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Exam questions · Maths

Further Algebra

  • 30 exam questions
  • 91 marks
  • 45 quick checks

Solving Quadratic Equations

Just this lesson
  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\(x = 6\)
    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\(x = 24.5\)
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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\((x + 2)(x - 3)\)
    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 \(x = 6\)
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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 \(x = -3\)
    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 \(x = -3\)

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

Simultaneous Equations

Just this lesson
  1. 1 Use [2 marks]

    The graph shows the straight lines \(A\) and \(B\). Use the graph to solve the simultaneous equations \(y = x - 1\) and \(2x + y = 11\).

    Two straight lines, A and B, that cross at the point (4, 3).
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    Model answer

    The lines cross at \((4, 3)\), so \(x = 4\) and \(y = 3\).

    Mark scheme

    • \(x = 4\) — B1
    • \(y = 3\) — B1
  2. 2 Solve [3 marks]

    Solve the simultaneous equations \(x + 2y = 11\) and \(x - y = 2\).

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    Model answer

    Subtracting gives \(3y = 9\), so \(y = 3\). Then \(x = 5\). Check: \(5 - 3 = 2\).

    Mark scheme

    • \(3y = 9\) or another correct elimination — M1
    • \(y = 3\) — A1
    • \(x = 5\) — A1
  3. 3 Solve [3 marks]

    Solve the simultaneous equations \(4x + y = 19\) and \(3x - y = 9\).

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    Model answer

    Adding gives \(7x = 28\), so \(x = 4\). Then \(16 + y = 19\), so \(y = 3\). Check: \(12 - 3 = 9\).

    Mark scheme

    • \(7x = 28\) — M1
    • \(x = 4\) — A1
    • \(y = 3\) — A1
  4. 4 Solve [4 marks]

    Solve the simultaneous equations \(4x + 3y = 17\) and \(2x - y = 1\).

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    Model answer

    From the second equation \(y = 2x - 1\). Substituting gives \(4x + 3(2x - 1) = 17\), so \(10x - 3 = 17\) and \(x = 2\). Then \(y = 3\).

    Mark scheme

    • \(y = 2x - 1\), or the equations made to match — M1
    • \(4x + 3(2x - 1) = 17\) — M1
    • \(x = 2\) — A1
    • \(y = 3\) — A1
  5. 5 Calculate [4 marks]

    2 large boxes and 3 small boxes have a total mass of 29 kg. 1 large box and 2 small boxes have a total mass of 17 kg. Calculate the mass of a large box and the mass of a small box.

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    Model answer

    \(2l + 3s = 29\) and \(l + 2s = 17\). Doubling the second gives \(2l + 4s = 34\). Subtracting the first gives \(s = 5\). Then \(l = 17 - 10 = 7\). A large box is 7 kg and a small box is 5 kg.

    Mark scheme

    • Two correct equations — M1
    • A correct elimination step — M1
    • Small box 5 kg — A1
    • Large box 7 kg — A1
  6. 6 Solve [4 marks]

    Solve the simultaneous equations \(y = x^2\) and \(y = 3x + 4\).

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    Model answer

    Setting them equal gives \(x^2 = 3x + 4\), so \(x^2 - 3x - 4 = 0\) and \((x - 4)(x + 1) = 0\). So \(x = 4\) or \(x = -1\), and \(y = 16\) or \(y = 1\). The solutions are \((4, 16)\) and \((-1, 1)\).

    Mark scheme

    • \(x^2 = 3x + 4\) — M1
    • \((x - 4)(x + 1)\) — M1
    • \(x = 4\) and \(x = -1\) — A1
    • \((4, 16)\) and \((-1, 1)\) — A1

Quick check

  1. 1

    Solve \(x + y = 9\) and \(x - y = 1\).

    1. A\(x = 4\), \(y = 5\)
    2. B\(x = 8\), \(y = 1\)
    3. C\(x = 5\), \(y = 4\)
    4. D\(x = 10\), \(y = -1\)
    Show answerHide answer

    C: \(x = 5\), \(y = 4\)

    Adding gives \(2x = 10\), so \(x = 5\) and \(y = 4\).

  2. 2

    When do you add the two equations in elimination?

