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N8Surds

Edexcel GCSE Maths (1MA1), Higher tier · Number

Practise Surds. 3 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

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Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy3 marks
(a) Write \(\sqrt{18}\) in the form \(k\sqrt{2}\), where \(k\) is an integer.[1]
(b) Simplify \(\sqrt{32} - \sqrt{18}\)
Give your answer in the form \(a\sqrt{2}\).[2]
Show the answer and mark scheme
(a) Answer: \(3\sqrt{2}\)
  • B1 for \(3\sqrt{2}\) or \(k = 3\)

Worked solution: \(\sqrt{18} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}\)

(b) Answer: \(\sqrt{2}\)
  • M1 for simplifying one surd correctly, e.g. \(\sqrt{18} = 3\sqrt{2}\)
  • A1 for \(\sqrt{2}\) cao

Worked solution: \(\sqrt{32} = 4\sqrt{2}\) and \(\sqrt{18} = 3\sqrt{2}\)
\(4\sqrt{2} - 3\sqrt{2} = \sqrt{2}\)

Question 2Medium3 marks
Show that \((3 + \sqrt{5})^2 - (3 - \sqrt{5})^2 = 12\sqrt{5}\).[3]
Show the answer and mark scheme
Answer: \((14 + 6\sqrt{5}) - (14 - 6\sqrt{5}) = 12\sqrt{5}\)
  • M1 for expanding one bracket correctly, e.g. \((3 + \sqrt{5})^2 = 9 + 3\sqrt{5} + 3\sqrt{5} + 5 = 14 + 6\sqrt{5}\)
  • M1 for both expansions with the subtraction correctly handled: \(14 + 6\sqrt{5} - 14 + 6\sqrt{5}\)
  • C1 for \(12\sqrt{5}\) from fully correct working

Worked solution: \((3 + \sqrt{5})^2 = 9 + 6\sqrt{5} + 5 = 14 + 6\sqrt{5}\)
\((3 - \sqrt{5})^2 = 9 - 6\sqrt{5} + 5 = 14 - 6\sqrt{5}\)
\((14 + 6\sqrt{5}) - (14 - 6\sqrt{5}) = 12\sqrt{5}\)

Question 3Hard7 marks
A rectangle has length \((5 + \sqrt{3})\) cm.
The area of the rectangle is 22 cm2.
(a) Show that the width of the rectangle is \((5 - \sqrt{3})\) cm.[3]
(b) Work out the perimeter of the rectangle.[1]
(c) Work out the length of a diagonal of the rectangle.
Give your answer as a surd in its simplest form.[3]
Show the answer and mark scheme
(a) Answer: \(\frac{22}{5 + \sqrt{3}} \times \frac{5 - \sqrt{3}}{5 - \sqrt{3}} = \frac{22(5 - \sqrt{3})}{25 - 3} = 5 - \sqrt{3}\)
  • M1 for width \(= \frac{22}{5 + \sqrt{3}}\)
  • M1 for multiplying numerator and denominator by \(5 - \sqrt{3}\)
  • C1 for denominator \(25 - 3 = 22\), giving \(5 - \sqrt{3}\)

Worked solution: Width \(= \frac{22}{5 + \sqrt{3}} = \frac{22(5 - \sqrt{3})}{(5 + \sqrt{3})(5 - \sqrt{3})} = \frac{22(5 - \sqrt{3})}{25 - 3} = 5 - \sqrt{3}\) cm.

(b) Answer: 20 cm
  • B1 for 20

Worked solution: \(2(5 + \sqrt{3}) + 2(5 - \sqrt{3}) = 20\) cm.

(c) Answer: \(2\sqrt{14}\) cm
  • M1 for \((5 + \sqrt{3})^2 + (5 - \sqrt{3})^2\)
  • M1 for \(28 + 10\sqrt{3} + 28 - 10\sqrt{3} = 56\)
  • A1 for \(2\sqrt{14}\)

Worked solution: \((5 + \sqrt{3})^2 = 28 + 10\sqrt{3}\) and \((5 - \sqrt{3})^2 = 28 - 10\sqrt{3}\).
\(d^2 = 56\), so \(d = \sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14}\) cm.

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