OpenRevise

Exam questions · Maths · Further Algebra

Surds

  • 6 exam questions
  • 17 marks
  • 9 quick checks
  1. 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
  2. 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
  3. 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
  4. 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
  5. 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
  6. 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

  1. 1

    What is \(\sqrt{5} \times \sqrt{5}\)?

    1. A\(5\)
    2. B\(\sqrt{10}\)
    3. C\(25\)
    4. D\(2\sqrt{5}\)
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    A: \(5\)

    A root times itself gives the number: \(\sqrt{a} \times \sqrt{a} = a\).

  2. 2

    Simplify \(\sqrt{12}\).

    1. A\(3\sqrt{2}\)
    2. B\(6\)
    3. C\(4\sqrt{3}\)
    4. D\(2\sqrt{3}\)
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    D: \(2\sqrt{3}\)

    \(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\).

  3. 3

    Work out \(\sqrt{2} \times \sqrt{8}\).

    1. A\(\sqrt{10}\)
    2. B\(8\)
    3. C\(4\)
    4. D\(2\sqrt{2}\)
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    C: \(4\)

    \(\sqrt{2 \times 8} = \sqrt{16} = 4\).

  4. 4

    Work out \(3\sqrt{2} + 5\sqrt{2}\).

    1. A\(15\sqrt{2}\)
    2. B\(8\sqrt{2}\)
    3. C\(8\sqrt{4}\)
    4. D\(8\)
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    B: \(8\sqrt{2}\)

    Like surds add, as in \(3x + 5x = 8x\).

  5. 5

    Simplify \(\sqrt{50}\).

    1. A\(5\sqrt{2}\)
    2. B\(25\sqrt{2}\)
    3. C\(10\sqrt{5}\)
    4. D\(2\sqrt{5}\)
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    A: \(5\sqrt{2}\)

    \(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\).

  6. 6

    Rationalise the denominator of \(\dfrac{6}{\sqrt{3}}\).

    1. A\(6\sqrt{3}\)
    2. B\(2\)
    3. C\(\sqrt{3}\)
    4. D\(2\sqrt{3}\)
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    D: \(2\sqrt{3}\)

    \(\dfrac{6\sqrt{3}}{3} = 2\sqrt{3}\).

  7. 7

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

    1. A\(11\)
    2. B\(9\)
    3. C\(7\)
    4. D\(9 - 2\sqrt{2}\)
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    C: \(7\)

    This is a difference of two squares: \(9 - 2 = 7\).

  8. 8

    Write \(\sqrt{48} + \sqrt{3}\) in the form \(a\sqrt{3}\).

    1. A\(7\sqrt{3}\)
    2. B\(5\sqrt{3}\)
    3. C\(\sqrt{51}\)
    4. D\(4\sqrt{6}\)
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    B: \(5\sqrt{3}\)

    \(\sqrt{48} = 4\sqrt{3}\), and \(4\sqrt{3} + \sqrt{3} = 5\sqrt{3}\).

  9. 9

    Rationalise the denominator of \(\dfrac{1}{2 + \sqrt{3}}\).

    1. A\(2 - \sqrt{3}\)
    2. B\(2 + \sqrt{3}\)
    3. C\(\dfrac{1}{2}\)
    4. D\(\dfrac{2 - \sqrt{3}}{7}\)
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    A: \(2 - \sqrt{3}\)

    Multiply top and bottom by \(2 - \sqrt{3}\); the bottom becomes \(4 - 3 = 1\).