Rationalising the Denominator: Surds and Conjugates

Q: What is a conjugate?

For a + b it is a - b: the same terms with the middle sign flipped. Their product, a^2 - b, has no root.

Why roots in denominators are tidied away, how to rationalise a single surd, how conjugates use the difference of two squares to clear $a + \sqrt b$ , and how the same trick unlocks limits in calculus.

3 min read Supports 3 lessons

$\frac{1}{\sqrt2}$ is a perfectly correct number, but it is awkward. How big is it? Is it the same as $\frac{\sqrt2}{2}$ ? Rationalising rewrites a fraction so its denominator is a whole number, which makes it easier to compare, simplify and combine.

A single root

Multiply the top and bottom by the root itself. Since $\sqrt n \cdot \sqrt n = n$ , the root disappears from the bottom. Multiplying by $\frac{\sqrt n}{\sqrt n}$ is multiplying by $1$ , so the value does not change.

A single root

\frac{k}{\sqrt n} = \frac{k}{\sqrt n}\cdot\frac{\sqrt n}{\sqrt n} = \frac{k\sqrt n}{n}

$\frac{6}{\sqrt3}$
A root on the bottom.

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Conjugates

Multiplying $3 + \sqrt2$ by $\sqrt2$ does not help: you get $3\sqrt2 + 2$ , still with a root. Instead use its conjugate, $3 - \sqrt2$ : the same two terms with the sign between them flipped.

Conjugates

(a + \sqrt b)(a - \sqrt b) = a^2 - b

It is a difference of two squares, so the root cancels out.

$\frac{1}{3 + \sqrt2}$
The conjugate of the bottom is $3 - \sqrt2$ .

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Watch out

Multiply the top by the conjugate too. Changing only the bottom changes the value of the fraction.

Why calculus cares

The same trick works on numerators. $\frac{\sqrt{x + 4} - 2}{x}$ cannot be evaluated at $x = 0$ , where it gives $\frac00$ . Multiply top and bottom by the conjugate of the top:

\frac{\sqrt{x + 4} - 2}{x}\cdot\frac{\sqrt{x + 4} + 2}{\sqrt{x + 4} + 2} = \frac{x}{x(\sqrt{x + 4} + 2)} = \frac{1}{\sqrt{x + 4} + 2}

Key idea

Now $x = 0$ can be substituted, giving $\frac14$ . This is exactly how some limits are found in calculus.

Common questions

Why do we rationalise the denominator?: It gives a standard form that is easier to compare, simplify and add. $\frac{1}{\sqrt2}$ and $\frac{\sqrt2}{2}$ are equal, but the second is clearer.
What is a conjugate?: For $a + \sqrt b$ it is $a - \sqrt b$ : the same terms with the middle sign flipped. Their product, $a^2 - b$ , has no root.

Practise rationalising the denominator

The Rationalising & Conjugates lesson has worked examples and unlimited practice that adapts to you.

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Common questions

Why do we rationalise the denominator?

It gives a standard form that is easier to compare, simplify and add. $\frac{1}{\sqrt2}$ and $\frac{\sqrt2}{2}$ are equal, but the second is clearer.

What is a conjugate?

For $a + \sqrt b$ it is $a - \sqrt b$ : the same terms with the middle sign flipped. Their product, $a^2 - b$ , has no root.