    1. AWhen the terms in one letter are the same
    2. BWhen the terms in one letter are opposites
    3. CWhen there are no brackets
    4. DWhen one equation has a fraction
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    B: When the terms in one letter are opposites

    Opposites such as \(+3y\) and \(-3y\) cancel when added.

  3. 3

    Solve \(3x + 2y = 16\) and \(x + 2y = 8\).

    1. A\(x = 4\), \(y = 2\)
    2. B\(x = 2\), \(y = 4\)
    3. C\(x = 8\), \(y = 0\)
    4. D\(x = 4\), \(y = 4\)
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    A: \(x = 4\), \(y = 2\)

    Subtract: \(2x = 8\), so \(x = 4\). Then \(4 + 2y = 8\) gives \(y = 2\).

  4. 4

    Solve \(2x + 3y = 13\) and \(3x - y = 3\).

    1. A\(x = 3\), \(y = 2\)
    2. B\(x = 1\), \(y = 4\)
    3. C\(x = 4\), \(y = 1\)
    4. D\(x = 2\), \(y = 3\)
    Show answerHide answer

    D: \(x = 2\), \(y = 3\)

    Multiply the second equation by 3 and add: \(11x = 22\), so \(x = 2\), \(y = 3\).

  5. 5

    Solve \(y = 2x + 1\) and \(3x + y = 16\).

    1. A\(x = 3\), \(y = 5\)
    2. B\(x = 7\), \(y = 3\)
    3. C\(x = 3\), \(y = 7\)
    4. D\(x = 5\), \(y = 3\)
    Show answerHide answer

    C: \(x = 3\), \(y = 7\)

    \(3x + 2x + 1 = 16\), so \(x = 3\) and \(y = 7\).

  6. 6

    Two straight lines are parallel. How many solutions do their simultaneous equations have?

    1. AOne
    2. BNone
    3. CTwo
    4. DInfinitely many
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    B: None

    Parallel lines never meet, so there is no point on both.

  7. 7

    2 adult and 3 child tickets cost \(\pounds 19\). 3 adult and 1 child ticket cost \(\pounds 18\). What does an adult ticket cost?

    1. A\(\pounds 5\)
    2. B\(\pounds 3\)
    3. C\(\pounds 4\)
    4. D\(\pounds 6\)
    Show answerHide answer

    A: \(\pounds 5\)

    \(2a + 3c = 19\) and \(3a + c = 18\) give \(a = 5\), \(c = 3\).

  8. 8

    Solve \(y = x^2\) and \(y = x + 6\).

    1. A\((3, 9)\) only
    2. B\((2, 4)\) and \((-3, 9)\)
    3. C\((3, 9)\) and \((-3, 9)\)
    4. D\((3, 9)\) and \((-2, 4)\)
    Show answerHide answer

    D: \((3, 9)\) and \((-2, 4)\)

    \(x^2 = x + 6\) gives \((x - 3)(x + 2) = 0\).

  9. 9

    Solve \(x^2 + y^2 = 25\) and \(y = x + 1\).

    1. A\((4, 3)\) and \((-3, -4)\)
    2. B\((3, 4)\) only
    3. C\((3, 4)\) and \((-4, -3)\)
    4. D\((0, 1)\) and \((-1, 0)\)
    Show answerHide answer

    C: \((3, 4)\) and \((-4, -3)\)

    \(x^2 + (x + 1)^2 = 25\) gives \(x^2 + x - 12 = 0\), so \(x = 3\) or \(x = -4\).

Inequalities and Regions

Just this lesson
  1. 1 Write down [2 marks]

    \(n\) is an integer such that \(1 \leq 2n < 9\). Write down all the possible values of \(n\).

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    Model answer

    Dividing by 2 gives \(0.5 \leq n < 4.5\), so \(n = 1, 2, 3, 4\).

    Mark scheme

    • \(0.5 \leq n < 4.5\) or three correct values — M1
    • 1, 2, 3, 4 — A1
  2. 2 Write down [5 marks]

    The diagram shows a shaded region \(R\). (a) Write down the three inequalities that define \(R\). [3 marks] (b) Show that the point \((4, 5)\) is inside \(R\) or on its boundary. [2 marks]

    A shaded triangular region R bounded by a horizontal line, a steep line through the origin and a falling diagonal line.
    Show answerHide answer

    Model answer

    (a) \(y \geq 2\), \(y \leq 2x\) and \(x + y \leq 9\). (b) \(5 \geq 2\), \(5 \leq 2 \times 4 = 8\) and \(4 + 5 = 9 \leq 9\), so the point satisfies all three inequalities, and is on the boundary \(x + y = 9\).

