OpenRevise

Exam questions · Maths · Further Algebra

Surds

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

    Simplify \(\sqrt{98}\).

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    Model answer

    \(\sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}\).

    Mark scheme

    • \(\sqrt{49 \times 2}\) — M1
    • \(7\sqrt{2}\) — A1
  2. 2 Expand [3 marks]

    Expand and simplify \((1 + \sqrt{5})(3 - \sqrt{5})\).

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    Model answer

    \(3 - \sqrt{5} + 3\sqrt{5} - 5 = -2 + 2\sqrt{5}\).

    Mark scheme

    • At least three of the four terms correct — M1
    • \(3 - \sqrt{5} + 3\sqrt{5} - 5\) — M1
    • \(-2 + 2\sqrt{5}\) — A1
  3. 3 Write [3 marks]

    Write \(\sqrt{27} + \sqrt{12}\) in the form \(a\sqrt{3}\).

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    Model answer

    \(\sqrt{27} = 3\sqrt{3}\) and \(\sqrt{12} = 2\sqrt{3}\), so the sum is \(5\sqrt{3}\).

    Mark scheme

    • \(3\sqrt{3}\) or \(2\sqrt{3}\) seen — M1
    • Both correct — M1
    • \(5\sqrt{3}\) — A1
  4. 4 Rationalise [2 marks]

    Rationalise the denominator of \(\dfrac{15}{\sqrt{3}}\).

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    Model answer

    \(\dfrac{15\sqrt{3}}{3} = 5\sqrt{3}\).

    Mark scheme

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

    Show that \(\dfrac{1}{\sqrt{3} - 1} = \dfrac{\sqrt{3} + 1}{2}\).

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    Model answer

    Multiply the top and bottom by \(\sqrt{3} + 1\): the bottom is \(3 - 1 = 2\), so the fraction is \(\dfrac{\sqrt{3} + 1}{2}\).

    Mark scheme

    • Multiplies by \(\sqrt{3} + 1\) — M1
    • Denominator \(3 - 1 = 2\) — M1
    • States the result — A1
  6. 6 Calculate [4 marks]

    A right-angled triangle has shorter sides of length \(\sqrt{3}\) cm and \(\sqrt{6}\) cm. (a) Calculate the length of the hypotenuse. [2 marks] (b) Calculate the area of the triangle, in the form \(a\sqrt{2}\). [2 marks]

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    Model answer

    (a) \(\sqrt{3 + 6} = \sqrt{9} = 3\) cm. (b) \(\dfrac{1}{2} \times \sqrt{3} \times \sqrt{6} = \dfrac{1}{2}\sqrt{18} = \dfrac{1}{2} \times 3\sqrt{2} = \dfrac{3}{2}\sqrt{2}\) cm\(^2\).

    Mark scheme

    • (a) \(3 + 6 = 9\) — M1
    • (a) 3 cm — A1
    • (b) \(\dfrac{1}{2}\sqrt{18}\) or \(\dfrac{1}{2} \times \sqrt{3} \times \sqrt{6}\) — M1
    • (b) \(\dfrac{3}{2}\sqrt{2}\) — 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 \(4 - 3 = 1\).