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Surds

HigherHigher tier onlyAQAEdexcelOCR

Practise surds with this free Higher GCSE Maths worksheet from Virtus Academy. You'll work through simplifying and rationalising surds, building confidence for your exam, and every question comes with worked solutions in the mark scheme. Suitable for AQA, Edexcel and OCR. Simplify a surd like √12 to 2√3 by taking out the largest square factor.

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These worksheets and mark schemes are original, written for Virtus Academy and checked against the current AQA, Edexcel and OCR specifications. Every worksheet comes with a full mark scheme.

This is a Higher tier only topic, so there's no Foundation paper — only the Higher worksheet and mark scheme below.

Topic overview

A surd is a root that cannot be written exactly as a whole number or fraction, such as \(\sqrt{2}\) or \(\sqrt{15}\). Leaving an answer as a surd keeps it exact, whereas a rounded decimal loses accuracy.

Simplifying a surd means extracting any square factor. Since \(\sqrt{12} = \sqrt{4 \times 3}\) and \(\sqrt{4} = 2\), it simplifies to \(2\sqrt{3}\). Looking for the largest square factor does the job in one step.

Surds add and subtract only when the root part matches, exactly like collecting like terms. So \(3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5}\), but \(\sqrt{2} + \sqrt{3}\) cannot be combined at all.

Revision notes

Simplifying surds

Find the largest square number that divides into the value, then take its root outside.

For \(\sqrt{50}\): since \(50 = 25 \times 2\), this becomes \(5\sqrt{2}\). Using a smaller square factor still works but needs a second simplification.

Adding and multiplying

Add only when the surd parts match. Multiplication is easier: multiply the numbers outside and the values inside separately.

So \(2\sqrt{3} \times 4\sqrt{5} = 8\sqrt{15}\), and \(\sqrt{3} \times \sqrt{3} = 3\).

Rationalising the denominator

A surd should not be left on the bottom of a fraction. Multiply top and bottom by that surd to clear it.

For \(\frac{6}{\sqrt{2}}\): multiply both by \(\sqrt{2}\) to get \(\frac{6\sqrt{2}}{2} = 3\sqrt{2}\).

Key points

  • A surd is an exact root that cannot be simplified to a rational number.
  • Extract the largest square factor to simplify.
  • Add or subtract only matching surds.
  • Multiply the outside numbers and the inside values separately.
  • \(\sqrt{a} \times \sqrt{a} = a\).
  • Rationalise to remove a surd from the denominator.

Worked examples

Example 1

Simplify \(\sqrt{72}\).

Working

\[72 = 36 \times 2\]find the largest square factor
\[\sqrt{36} = 6\]take the root of the square factor
\[6\sqrt{2}\]write the simplified surd

Example 2

Simplify \(3\sqrt{7} + 5\sqrt{7}\).

Working

\[\text{The surd parts match}\]these are like terms
\[8\sqrt{7}\]add the numbers outside the root

Example 3

Rationalise \(\dfrac{10}{\sqrt{5}}\).

Working

\[\frac{10}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}}\]multiply top and bottom by the surd
\[\frac{10\sqrt{5}}{5}\]the denominator becomes a whole number
\[2\sqrt{5}\]simplify the fraction

Common mistakes

  • Adding surds with different roots.

    √2 + √3 cannot be combined. Only matching surd parts add.

  • Thinking \(\sqrt{a+b} = \sqrt{a} + \sqrt{b}\).

    √(9+16) is 5, but √9 + √16 is 7. The roots do not distribute over addition.

  • Not using the largest square factor.

    √50 as √(2×25) gives 5√2 in one step; using smaller factors needs more work.

  • Leaving a surd in the denominator.

    Answers should be rationalised unless the question says otherwise.

Exam tips

  • Learn the square numbers so you spot factors quickly.
  • Check whether the surd inside can be simplified further.
  • Treat matching surds like algebraic like terms.
  • Always rationalise the denominator in a final answer.

Key terms

Surd
An exact root that is not a whole number or fraction.
Rationalise
To remove a surd from a denominator.
Square factor
A factor that is a perfect square.
Exact value
An answer left in surd form rather than rounded.

Written and reviewed against the current AQA, Edexcel and OCR specifications. Spotted an error? Let us know.