    Mark scheme

    • (a) \(y \geq 2\) — B1
    • (a) \(y \leq 2x\) — B1
    • (a) \(x + y \leq 9\) — B1
    • (b) Substitutes \((4, 5)\) into at least two inequalities — M1
    • (b) All three satisfied, with a conclusion — A1
  3. 3 Solve [3 marks]

    Solve \(-2 < 3x + 1 \leq 10\).

    Show answerHide answer

    Model answer

    Subtract 1: \(-3 < 3x \leq 9\). Divide by 3: \(-1 < x \leq 3\).

    Mark scheme

    • \(-3 < 3x \leq 9\) — M1
    • \(-1 < x\) — A1
    • \(x \leq 3\) — A1
  4. 4 Write down [2 marks]

    A region is above the line \(y = 2\), below the line \(y = 5\) and to the right of the line \(x = 1\), and includes all three lines. Write down the three inequalities that define the region.

    Show answerHide answer

    Model answer

    The inequalities are \(y \geq 2\), \(y \leq 5\) and \(x \geq 1\).

    Mark scheme

    • Two correct inequalities — M1
    • \(y \geq 2\), \(y \leq 5\) and \(x \geq 1\) — A1
  5. 5 Solve [3 marks]

    Solve \(x^2 - 7x + 10 < 0\).

    Show answerHide answer

    Model answer

    \((x - 2)(x - 5) < 0\), with roots 2 and 5. The curve is below the axis between the roots, so \(2 < x < 5\).

    Mark scheme

    • \((x - 2)(x - 5)\) — M1
    • A sketch or a clear statement that the curve is below the axis between the roots — M1
    • \(2 < x < 5\) — A1
  6. 6 Solve [3 marks]

    Solve \(x^2 - x - 12 \leq 0\).

    Show answerHide answer

    Model answer

    \((x - 4)(x + 3) \leq 0\), with roots \(-3\) and 4. The curve is below or on the axis between the roots, so \(-3 \leq x \leq 4\).

    Mark scheme

    • \((x - 4)(x + 3)\) — M1
    • A sketch or a clear statement that the curve is below the axis between the roots — M1
    • \(-3 \leq x \leq 4\) — A1

Quick check

  1. 1

    What does a dashed boundary line mean on a graph of an inequality?

    1. AThe line is included
    2. BThe inequality has no solutions
    3. CThe region is below the line
    4. DThe line itself is not included
    Show answerHide answer

    D: The line itself is not included

    A dashed line goes with \(<\) or \(>\).

  2. 2

    Which inequality describes the region to the right of the line \(x = 1\), including the line?

    1. A\(x \leq 1\)
    2. B\(y \geq 1\)
    3. C\(x \geq 1\)
    4. D\(x > 1\)
    Show answerHide answer

    C: \(x \geq 1\)

    To the right means larger \(x\), and the line is included.

  3. 3

    Which inequality describes the region on or below the line \(y = 2x\)?

    1. A\(y \geq 2x\)
    2. B\(y \leq 2x\)
    3. C\(y < 2x\)
    4. D\(x \leq 2y\)
    Show answerHide answer

    B: \(y \leq 2x\)

    Below the line means smaller \(y\), and the line is included.

  4. 4

    Which integers satisfy \(-2 < x \leq 3\)?

    1. A\(-1, 0, 1, 2, 3\)
    2. B\(-2, -1, 0, 1, 2, 3\)
    3. C\(-1, 0, 1, 2\)
    4. D\(-2, -1, 0, 1, 2\)
    Show answerHide answer

    A: \(-1, 0, 1, 2, 3\)

    \(-2\) is not included but 3 is.

  5. 5

    The origin is tested in \(x + y \leq 6\). What does this show?

    1. AThe origin is not in the region
    2. BThe line is dashed
    3. CThe inequality has no solutions
    4. DThe origin is in the region, so shade that side
    Show answerHide answer

    D: The origin is in the region, so shade that side

    \(0 + 0 \leq 6\) is true.

