OpenRevise

Exam questions · Maths · Further Algebra

Surds

  • 6 exam questions
  • 17 marks
  • 9 quick checks
  1. 1 Simplify [2 marks]

    Write \(\sqrt{45}\) in the form \(a\sqrt{5}\), where \(a\) is an integer.

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    Model answer

    \(\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}\).

    Mark scheme

    • \(\sqrt{9 \times 5}\) — M1
    • \(3\sqrt{5}\) — A1
  2. 2 Expand [3 marks]

    (a) Expand and simplify \((\sqrt{5} + 2)^2\). [2 marks] (b) Show that \((\sqrt{5} + 2)(\sqrt{5} - 2) = 1\). [1 mark]

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    Model answer

    (a) \(5 + 2\sqrt{5} + 2\sqrt{5} + 4 = 9 + 4\sqrt{5}\). (b) \(5 - 2\sqrt{5} + 2\sqrt{5} - 4 = 1\).

    Mark scheme

    • (a) Three of the four terms correct — M1
    • (a) \(9 + 4\sqrt{5}\) — A1
    • (b) \(5 - 4 = 1\) shown — Q1
  3. 3 Simplify [3 marks]

    Write \(\sqrt{200} - \sqrt{50}\) in the form \(a\sqrt{2}\).

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    Model answer

    \(\sqrt{200} = 10\sqrt{2}\) and \(\sqrt{50} = 5\sqrt{2}\), so the difference is \(5\sqrt{2}\).

    Mark scheme

    • \(10\sqrt{2}\) or \(5\sqrt{2}\) seen — M1
    • Both correct — M1
    • \(5\sqrt{2}\) — A1
  4. 4 Rationalise [2 marks]

    Rationalise the denominator of \(\dfrac{10}{\sqrt{5}}\).

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    Model answer

    \(\dfrac{10\sqrt{5}}{5} = 2\sqrt{5}\).

    Mark scheme

    • Multiplies the top and bottom by \(\sqrt{5}\) — M1
    • \(2\sqrt{5}\) — A1
  5. 5 Show that [3 marks]

    Show that \(\dfrac{4}{\sqrt{7} - \sqrt{3}} = \sqrt{7} + \sqrt{3}\).

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    Model answer

    Multiply the top and bottom by \(\sqrt{7} + \sqrt{3}\): the bottom becomes \(7 - 3 = 4\), and the fraction is \(\dfrac{4(\sqrt{7} + \sqrt{3})}{4} = \sqrt{7} + \sqrt{3}\).

    Mark scheme

    • Multiplies by \(\sqrt{7} + \sqrt{3}\) — M1
    • Denominator \(7 - 3 = 4\) — M1
    • Completes with a conclusion — Q1
  6. 6 Work out [4 marks]

    A rectangle has length \((3 + \sqrt{2})\) cm and width \((3 - \sqrt{2})\) cm. (a) Show that the area of the rectangle is 7 cm\(^2\). [2 marks] (b) Work out the perimeter of the rectangle. [2 marks]

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    Model answer

    (a) \((3 + \sqrt{2})(3 - \sqrt{2}) = 9 - 2 = 7\). (b) The perimeter is \(2 \times ((3 + \sqrt{2}) + (3 - \sqrt{2})) = 2 \times 6 = 12\) cm.

    Mark scheme

    • (a) \((3 + \sqrt{2})(3 - \sqrt{2})\) expanded — M1
    • (a) \(9 - 2 = 7\) — Q1
    • (b) \(2(3 + \sqrt{2} + 3 - \sqrt{2})\) — M1
    • (b) 12 cm — 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 \(4 - 3 = 1\).