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Exam questions · Maths

Further Algebra

  • 30 exam questions
  • 90 marks
  • 45 quick checks

Solving Quadratic Equations

Just this lesson
  1. 1 Solve [3 marks]

    Solve \(x^2 - 3x - 10 = 0\).

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    Model answer

    \((x - 5)(x + 2) = 0\), so \(x = 5\) or \(x = -2\).

    Mark scheme

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

    Solve \(x^2 = 5x + 14\).

    Show answerHide answer

    Model answer

    Rearrange to \(x^2 - 5x - 14 = 0\), so \((x - 7)(x + 2) = 0\) and \(x = 7\) or \(x = -2\).

    Mark scheme

    • \(x^2 - 5x - 14 = 0\) — M1
    • \((x - 7)(x + 2)\) — M1
    • \(x = 7\) and \(x = -2\) — A1
  3. 3 Show that [4 marks]

    The diagram shows a rectangle with area 24 cm\(^2\). (a) Show that \(x^2 + 4x - 21 = 0\). [2 marks] (b) Work out the value of \(x\). [2 marks]

    A rectangle with length (x + 3) cm, width (x + 1) cm and area 24 square centimetres.
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    Model answer

    (a) \((x + 3)(x + 1) = 24\), so \(x^2 + 4x + 3 = 24\) and \(x^2 + 4x - 21 = 0\). (b) \((x + 7)(x - 3) = 0\), so \(x = -7\) or \(x = 3\). A length cannot be negative, so \(x = 3\).

    Mark scheme

    • (a) \((x + 3)(x + 1) = 24\) — M1
    • (a) \(x^2 + 4x - 21 = 0\) shown — Q1
    • (b) \((x + 7)(x - 3)\) — M1
    • (b) \(x = 3\) with \(x = -7\) rejected — A1
  4. 4 Solve [3 marks]

    (a) Solve \(x^2 - 100 = 0\). [1 mark] (b) Solve \(x^2 - 8x = 0\). [2 marks]

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    Model answer

    (a) \(x = 10\) or \(x = -10\). (b) \(x(x - 8) = 0\), so \(x = 0\) or \(x = 8\).

    Mark scheme

    • (a) \(x = \pm 10\) — B1
    • (b) \(x(x - 8)\) — M1
    • (b) \(x = 0\) and \(x = 8\) — A1
  5. 5 Solve [3 marks]

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

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    Model answer

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

    Mark scheme

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

    Solve \(x^2 - 6x + 2 = 0\). Give your answers in the form \(a \pm \sqrt{b}\).

    Show answerHide answer

    Model answer

    Complete the square: \((x - 3)^2 - 9 + 2 = 0\), so \((x - 3)^2 = 7\) and \(x = 3 \pm \sqrt{7}\).

    Mark scheme

    • \((x - 3)^2\) seen — M1
    • \((x - 3)^2 = 7\) — M1
    • \(3 \pm \sqrt{7}\) — 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 \(x = -3\)

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

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 \(y = 3x - 2\) and \(x + y = 6\).

    Two straight lines, A and B, that cross at the point (2, 4).
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    Model answer

    The lines cross at \((2, 4)\), so \(x = 2\) and \(y = 4\).

    Mark scheme

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

    Solve the simultaneous equations \(2x + y = 13\) and \(x - y = 2\).

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    Model answer

    Adding gives \(3x = 15\), so \(x = 5\). Then \(y = 13 - 10 = 3\). Check: \(5 - 3 = 2\).

    Mark scheme

    • \(3x = 15\) or another correct elimination — M1
    • \(x = 5\) — A1
    • \(y = 3\) — A1
  3. 3 Solve [3 marks]

    Solve the simultaneous equations \(5x + 2y = 16\) and \(3x + 2y = 8\).

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    Model answer

    Subtracting gives \(2x = 8\), so \(x = 4\). Then \(20 + 2y = 16\), so \(y = -2\). Check: \(12 - 4 = 8\).

    Mark scheme

    • \(2x = 8\) — M1
    • \(x = 4\) — A1
    • \(y = -2\) — A1
  4. 4 Solve [4 marks]

    Solve the simultaneous equations \(3x + 2y = 18\) and \(2x - y = 5\).

