Rationalising the denominator: surds made simple

Look for square factors: $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$.

• $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$
• $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$
• $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$

• $\dfrac{6}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}$
• $\dfrac{5}{2\sqrt{5}} = \dfrac{5\sqrt{5}}{2 \times 5} = \dfrac{\sqrt{5}}{2}$ (multiply by $\sqrt{5}$ only, not $2\sqrt{5}$)
• $\dfrac{10}{\sqrt{5}} = \dfrac{10\sqrt{5}}{5} = 2\sqrt{5}$

Multiply top and bottom by the same expression with the sign flipped (the conjugate). The bottom becomes a difference of two squares, and the surd disappears:

This uses $(a + b)(a - b) = a^2 - b^2$ from expanding brackets.

Before calculators, dividing by $\sqrt{2} \approx 1.414$ by hand was hard, but $\tfrac{\sqrt{2}}{2}$ is just half of 1.414. Today it's mainly about writing answers in a standard, simplified form so they're easy to compare and combine. Exact answers to quadratics often come out as surds; our quadratic equation solver shows them in surd form.

• Multiplying only the bottom by the surd.
• Multiplying by the whole denominator, like $2\sqrt{5}$, when $\sqrt{5}$ is enough.
• Using the same sign instead of the conjugate for $a + \sqrt{b}$.
• Not simplifying at the end: $\tfrac{6\sqrt{3}}{3}$ should become $2\sqrt{3}$.

1. Rationalise $\dfrac{3}{\sqrt{6}}$.
2. Simplify $\sqrt{48}$.
3. Rationalise $\dfrac{2}{\sqrt{3} + 1}$.
4. Rationalise $\dfrac{\sqrt{2}}{\sqrt{8}}$.

Answers: $\tfrac{\sqrt{6}}{2}$, $4\sqrt{3}$, $\sqrt{3} - 1$, $\tfrac{1}{2}$ (simplify first: $\sqrt{8} = 2\sqrt{2}$).

Show that $\dfrac{12}{\sqrt{3}}$ can be written as $4\sqrt{3}$.

$\dfrac{12}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{12\sqrt{3}}{3} = 4\sqrt{3}$. In a "show that" question, write every step.

Write $\dfrac{\sqrt{2} + 1}{\sqrt{2} - 1}$ in the form $a + b\sqrt{2}$.

Multiply top and bottom by $\sqrt{2} + 1$:

So $a = 3$ and $b = 2$.