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Exam questions · Maths · Further Algebra

Solving Quadratic Equations

  • 6 exam questions
  • 19 marks
  • 9 quick checks
  1. 1 Solve [3 marks]

    Solve \(x^2 + x - 12 = 0\).

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    Model answer

    \((x + 4)(x - 3) = 0\), so \(x = -4\) or \(x = 3\).

    Mark scheme

    • \((x + 4)(x - 3)\) — M1
    • \(x = -4\) — A1
    • \(x = 3\) — A1
  2. 2 Solve [3 marks]

    Solve \(x^2 = 10 - 3x\).

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    Model answer

    Rearrange to \(x^2 + 3x - 10 = 0\), so \((x + 5)(x - 2) = 0\) and \(x = -5\) or \(x = 2\).

    Mark scheme

    • \(x^2 + 3x - 10 = 0\) — M1
    • \((x + 5)(x - 2)\) — M1
    • \(x = -5\) and \(x = 2\) — A1
  3. 3 Show that [4 marks]

    The diagram shows a right-angled triangle. The sides are \(x\) cm and \((x + 2)\) cm, and the hypotenuse is 10 cm. (a) Show that \(x^2 + 2x - 48 = 0\). [2 marks] (b) Calculate the length of the shortest side of the triangle. [2 marks]

    A right-angled triangle with shorter sides x cm and (x + 2) cm and hypotenuse 10 cm.
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    Model answer

    (a) By Pythagoras, \(x^2 + (x + 2)^2 = 100\), so \(2x^2 + 4x + 4 = 100\), which gives \(x^2 + 2x - 48 = 0\). (b) \((x + 8)(x - 6) = 0\), so \(x = 6\) as a length is positive. The shortest side is 6 cm.

    Mark scheme

    • (a) \(x^2 + (x + 2)^2 = 100\) — M1
    • (a) \(x^2 + 2x - 48 = 0\) shown — A1
    • (b) \((x + 8)(x - 6)\) — M1
    • (b) 6 cm — A1
  4. 4 Solve [3 marks]

    (a) Solve \(x^2 - 81 = 0\). [1 mark] (b) Solve \(x^2 + 9x = 0\). [2 marks]

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    Model answer

    (a) \(x = 9\) or \(x = -9\). (b) \(x(x + 9) = 0\), so \(x = 0\) or \(x = -9\).

    Mark scheme

    • (a) \(x = \pm 9\) — B1
    • (b) \(x(x + 9)\) — M1
    • (b) \(x = 0\) and \(x = -9\) — A1
  5. 5 Solve [3 marks]

    Solve \(2x^2 - x - 10 = 0\).

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    Model answer

    \(2x^2 - 5x + 4x - 10 = x(2x - 5) + 2(2x - 5) = (2x - 5)(x + 2) = 0\), so \(x = \dfrac{5}{2}\) or \(x = -2\).

    Mark scheme

    • \((2x - 5)(x + 2)\) — M1
    • \(x = \dfrac{5}{2}\) — A1
    • \(x = -2\) — A1
  6. 6 Solve [3 marks]

    Use the quadratic formula to solve \(x^2 + 3x - 5 = 0\). Give your answers in surd form.

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    Model answer

    \(x = \dfrac{-3 \pm \sqrt{9 + 20}}{2} = \dfrac{-3 \pm \sqrt{29}}{2}\).

    Mark scheme

    • \(\dfrac{-3 \pm \sqrt{3^2 - 4 \times 1 \times (-5)}}{2}\) — M1
    • \(\sqrt{29}\) seen — M1
    • \(\dfrac{-3 \pm \sqrt{29}}{2}\) — A1

Quick check

  1. 1

    Solve \((x - 2)(x - 3) = 0\).

    1. A\(x = -2\) or \(x = -3\)
    2. B\(x = 2\) or \(x = 3\)
    3. C\(x = 2\) or \(x = -3\)
    4. D\(x = 6\)
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    B: \(x = 2\) or \(x = 3\)

    Each bracket can be zero: \(x - 2 = 0\) or \(x - 3 = 0\).

  2. 2

    Solve \(x^2 - 49 = 0\).

    1. A\(x = 7\) or \(x = -7\)
    2. B\(x = 7\) only
    3. C\(x = 49\)
    4. D\(x = 24.5\)
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    A: \(x = 7\) or \(x = -7\)

    \(x^2 = 49\) has two square roots.

  3. 3

    Solve \(x^2 - 6x = 0\).

    1. A\(x = 6\) only
    2. B\(x = 0\) or \(x = -6\)
    3. C\(x = 3\)
    4. D\(x = 0\) or \(x = 6\)
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    D: \(x = 0\) or \(x = 6\)

    \(x(x - 6) = 0\), so \(x = 0\) or \(x = 6\).

  4. 4

    Factorise \(x^2 - 5x + 6\).

    1. A\((x + 2)(x + 3)\)
    2. B\((x - 1)(x - 6)\)
    3. C\((x - 2)(x - 3)\)
    4. D\((x + 2)(x - 3)\)
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    C: \((x - 2)(x - 3)\)

    The numbers multiply to 6 and add to \(-5\): \(-2\) and \(-3\).

  5. 5

    What is the first step in solving \(x^2 = 3x + 10\) by factorising?

    1. ADivide both sides by \(x\)
    2. BRearrange to \(x^2 - 3x - 10 = 0\)
    3. CTake the square root of both sides
    4. DFactorise the right-hand side
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    B: Rearrange to \(x^2 - 3x - 10 = 0\)

    One side must be zero before you factorise.

  6. 6

    Solve \(x^2 + 5x + 6 = 0\).

    1. A\(x = -2\) or \(x = -3\)
    2. B\(x = 2\) or \(x = 3\)
    3. C\(x = -1\) or \(x = -6\)
    4. D\(x = 5\) or \(x = 6\)
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    A: \(x = -2\) or \(x = -3\)

    \((x + 2)(x + 3) = 0\).

  7. 7

    Solve \(2x^2 + 7x + 3 = 0\).

    1. A\(x = \dfrac{1}{2}\) or \(x = 3\)
    2. B\(x = -2\) or \(x = -3\)
    3. C\(x = -\dfrac{1}{3}\) or \(x = -2\)
    4. D\(x = -\dfrac{1}{2}\) or \(x = -3\)
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    D: \(x = -\dfrac{1}{2}\) or \(x = -3\)

    \((2x + 1)(x + 3) = 0\).

  8. 8

    What is the value of \(b^2 - 4ac\) for \(x^2 + 2x - 8 = 0\)?

    1. A\(-28\)
    2. B\(32\)
    3. C\(36\)
    4. D\(4\)
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    C: \(36\)

    \(4 - 4 \times 1 \times (-8) = 4 + 32 = 36\).

  9. 9

    Use the quadratic formula to solve \(2x^2 + 5x - 3 = 0\).

    1. A\(x = -\dfrac{1}{2}\) or \(x = 3\)
    2. B\(x = \dfrac{1}{2}\) or \(x = -3\)
    3. C\(x = \dfrac{1}{2}\) or \(x = 3\)
    4. D\(x = 2\) or \(x = -3\)
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    B: \(x = \dfrac{1}{2}\) or \(x = -3\)

    \(x = \dfrac{-5 \pm \sqrt{49}}{4} = \dfrac{-5 \pm 7}{4}\).