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    Model answer

    From the second equation \(y = 2x - 5\). Substituting gives \(3x + 2(2x - 5) = 18\), so \(7x - 10 = 18\) and \(x = 4\). Then \(y = 3\).

    Mark scheme

    • \(y = 2x - 5\), or the equations made to match — M1
    • \(3x + 2(2x - 5) = 18\) — M1
    • \(x = 4\) — A1
    • \(y = 3\) — A1
  5. 5 Work out [4 marks]

    3 pens and 2 pencils cost 190p. 2 pens and 1 pencil cost 110p. Work out the cost of one pen and the cost of one pencil.

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    Model answer

    \(3p + 2q = 190\) and \(2p + q = 110\). Then \(q = 110 - 2p\), so \(3p + 220 - 4p = 190\) and \(p = 30\), \(q = 50\). A pen costs 30p and a pencil costs 50p.

    Mark scheme

    • Two correct equations — M1
    • A correct method to eliminate a letter — M1
    • Pen 30p — A1
    • Pencil 50p — A1
  6. 6 Solve [4 marks]

    Solve the simultaneous equations \(x^2 + y^2 = 20\) and \(y = 2x\).

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    Model answer

    Substituting gives \(x^2 + 4x^2 = 20\), so \(5x^2 = 20\), \(x^2 = 4\) and \(x = 2\) or \(x = -2\). Then \(y = 4\) or \(y = -4\). The solutions are \((2, 4)\) and \((-2, -4)\).

    Mark scheme

    • \(x^2 + (2x)^2 = 20\) — M1
    • \(x = 2\) and \(x = -2\) — A1
    • One correct \(y\)-value — M1
    • \((2, 4)\) and \((-2, -4)\) — 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
    Show answerHide answer

    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\(x = 4\), \(y = 4\)
    Show answerHide answer

    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
    Show answerHide answer

    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]

    Write down all the integers that satisfy \(2 < x \leq 6\).

    Show answerHide answer

    Model answer

    The integers are 3, 4, 5, 6.

    Mark scheme

    • At least three correct and no more than one wrong — M1
    • 3, 4, 5, 6 — 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) How many points with integer coordinates are inside \(R\) or on its boundary? [2 marks]

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

    Model answer

    (a) \(y \geq 1\), \(y \leq x\) and \(x \leq 5\). (b) For \(x = 1, 2, 3, 4, 5\) there are \(1, 2, 3, 4, 5\) points, so the total is 15.

    Mark scheme

    • (a) \(y \geq 1\) — B1
    • (a) \(y \leq x\) — B1
    • (a) \(x \leq 5\) — B1
    • (b) Counts by columns, such as 1, 2, 3, 4, 5 — M1
    • (b) 15 — A1
  3. 3 Solve [2 marks]

    Solve \(5x - 3 \leq 17\).

    Show answerHide answer

    Model answer

    \(5x \leq 20\), so \(x \leq 4\).

    Mark scheme

    • \(5x \leq 20\) — M1
    • \(x \leq 4\) — A1
  4. 4 Decide [2 marks]

    Here are two points: \((1, 3)\) and \((2, 6)\). For each point, say whether it satisfies the inequality \(y \leq 2x + 1\). You must show your working.

    Show answerHide answer

    Model answer

    For \((1, 3)\): \(2 \times 1 + 1 = 3\) and \(3 \leq 3\), so it does. For \((2, 6)\): \(2 \times 2 + 1 = 5\) and \(6 \leq 5\) is false, so it does not.

    Mark scheme

    • \((1, 3)\) satisfies it, with \(2 \times 1 + 1 = 3\) — B1
    • \((2, 6)\) does not satisfy it, with \(2 \times 2 + 1 = 5\) — B1
  5. 5 Solve [3 marks]

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

    Show answerHide answer

    Model answer

    \(x^2 + x - 12 = (x + 4)(x - 3)\), with roots \(-4\) and 3. The curve is above the \(x\)-axis outside the roots, so \(x < -4\) or \(x > 3\).

