IGCSE · Practice questions · Surds: simplifying and rationalising

Exact answers, simplest form.

Extended. Six original Cambridge-style questions on surds: simplifying, collecting, expanding and rationalising the denominator.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Largest square out, surds off the bottom.

Simplify: split off the largest square factor: \(\sqrt{98} = 7\sqrt{2}\).
Collect only like surds, after simplifying each one.
Expand every term; \(\sqrt{a} \times \sqrt{a} = a\).
Rationalise \(\dfrac{k}{\sqrt{a}}\) by multiplying by \(\dfrac{\sqrt{a}}{\sqrt{a}}\).
Conjugate: for \(p + \sqrt{q}\) on the bottom, multiply by \(p - \sqrt{q}\).

Recall

[2 marks]

Simplify (a) \(\sqrt{45}\) (b) \(\sqrt{98}\).

(a) \(\sqrt{9}\sqrt{5} = 3\sqrt{5}\) ✓
(b) \(\sqrt{49}\sqrt{2} = 7\sqrt{2}\) ✓

Calculation

[2 marks]

Simplify \(\sqrt{12} + \sqrt{27} - \sqrt{48}\).

\(2\sqrt{3} + 3\sqrt{3} - 4\sqrt{3}\) ✓
\(= \sqrt{3}\) ✓

Calculation

[2 marks]

Expand and simplify \((5 - \sqrt{2})(3 + \sqrt{2})\).

\(15 + 5\sqrt{2} - 3\sqrt{2} - 2\) ✓
\(= 13 + 2\sqrt{2}\) ✓

Calculation

[2 marks]

Rationalise the denominator and simplify (a) \(\dfrac{15}{\sqrt{5}}\) (b) \(\dfrac{2}{\sqrt{8}}\).

(a) \(\dfrac{15\sqrt{5}}{5} = 3\sqrt{5}\) ✓
(b) \(\sqrt{8} = 2\sqrt{2}\), so \(\dfrac{2}{2\sqrt{2}} = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}\) ✓

Analysis

[3 marks]

Rationalise the denominator of \(\dfrac{6}{\sqrt{5} - \sqrt{2}}\). Give your answer in its simplest form.

Multiply top and bottom by \(\sqrt{5} + \sqrt{2}\). ✓
Bottom: \((\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2}) = 5 - 2 = 3\). ✓
\(\dfrac{6(\sqrt{5} + \sqrt{2})}{3} = 2\sqrt{5} + 2\sqrt{2}\) ✓

Analysis

[4 marks]

A right-angled triangle has shorter sides \(\sqrt{12}\) cm and \(\sqrt{24}\) cm.

(a) Find the length of the hypotenuse.
(b) Find the area in the form \(a\sqrt{2}\) cm².
(c) Find the perimeter, with each surd in its simplest form.

(a) \(h^{2} = 12 + 24 = 36\), so \(h = 6\) cm. ✓
(b) Area \(= \dfrac{1}{2}\sqrt{12}\sqrt{24} = \dfrac{1}{2}\sqrt{288} = \dfrac{1}{2} \times 12\sqrt{2} = 6\sqrt{2}\) cm². ✓
(c) \(\sqrt{12} = 2\sqrt{3}\) and \(\sqrt{24} = 2\sqrt{6}\). ✓
Perimeter \(= 6 + 2\sqrt{3} + 2\sqrt{6}\) cm. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.