Exam questions · Maths · Further Algebra
Surds
- 6 exam questions
- 17 marks
- 9 quick checks
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1 Simplify [2 marks]
Simplify \(\sqrt{75}\).
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Model answer
\(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\).
Mark scheme
- \(\sqrt{25 \times 3}\) — M1
- \(5\sqrt{3}\) — A1
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2 Show that [3 marks]
Show that \(\sqrt{18} + \sqrt{8} = 5\sqrt{2}\).
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Model answer
\(\sqrt{18} = 3\sqrt{2}\) and \(\sqrt{8} = 2\sqrt{2}\), so the sum is \(3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2}\).
Mark scheme
- \(\sqrt{18} = 3\sqrt{2}\) or \(\sqrt{8} = 2\sqrt{2}\) — M1
- Both simplified correctly — M1
- \(3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2}\) with a conclusion — C1
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3 Expand [3 marks]
Expand and simplify \((2 + \sqrt{3})(4 - \sqrt{3})\).
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Model answer
\(8 - 2\sqrt{3} + 4\sqrt{3} - 3 = 5 + 2\sqrt{3}\).
Mark scheme
- At least three of the four terms correct — M1
- \(8 - 2\sqrt{3} + 4\sqrt{3} - 3\) — M1
- \(5 + 2\sqrt{3}\) — A1
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4 Rationalise [2 marks]
Rationalise the denominator of \(\dfrac{14}{\sqrt{7}}\). Give your answer in its simplest form.
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Model answer
\(\dfrac{14}{\sqrt{7}} \times \dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac{14\sqrt{7}}{7} = 2\sqrt{7}\).
Mark scheme
- Multiplies the top and bottom by \(\sqrt{7}\) — M1
- \(2\sqrt{7}\) — A1
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5 Rationalise [3 marks]
Write \(\dfrac{6}{3 + \sqrt{3}}\) in the form \(a - \sqrt{b}\), where \(a\) and \(b\) are integers.
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Model answer
Multiply the top and bottom by \(3 - \sqrt{3}\): \(\dfrac{6(3 - \sqrt{3})}{9 - 3} = \dfrac{6(3 - \sqrt{3})}{6} = 3 - \sqrt{3}\).
Mark scheme
- Multiplies by \(3 - \sqrt{3}\) — M1
- Denominator \(9 - 3 = 6\) — M1
- \(3 - \sqrt{3}\) — A1
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6 Work out [4 marks]
A right-angled triangle has shorter sides of length \(\sqrt{2}\) cm and \(\sqrt{18}\) cm. (a) Work out the area of the triangle. (2 marks) (b) Work out the length of the hypotenuse. Give your answer in the form \(a\sqrt{b}\). (2 marks)
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Model answer
(a) \(\dfrac{1}{2} \times \sqrt{2} \times \sqrt{18} = \dfrac{1}{2} \times \sqrt{36} = 3\) cm\(^2\). (b) \(\sqrt{2 + 18} = \sqrt{20} = 2\sqrt{5}\) cm.
Mark scheme
- (a) \(\dfrac{1}{2} \times \sqrt{2} \times \sqrt{18}\) — M1
- (a) 3 cm\(^2\) — A1
- (b) \(\sqrt{20}\) — M1
- (b) \(2\sqrt{5}\) — A1
Quick check
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1
What is \(\sqrt{5} \times \sqrt{5}\)?
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A: \(5\)
A root times itself gives the number: \(\sqrt{a} \times \sqrt{a} = a\).
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2
Simplify \(\sqrt{12}\).
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D: \(2\sqrt{3}\)
\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\).
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3
Work out \(\sqrt{2} \times \sqrt{8}\).
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C: \(4\)
\(\sqrt{2 \times 8} = \sqrt{16} = 4\).
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4
Work out \(3\sqrt{2} + 5\sqrt{2}\).
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B: \(8\sqrt{2}\)
Like surds add, as in \(3x + 5x = 8x\).
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5
Simplify \(\sqrt{50}\).
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A: \(5\sqrt{2}\)
\(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\).
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6
Rationalise the denominator of \(\dfrac{6}{\sqrt{3}}\).
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D: \(2\sqrt{3}\)
\(\dfrac{6\sqrt{3}}{3} = 2\sqrt{3}\).
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7
Expand and simplify \((3 + \sqrt{2})(3 - \sqrt{2})\).
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C: \(7\)
This is a difference of two squares: \(9 - 2 = 7\).
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8
Write \(\sqrt{48} + \sqrt{3}\) in the form \(a\sqrt{3}\).
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B: \(5\sqrt{3}\)
\(\sqrt{48} = 4\sqrt{3}\), and \(4\sqrt{3} + \sqrt{3} = 5\sqrt{3}\).
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9
Rationalise the denominator of \(\dfrac{1}{2 + \sqrt{3}}\).
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A: \(2 - \sqrt{3}\)
Multiply top and bottom by \(2 - \sqrt{3}\); the bottom becomes \(4 - 3 = 1\).