    Mark scheme

    • \((x + 4)(x - 3)\) or the roots \(-4\) and 3 — M1
    • A sketch or a clear statement of which regions are above the axis — M1
    • \(x < -4\) or \(x > 3\) — A1
  6. 6 Solve [3 marks]

    (a) Solve \(x^2 < 25\). [1 mark] (b) Solve \(x^2 - 3x - 10 \leq 0\). [2 marks]

    Show answerHide answer

    Model answer

    (a) \(-5 < x < 5\). (b) \((x - 5)(x + 2) \leq 0\), so \(-2 \leq x \leq 5\).

    Mark scheme

    • (a) \(-5 < x < 5\) — B1
    • (b) Roots \(-2\) and 5 — M1
    • (b) \(-2 \leq x \leq 5\) — 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]

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

    Show answerHide answer

    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]

    Show answerHide answer

    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 \(a\sqrt{2}\).

    Show answerHide answer

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

    Show answerHide answer

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

    Show answerHide answer

    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]

    Show answerHide answer

    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\(2\sqrt{5}\)
    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\(2\sqrt{3}\)
    Show answerHide answer

    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\(9 - 2\sqrt{2}\)
    Show answerHide answer

    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 Show that [3 marks]

    Show that the sum of three consecutive even numbers is a multiple of 6.

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    Model answer

    Let the numbers be \(2n\), \(2n + 2\) and \(2n + 4\). Their sum is \(6n + 6 = 6(n + 1)\), which is a multiple of 6.

    Mark scheme

    • \(2n\), \(2n + 2\), \(2n + 4\) — M1
    • \(6n + 6\) — M1
    • \(6(n + 1)\) with a conclusion — Q1
  2. 2 Simplify [3 marks]

    Simplify \(\dfrac{x^2 + 5x}{x^2 - 25}\).

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    Model answer

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

    Mark scheme

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

    Solve \(\dfrac{x - 2}{3} + \dfrac{x + 1}{2} = 4\).

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    Model answer

    Multiply every term by 6: \(2(x - 2) + 3(x + 1) = 24\). Then \(5x - 1 = 24\), so \(x = 5\).

    Mark scheme

    • \(2(x - 2) + 3(x + 1) = 24\) — M1
    • \(5x - 1 = 24\) — M1
    • \(x = 5\) — A1
  4. 4 Show that [2 marks]

    Edith says, “\(n^2 > n\) for every number \(n\).” Show that Edith is wrong.

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    Model answer

    When \(n = 1\), \(n^2 = 1\) and \(n = 1\), so \(n^2\) is not greater than \(n\). This counter-example shows she is wrong. (\(n = 0.5\) also works.)

    Mark scheme

    • A valid counter-example, such as \(n = 1\) or \(n = 0.5\) — M1
    • Substitutes and shows the statement is false — Q1
  5. 5 Prove [3 marks]

    Prove that \((2n + 1)^2 - (2n - 1)^2\) is a multiple of 8 for every positive integer \(n\).

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    Model answer

    \((2n + 1)^2 = 4n^2 + 4n + 1\) and \((2n - 1)^2 = 4n^2 - 4n + 1\). The difference is \(8n\), which is a multiple of 8.

    Mark scheme

    • \(4n^2 + 4n + 1\) or \(4n^2 - 4n + 1\) — M1
    • \(8n\) — M1
    • States that \(8n\) is a multiple of 8 — Q1
  6. 6 Write [3 marks]

    Write \(\dfrac{2}{x + 1} - \dfrac{1}{x - 2}\) as a single fraction, in its simplest form.

    Show answerHide answer

    Model answer

    The common denominator is \((x + 1)(x - 2)\), so the fraction is \(\dfrac{2(x - 2) - (x + 1)}{(x + 1)(x - 2)} = \dfrac{x - 5}{(x + 1)(x - 2)}\).

    Mark scheme

    • Common denominator \((x + 1)(x - 2)\) — M1
    • \(2(x - 2) - (x + 1)\) — M1
    • \(\dfrac{x - 5}{(x + 1)(x - 2)}\) — 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
    Show answerHide answer

    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\(n^2\)
    Show answerHide answer

    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\(\dfrac{5x + 6}{3x + 2}\)
    Show answerHide answer

    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: \(4n\)

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