  6. 6

    How many points with whole-number coordinates satisfy \(x \geq 1\), \(y \geq 1\) and \(x + y \leq 6\)?

    1. A\(10\)
    2. B\(21\)
    3. C\(15\)
    4. D\(25\)
    Show answerHide answer

    C: \(15\)

    The columns \(x = 1, 2, 3, 4, 5\) have \(5, 4, 3, 2, 1\) points.

  7. 7

    Which kind of boundary line goes with the inequality \(y > 3\)?

    1. AA solid line
    2. BA dashed line
    3. CA curved line
    4. DNo line is drawn
    Show answerHide answer

    B: A dashed line

    Strict inequalities do not include the boundary.

  8. 8

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

    1. A\(-2 < x < 3\)
    2. B\(x < -2\) or \(x > 3\)
    3. C\(-3 < x < 2\)
    4. D\(x < 3\)
    Show answerHide answer

    A: \(-2 < x < 3\)

    The roots are \(-2\) and 3, and the curve is below the axis between them.

  9. 9

    Solve \(x^2 > 9\).

    1. A\(-3 < x < 3\)
    2. B\(x > 3\)
    3. C\(x > 9\)
    4. D\(x < -3\) or \(x > 3\)
    Show answerHide answer

    D: \(x < -3\) or \(x > 3\)

    The curve is above the axis outside the roots \(-3\) and 3.

Surds

Just this lesson
  1. 1 Simplify [2 marks]

    Simplify \(\sqrt{98}\).

    Show answerHide answer

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

    Show answerHide answer

    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 \(a\sqrt{3}\).

    Show answerHide answer

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

    Show answerHide answer

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

    Show answerHide answer

    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]

    Show answerHide answer

    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\(2\sqrt{5}\)
    Show answerHide answer

    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\(2\sqrt{3}\)
    Show answerHide answer

    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\(2\sqrt{2}\)
    Show answerHide answer

    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\(8\)
    Show answerHide answer

    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\(2\sqrt{5}\)
    Show answerHide answer

    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\(4\sqrt{6}\)
    Show answerHide answer

    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\(\dfrac{2 - \sqrt{3}}{7}\)
    Show answerHide answer

    A: \(2 - \sqrt{3}\)

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

Algebraic Fractions and Proof

Just this lesson
  1. 1 Prove [3 marks]

    Prove that the sum of any three consecutive odd numbers is a multiple of 3.

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    Model answer

    Let the numbers be \(2n + 1\), \(2n + 3\) and \(2n + 5\). Their sum is \(6n + 9 = 3(2n + 3)\), which is a multiple of 3.

    Mark scheme

    • \(2n + 1\), \(2n + 3\), \(2n + 5\) — M1
    • \(6n + 9\) — M1
    • \(3(2n + 3)\) with a conclusion — A1
  2. 2 Simplify [3 marks]

    Simplify \(\dfrac{3x^2 + 6x}{x^2 - 4}\).

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    Model answer

    \(\dfrac{3x(x + 2)}{(x - 2)(x + 2)} = \dfrac{3x}{x - 2}\).

    Mark scheme

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

    Solve \(\dfrac{x + 2}{5} + \dfrac{x - 3}{2} = 1\).

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    Model answer

    Multiply every term by 10: \(2(x + 2) + 5(x - 3) = 10\). Then \(7x - 11 = 10\), so \(x = 3\).

    Mark scheme

    • \(2(x + 2) + 5(x - 3) = 10\) — M1
    • \(7x - 11 = 10\) — M1
    • \(x = 3\) — A1
  4. 4 Show that [2 marks]

    Show that the statement “\(n^2 + n + 11\) is always prime” is not true.

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    Model answer

    When \(n = 11\), \(121 + 11 + 11 = 143 = 11 \times 13\), which is not prime. So the statement is not true.

    Mark scheme

    • A counter-example, such as \(n = 11\), substituted — M1
    • \(143 = 11 \times 13\) with a conclusion — A1
  5. 5 Prove [3 marks]

    Prove that \((n + 2)^2 - n^2\) is a multiple of 4 for every integer \(n\).

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    Model answer

    \((n + 2)^2 - n^2 = n^2 + 4n + 4 - n^2 = 4n + 4 = 4(n + 1)\), which is a multiple of 4.

