply by \(\dfrac{\sqrt{b}}{\sqrt{b}}\):
Binomial denominator — multiply by the conjugate:
How to rationalise a denominator
- Choose the multiplier: a single surd \(\sqrt{b}\) needs \(\dfrac{\sqrt{b}}{\sqrt{b}}\); a binomial \(a\pm\sqrt{b}\) needs its conjugate \(a\mp\sqrt{b}\).
- Multiply: multiply numerator and denominator by that multiplier, expanding with \(\sqrt{b}\,\sqrt{b}=b\) or \((a+\sqrt{b})(a-\sqrt{b})=a^2-b\).
- Simplify: simplify any surd in the numerator and cancel common factors for the simplest exact form.
Multiply top and bottom by \(\sqrt{3}\):
| \(\dfrac{6}{\sqrt{3}}\) | \(=\) | \(\dfrac{6}{\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}\) |
| \(=\) | \(\dfrac{6\sqrt{3}}{3}\) |
Simplify the fraction \(\dfrac{6}{3}=2\):
| \(\dfrac{6\sqrt{3}}{3}\) | \(=\) | \(2\sqrt{3}\) |
\(\dfrac{6}{\sqrt{3}}=2\sqrt{3}\).
Multiply top and bottom by \(\sqrt{6}\):
| \(\dfrac{\sqrt{10}}{\sqrt{6}}\) | \(=\) | \(\dfrac{\sqrt{10}}{\sqrt{6}}\times\dfrac{\sqrt{6}}{\sqrt{6}}\) |
| \(=\) | \(\dfrac{\sqrt{60}}{6}\) |
Simplify \(\sqrt{60}=\sqrt{4\times 15}=2\sqrt{15}\):
| \(\dfrac{\sqrt{60}}{6}\) | \(=\) | \(\dfrac{2\sqrt{15}}{6}\) |
| \(=\) | \(\dfrac{\sqrt{15}}{3}\) |
\(\dfrac{\sqrt{10}}{\sqrt{6}}=\dfrac{\sqrt{15}}{3}\).
Multiply by the conjugate \(3-\sqrt{5}\):
| \(\dfrac{4}{3+\sqrt{5}}\) | \(=\) | \(\dfrac{4}{3+\sqrt{5}}\times\dfrac{3-\sqrt{5}}{3-\sqrt{5}}\) |
Denominator is a difference of squares:
| \((3+\sqrt{5})(3-\sqrt{5})\) | \(=\) | \(3^{2}-(\sqrt{5})^{2}\) |
| \(=\) | \(9-5=4\) |
So the fraction becomes:
| \(=\) | \(\dfrac{4(3-\sqrt{5})}{4}\) | |
| \(=\) | \(3-\sqrt{5}\) |
\(\dfrac{4}{3+\sqrt{5}}=3-\sqrt{5}\).
Multiply by the conjugate \(2+\sqrt{3}\):
| \(\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\) | \(=\) | \(\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\times\dfrac{2+\sqrt{3}}{2+\sqrt{3}}\) |
Denominator (difference of squares):
| \((2-\sqrt{3})(2+\sqrt{3})\) | \(=\) | \(2^{2}-(\sqrt{3})^{2}=4-3=1\) |
Numerator (expand the square):
| \((2+\sqrt{3})^{2}\) | \(=\) | \(4+2\cdot 2\sqrt{3}+3\) |
| \(=\) | \(7+4\sqrt{3}\) |
Divide by \(1\):
| \(=\) | \(\dfrac{7+4\sqrt{3}}{1}=7+4\sqrt{3}\) |
\(\dfrac{2+\sqrt{3}}{2-\sqrt{3}}=7+4\sqrt{3}\).
Common pitfalls
Frequently asked questions
What does rationalising the denominator mean?
It means rewriting a fraction so there is no surd in the denominator, without changing its value, by multiplying the top and bottom by a suitable surd or conjugate.
How do you rationalise \(\dfrac{a}{\sqrt{b}}\)?
Multiply numerator and denominator by \(\sqrt{b}\): \(\dfrac{a}{\sqrt{b}}\times\dfrac{\sqrt{b}}{\sqrt{b}}=\dfrac{a\sqrt{b}}{b}\), then simplify.
What is a conjugate and when do you use it?
The conjugate of \(a+\sqrt{b}\) is \(a-\sqrt{b}\). Use it when the denominator is a binomial surd, because \((a+\sqrt{b})(a-\sqrt{b})=a^2-b\) is rational.
Why does multiplying by the surd not change the value?
Because \(\dfrac{\sqrt{b}}{\sqrt{b}}=1\), and multiplying any number by \(1\) leaves it unchanged — only the form of the fraction changes.
Do you always have to simplify after rationalising?
Yes, give the simplest exact form: reduce any surd in the numerator and cancel common factors, e.g. \(\dfrac{6\sqrt{3}}{3}=2\sqrt{3}\).