OpenRevise

Exam questions · Maths · Further Algebra

Solving Quadratic Equations

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

    Solve \(x^2 + 7x + 12 = 0\).

    Show answerHide answer

    Model answer

    \(x^2 + 7x + 12 = (x + 3)(x + 4) = 0\), so \(x = -3\) or \(x = -4\).

    Mark scheme

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

    Solve \(x^2 = 6x - 8\).

    Show answerHide answer

    Model answer

    Rearrange to \(x^2 - 6x + 8 = 0\). Factorising gives \((x - 2)(x - 4) = 0\), so \(x = 2\) or \(x = 4\).

    Mark scheme

    • \(x^2 - 6x + 8 = 0\) — M1
    • \((x - 2)(x - 4)\) — M1
    • \(x = 2\) and \(x = 4\) — A1
  3. 3 Show that [4 marks]

    The diagram shows a rectangle. The area of the rectangle is 18 cm\(^2\). (a) Show that \(x^2 + 3x - 28 = 0\). (2 marks) (b) Hence work out the value of \(x\). (2 marks)

    A rectangle with length (x + 5) cm, width (x minus 2) cm and area 18 square centimetres.
    Show answerHide answer

    Model answer

    (a) \((x + 5)(x - 2) = 18\), so \(x^2 + 3x - 10 = 18\) and \(x^2 + 3x - 28 = 0\). (b) \((x + 7)(x - 4) = 0\), so \(x = -7\) or \(x = 4\). A length cannot be negative, so \(x = 4\).

    Mark scheme

    • (a) \((x + 5)(x - 2) = 18\) or \(x^2 + 3x - 10 = 18\) — M1
    • (a) \(x^2 + 3x - 28 = 0\) shown — C1
    • (b) \((x + 7)(x - 4)\) — M1
    • (b) \(x = 4\) with \(x = -7\) rejected — A1
  4. 4 Solve [4 marks]

    (a) Solve \(x^2 - 36 = 0\). (2 marks) (b) Solve \(x^2 - 5x = 0\). (2 marks)

    Show answerHide answer

    Model answer

    (a) \(x^2 = 36\), so \(x = 6\) or \(x = -6\). (b) \(x(x - 5) = 0\), so \(x = 0\) or \(x = 5\).

    Mark scheme

    • (a) \(x^2 = 36\) or \((x - 6)(x + 6)\) — M1
    • (a) \(x = 6\) and \(x = -6\) — A1
    • (b) \(x(x - 5)\) — M1
    • (b) \(x = 0\) and \(x = 5\) — A1
  5. 5 Solve [3 marks]

    Solve \(2x^2 - 5x - 3 = 0\).

    Show answerHide answer

    Model answer

    \(2x^2 - 6x + x - 3 = 2x(x - 3) + (x - 3) = (2x + 1)(x - 3) = 0\), so \(x = -\dfrac{1}{2}\) or \(x = 3\).

    Mark scheme

    • \(2x^2 - 6x + x - 3\) or \((2x + 1)(x - 3)\) — M1
    • \(x = -\dfrac{1}{2}\) — A1
    • \(x = 3\) — A1
  6. 6 Solve [3 marks]

    Solve \(x^2 + 4x - 3 = 0\). Give your solutions in the form \(p \pm \sqrt{q}\), where \(p\) and \(q\) are integers.

    Show answerHide answer

    Model answer

    Using the formula, \(x = \dfrac{-4 \pm \sqrt{16 + 12}}{2} = \dfrac{-4 \pm \sqrt{28}}{2} = \dfrac{-4 \pm 2\sqrt{7}}{2} = -2 \pm \sqrt{7}\).

    Mark scheme

    • \(\dfrac{-4 \pm \sqrt{4^2 - 4 \times 1 \times (-3)}}{2}\) or \((x + 2)^2 - 7\) — M1
    • \(\sqrt{28} = 2\sqrt{7}\) seen — M1
    • \(-2 \pm \sqrt{7}\) — 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\(x = 6\)
    Show answerHide answer

    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\(x = 24.5\)
    Show answerHide answer

    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 \(x = 6\)
    Show answerHide answer

    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\((x + 2)(x - 3)\)
    Show answerHide answer

    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
    Show answerHide answer

    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 \(x = 6\)
    Show answerHide answer

    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 \(x = -3\)
    Show answerHide answer

    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\(4\)
    Show answerHide answer

    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 \(x = -3\)
    Show answerHide answer

    B: \(x = \dfrac{1}{2}\) or \(x = -3\)

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