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How do you simplify surds, calculate with them, and rationalise a denominator?

Simplify surds, carry out the four operations with surds, expand brackets containing surds, and rationalise the denominator of a fraction (Higher tier).

A CCEA GCSE Mathematics Higher answer on surds, covering simplifying surds, the four operations, expanding brackets containing surds, and rationalising the denominator, including with conjugates.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. What a surd is
  3. Simplifying surds
  4. Adding, subtracting and multiplying
  5. Rationalising the denominator
  6. Why surds matter

What this dot point is asking

Surds are a Higher-tier CCEA Number topic that keeps answers exact rather than rounded. A surd is a root that cannot be written exactly as a fraction. You must simplify surds, add, subtract, multiply and divide them, expand brackets containing them, and rationalise denominators, including those of the form a+ba + \sqrt{b}. Surds appear on the non-calculator work and feed directly into Pythagoras, exact-value trigonometry and the quadratic formula, so they are a high-value Higher topic.

What a surd is

An irrational number cannot be written as an exact fraction; its decimal neither terminates nor recurs. The square root of any whole number that is not a perfect square is irrational, so 2,3\sqrt{2}, \sqrt{3} and 5\sqrt{5} are surds, but 9=3\sqrt{9} = 3 is not. Leaving an answer as a surd keeps it exact, which is why CCEA asks for answers "in surd form" or "in the form aba\sqrt{b}".

Simplifying surds

The key move is to split out a square factor.

So 72=36×2=62\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}. Choosing the largest square factor finishes in one step; using a smaller one, such as 72=4×18=218\sqrt{72} = \sqrt{4 \times 18} = 2\sqrt{18}, means you must simplify again.

Adding, subtracting and multiplying

Surds behave like algebraic terms: only like surds combine.

For addition and subtraction, simplify first so that matching surds appear, then add the coefficients: 12+27=23+33=53\sqrt{12} + \sqrt{27} = 2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}. For multiplication, multiply coefficients together and surds together: 25×32=6102\sqrt{5} \times 3\sqrt{2} = 6\sqrt{10}. When you multiply a surd by itself the root disappears: 7×7=7\sqrt{7} \times \sqrt{7} = 7. Expanding brackets follows the usual rules, for example (2+3)(1+3)=2+23+3+3=5+33(2 + \sqrt{3})(1 + \sqrt{3}) = 2 + 2\sqrt{3} + \sqrt{3} + 3 = 5 + 3\sqrt{3}.

Rationalising the denominator

Convention says a final answer should not have a surd in the denominator.

Why surds matter

Surds are how CCEA asks for exact answers in Pythagoras, in trigonometry with the special angles such as sin60=32\sin 60^\circ = \tfrac{\sqrt{3}}{2}, and in the quadratic formula when the discriminant is not a perfect square. Keeping a value as 525\sqrt{2} rather than 7.077.07\ldots avoids rounding error that would otherwise build up through a multi-step problem, and the mark schemes specifically reward exact surd answers where they are requested.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20193 marksSimplify 7512\sqrt{75} - \sqrt{12}, giving your answer in the form aba\sqrt{b}. (Higher, non-calculator.)
Show worked answer →

Simplify each surd by taking out the largest square factor.

75=25×3=53\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3} and 12=4×3=23\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}.

Both now have 3\sqrt{3}, so subtract the coefficients: 5323=335\sqrt{3} - 2\sqrt{3} = 3\sqrt{3}.

A mark is given for each correct simplification and a mark for 333\sqrt{3}. Writing 7512=63\sqrt{75} - \sqrt{12} = \sqrt{63} is wrong, because surds only combine when the number under the root is the same.

CCEA 20213 marksRationalise the denominator of 62\dfrac{6}{\sqrt{2}} and simplify fully. (Higher, non-calculator.)
Show worked answer →

Multiply the top and bottom by 2\sqrt{2} to clear the surd from the denominator.

62×22=622\dfrac{6}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{6\sqrt{2}}{2}.

Simplify the fraction: 62=3\dfrac{6}{2} = 3, so the answer is 323\sqrt{2}.

Marks are for multiplying by 22\tfrac{\sqrt{2}}{\sqrt{2}}, for 622\tfrac{6\sqrt{2}}{2}, and for the simplified 323\sqrt{2}. Stopping at 622\tfrac{6\sqrt{2}}{2} without cancelling loses the final mark.

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