Rationalising the Denominator
Rationalising the denominator rewrites a fraction so that no surd is left on the bottom. For \(\dfrac{9}{2\sqrt{3}}\) you multiply top and bottom by \(\sqrt{3}\); for a binomial denominator such as \(\sqrt{5}-\sqrt{2}\) you multiply by its conjugate \(\sqrt{5}+\sqrt{2}\). This is NSW Year 10 Advanced maths work (NESA outcome MA5‑IND‑P‑02).
You will rationalise single-surd and binomial denominators, add fractions whose denominators are conjugates, write answers in the form \(a+b\sqrt{n}\) and find unknown integers. Harder questions use the reciprocal of a surd expression, add a long chain of fractions that cancel in pairs, find an exact width from an area and a gradient that contains a surd, and solve linear equations with a surd coefficient. It is part of Year 10 Advanced Surds.
Every question with a fully worked solution.
- Rationalising the Denominator - Video - Rationalising the Denominator Watch
Theory
To rationalise the denominator, multiply the top and bottom of the fraction by a surd expression that makes the bottom rational: the surd itself, or the conjugate for a binomial. This Year 10 Advanced maths page covers single-surd and binomial denominators, negative denominators, binomials on top, sums of fractions and equations with surd coefficients.
A fraction has a rational denominator when there is no surd on the bottom. \(\dfrac{5}{\sqrt{3}}\) and \(\dfrac{5\sqrt{3}}{3}\) are equal, but the second form is the standard way to write the answer, and it is easier to add, compare and estimate.
To rationalise, multiply by a fraction equal to \(1\), such as \(\dfrac{\sqrt{3}}{\sqrt{3}}\). This changes the look of the fraction but not its value. It works because \(\sqrt{3}\times\sqrt{3} = 3\).
For a binomial denominator such as \(4+\sqrt{3}\), multiplying by \(\sqrt{3}\) does not remove the surd. Use the conjugate \(4-\sqrt{3}\): the same two terms with the middle sign changed. The product is a difference of two squares, \((4+\sqrt{3})(4-\sqrt{3}) = 16-3 = 13\), which is rational.
Whatever you multiply the bottom by, multiply the top by the same thing, then simplify the result. Rationalising is used throughout Advanced maths in Years 11 and 12.
| Denominator | Multiply top and bottom by | New denominator |
|---|---|---|
| \(k\sqrt{b}\) | \(\sqrt{b}\) | \(kb\) |
| \(p+\sqrt{q}\) | \(p-\sqrt{q}\) | \(p^2-q\) |
| \(\sqrt{p}-\sqrt{q}\) | \(\sqrt{p}+\sqrt{q}\) | \(p-q\) |
| \(m\sqrt{p}+n\sqrt{q}\) | \(m\sqrt{p}-n\sqrt{q}\) | \(m^2p-n^2q\) |
How to rationalise a denominator
- Simplify any surds first, and cancel common factors.
- Choose the multiplier: the surd for a single-term denominator, the conjugate for a binomial.
- Multiply the top and the bottom by it. Expand the top; the bottom is a difference of two squares.
- Simplify: cancel only a factor common to every term on top and the bottom. If the new denominator is negative, move the sign to the front.
Other question types
- Adding fractions: rationalise each fraction, then add. With conjugate denominators, the surds often cancel and the answer is rational.
- Form \(a+b\sqrt{n}\): rationalise, divide every term by the denominator, then read off \(a\) and \(b\).
- Pronumerals: the conjugate of \(\sqrt{x+4}-\sqrt{x}\) is \(\sqrt{x+4}+\sqrt{x}\), and their product is \((x+4)-x = 4\).
- Equations: collect the \(x\) terms, factorise out \(x\), divide, then rationalise (Example 4).
- Measurement and gradients: width \(=\) area \(\div\) length and gradient \(= \dfrac{\text{rise}}{\text{run}}\) give fractions with surds; rationalise before adding or comparing.
- Long sums: if each term rationalises to a difference such as \(\sqrt{n+1}-\sqrt{n}\), the middle terms cancel in pairs, leaving only the first and last.