    Mark scheme

    • \(n^2 + 4n + 4\) — M1
    • \(4n + 4\) — M1
    • \(4(n + 1)\) with a conclusion — A1
  6. 6 Simplify [3 marks]

    Simplify \(\dfrac{x^2 - 3x - 10}{x^2 - 4}\).

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    Model answer

    \(\dfrac{(x - 5)(x + 2)}{(x - 2)(x + 2)} = \dfrac{x - 5}{x - 2}\).

    Mark scheme

    • \((x - 5)(x + 2)\) or \((x - 2)(x + 2)\) — M1
    • Both factorised — M1
    • \(\dfrac{x - 5}{x - 2}\) — A1

Quick check

  1. 1

    What may be cancelled in an algebraic fraction?

    1. AAny terms that appear on the top and bottom
    2. BFactors that multiply the whole top and the whole bottom
    3. COnly numbers
    4. DOnly letters
    Show answerHide answer

    B: Factors that multiply the whole top and the whole bottom

    Terms that are added or subtracted cannot be cancelled.

  2. 2

    Simplify \(\dfrac{x^2 - 9}{x + 3}\).

    1. A\(x - 3\)
    2. B\(x + 3\)
    3. C\(x - 9\)
    4. D\(\dfrac{x - 9}{1}\)
    Show answerHide answer

    A: \(x - 3\)

    \(\dfrac{(x - 3)(x + 3)}{x + 3} = x - 3\).

  3. 3

    Which statement about \(\dfrac{x + 3}{3}\) is correct?

    1. AIt simplifies to \(x\)
    2. BIt simplifies to \(x + 1\)
    3. CIt simplifies to 1
    4. DThe 3s cannot be cancelled because the 3 on top is added
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    D: The 3s cannot be cancelled because the 3 on top is added

    Only factors can be cancelled, not terms.

  4. 4

    Solve \(\dfrac{x - 1}{3} + \dfrac{x + 2}{6} = 2\).

    1. A\(x = 3\)
    2. B\(x = 5\)
    3. C\(x = 4\)
    4. D\(x = 6\)
    Show answerHide answer

    C: \(x = 4\)

    Multiply by 6: \(2(x - 1) + (x + 2) = 12\), so \(3x = 12\).

  5. 5

    Which expression is an odd number for any whole number \(n\)?

    1. A\(2n\)
    2. B\(2n + 1\)
    3. C\(n + 1\)
    4. D\(n^2\)
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    B: \(2n + 1\)

    \(2n\) is even, so adding 1 makes it odd.

  6. 6

    What is the sum of three consecutive whole numbers \(n\), \(n + 1\) and \(n + 2\)?

    1. A\(3n + 3\)
    2. B\(3n\)
    3. C\(3n + 2\)
    4. D\(n + 3\)
    Show answerHide answer

    A: \(3n + 3\)

    \(n + n + 1 + n + 2 = 3n + 3 = 3(n + 1)\).

  7. 7

    Which value of \(n\) is a counter-example to “\(n^2 + n + 1\) is always prime”?

    1. A\(n = 1\)
    2. B\(n = 2\)
    3. C\(n = 3\)
    4. D\(n = 4\)
    Show answerHide answer

    D: \(n = 4\)

    \(16 + 4 + 1 = 21 = 3 \times 7\), which is not prime.

  8. 8

    Simplify \(\dfrac{x^2 + 5x + 6}{x^2 + 3x + 2}\).

    1. A\(\dfrac{x + 2}{x + 1}\)
    2. B\(\dfrac{x + 3}{x + 2}\)
    3. C\(\dfrac{x + 3}{x + 1}\)
    4. D\(\dfrac{5x + 6}{3x + 2}\)
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    C: \(\dfrac{x + 3}{x + 1}\)

    \(\dfrac{(x + 2)(x + 3)}{(x + 1)(x + 2)} = \dfrac{x + 3}{x + 1}\).

  9. 9

    Expand and simplify \((n + 1)^2 - (n - 1)^2\).

    1. A\(2\)
    2. B\(4n\)
    3. C\(2n\)
    4. D\(4n + 2\)
    Show answerHide answer

    B: \(4n\)

    \(n^2 + 2n + 1 - n^2 + 2n - 1 = 4n\).