Rationalising the denominator

Remove surds from the denominator of a fraction, using the surd itself or its conjugate, and write answers in simplest exact form.

Why rationalise?

To the denominator means to rewrite a fraction with no surd on the bottom. It gives a standard form for exact answers, makes fractions easier to compare and add, and is often asked for directly.

Denominator $\sqrt{b}$

Example 1●

Rationalise $\dfrac{6}{\sqrt{3}}$.

$= \dfrac{6}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3}$

√3 × √3 = 3, which is rational.

$= 2\sqrt{3}$

Example 2●●

Rationalise $\dfrac{5}{2\sqrt{10}}$.

$= \dfrac{5\sqrt{10}}{2 \times 10} = \dfrac{5\sqrt{10}}{20}$

Only the surd part needs removing, so multiply by √10 over √10.

$= \dfrac{\sqrt{10}}{4}$

Denominator $a + \sqrt{b}$: use the conjugate

Multiplying by $\sqrt{b}$ won't clear $a + \sqrt{b}$. Instead multiply by the conjugate $a - \sqrt{b}$, because $(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b$ is rational.

Example 3●●

Rationalise $\dfrac{4}{3 + \sqrt{5}}$.

$= \dfrac{4}{3 + \sqrt{5}} \times \dfrac{3 - \sqrt{5}}{3 - \sqrt{5}}$

The conjugate of 3 + √5 is 3 − √5.

$= \dfrac{4(3 - \sqrt{5})}{4}$

Bottom: 3² − (√5)² = 9 − 5 = 4.

$= 3 - \sqrt{5}$

Example 4●●●

Write $\dfrac{\sqrt{3}}{\sqrt{3} - 1} + \dfrac{1}{\sqrt{3}}$ with a rational denominator.

$\dfrac{\sqrt{3}}{\sqrt{3} - 1} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} = \dfrac{3 + \sqrt{3}}{2}$

$\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$

$\dfrac{3(3 + \sqrt{3}) + 2\sqrt{3}}{6} = \dfrac{9 + 5\sqrt{3}}{6}$

Common denominator 6.

Flashcards

Rationalise k/√b

Multiply top and bottom by $\sqrt{b}$.

Rationalise k/(a + √b)

Multiply top and bottom by the conjugate $a - \sqrt{b}$.

Why it works

You multiply by $1$, so the value doesn't change.

Practice

Rationalise $\dfrac{1}{\sqrt{2}}$.
To rationalise $\dfrac{2}{\sqrt{7} - 2}$, multiply top and bottom by:
Rationalise $\dfrac{2}{\sqrt{6} - 2}$. (Type something like 3+sqrt(2).)
Rationalise $\dfrac{3}{\sqrt{6}}$.
Rationalise $\dfrac{10}{\sqrt{5}}$. (Type something like 3sqrt(2).)
$\dfrac{1}{\sqrt{3} + 1} =$
To rationalise $\dfrac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}$, multiply top and bottom by:
Rationalise $\dfrac{6}{3 - \sqrt{3}}$. (Type something like 2+sqrt(5).)