(i) \(\dfrac{9}{2\sqrt{3}}\)
(ii) \(\dfrac{6}{\sqrt{8}}\)
(i) Multiply the top and bottom by \(\sqrt{3}\) only. The \(2\) is already rational.
| \(\dfrac{9}{2\sqrt{3}}\times\dfrac{\sqrt{3}}{\sqrt{3}}\) | \(=\) | \(\dfrac{9\sqrt{3}}{6}\) |
| \(\) | \(=\) | \(\mathbf{\dfrac{3\sqrt{3}}{2}}\) |
(ii) Simplify the surd first: \(\sqrt{8} = 2\sqrt{2}\).
| \(\dfrac{6}{2\sqrt{2}}\) | \(=\) | \(\dfrac{3}{\sqrt{2}}\times\dfrac{\sqrt{2}}{\sqrt{2}}\) |
| \(\) | \(=\) | \(\mathbf{\dfrac{3\sqrt{2}}{2}}\) |
Multiplying by \(\sqrt{8}\) straight away also works, but gives bigger numbers to cancel.
(i) \(\dfrac{6}{\sqrt{5}-\sqrt{2}}\)
(ii) \(\dfrac{2}{1-\sqrt{3}}\)
(i) Multiply the top and bottom by the conjugate \(\sqrt{5}+\sqrt{2}\). The bottom becomes a difference of two squares.
| \(\dfrac{6(\sqrt{5}+\sqrt{2})}{(\sqrt{5})^2-(\sqrt{2})^2}\) | \(=\) | \(\dfrac{6(\sqrt{5}+\sqrt{2})}{3}\) |
| \(\) | \(=\) | \(\mathbf{2\sqrt{5}+2\sqrt{2}}\) |
(ii) The conjugate is \(1+\sqrt{3}\). This time the new denominator is negative.
| \(\dfrac{2(1+\sqrt{3})}{1-3}\) | \(=\) | \(\dfrac{2(1+\sqrt{3})}{-2}\) |
| \(\) | \(=\) | \(\mathbf{-1-\sqrt{3}}\) |
Check the sign: \(1-\sqrt{3} \lt 0\), so the original fraction is negative, and so is \(-1-\sqrt{3}\).
Multiply the top and bottom by \(\sqrt{5}+\sqrt{2}\). The top becomes a perfect square.
| \(\text{top}\) | \(=\) | \((\sqrt{5}+\sqrt{2})^2 = 5+2\sqrt{10}+2 = 7+2\sqrt{10}\) |
| \(\text{bottom}\) | \(=\) | \(5-2 = 3\) |
| \(\dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\) | \(=\) | \(\mathbf{\dfrac{7+2\sqrt{10}}{3}}\) |
Nothing cancels: \(3\) does not divide both \(7\) and \(2\). Do not cancel the \(3\) with only one term.
Collect the \(x\) terms and take out \(x\) as a common factor.
| \(\sqrt{3}\,x-x\) | \(=\) | \(4\) |
| \(x(\sqrt{3}-1)\) | \(=\) | \(4\) |
| \(x\) | \(=\) | \(\dfrac{4}{\sqrt{3}-1}\) |
Rationalise with the conjugate \(\sqrt{3}+1\):
| \(x\) | \(=\) | \(\dfrac{4(\sqrt{3}+1)}{3-1}\) |
| \(\) | \(=\) | \(2(\sqrt{3}+1)\) |
| \(\) | \(=\) | \(\boldsymbol{2\sqrt{3}+2}\) |
Check: \(x\approx 5.464\); \(\sqrt{3}\,x+2\approx 11.464\) and \(x+6\approx 11.464\).
Common pitfalls
Frequently asked questions
What does rationalise the denominator mean?
It means rewriting a fraction so that there is no surd on the bottom, without changing its value. For example \(\dfrac{5}{\sqrt{3}}\) is rewritten as \(\dfrac{5\sqrt{3}}{3}\).
How do you rationalise a denominator with two terms?
Multiply the top and bottom by the conjugate of the denominator, which has the same terms with the middle sign changed. The denominator becomes a difference of two squares, which has no surd.
Why do we rationalise denominators?
A rational denominator makes fractions easier to add, compare and estimate, and it gives one standard form for exact answers, so two correct answers look the same.
What is the conjugate of root 5 minus root 2?
\(\sqrt{5}+\sqrt{2}\). Multiplying them gives \(5-2 = 3\).
Can the answer have a negative denominator?
The new denominator can come out negative, for example \(1-3 = -2\). Divide every term on top by it, which changes their signs, so the final answer has no negative denominator.