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Exam questions · Maths

Further Algebra

  • 30 exam questions
  • 94 marks
  • 45 quick checks

Solving Quadratic Equations

Just this lesson
  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.
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    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)

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

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    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.

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    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 \(x = -3\)

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

Simultaneous Equations

Just this lesson
  1. 1 Use [2 marks]

    Here are the graphs of the straight lines \(A\) and \(B\). Use the graphs to solve the simultaneous equations \(y = x + 2\) and \(x + y = 10\).

    Two straight lines, A and B, that cross at the point (4, 6).
    Show answerHide answer

    Model answer

    The lines cross at \((4, 6)\), so \(x = 4\) and \(y = 6\).

    Mark scheme

    • \(x = 4\) — B1
    • \(y = 6\) — B1
  2. 2 Solve [3 marks]

    Solve the simultaneous equations \(x + y = 11\) and \(x - y = 3\).

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    Model answer

    Adding gives \(2x = 14\), so \(x = 7\). Then \(y = 11 - 7 = 4\). Check: \(7 - 4 = 3\).

    Mark scheme

    • Adds or eliminates \(y\) correctly — M1
    • \(x = 7\) — A1
    • \(y = 4\) — A1
  3. 3 Solve [3 marks]

    Solve the simultaneous equations \(3x + y = 14\) and \(2x - y = 6\).

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    Model answer

    Adding gives \(5x = 20\), so \(x = 4\). Then \(12 + y = 14\), so \(y = 2\). Check: \(8 - 2 = 6\).

    Mark scheme

    • \(5x = 20\) — M1
    • \(x = 4\) — A1
    • \(y = 2\) — A1
  4. 4 Solve [4 marks]

    Solve the simultaneous equations \(2x + 3y = 12\) and \(4x - y = 10\).

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    Model answer

    Multiply the second equation by 3: \(12x - 3y = 30\). Adding to the first gives \(14x = 42\), so \(x = 3\). Then \(6 + 3y = 12\), so \(y = 2\). Check: \(12 - 2 = 10\).

    Mark scheme

    • Multiplies to match the coefficients, such as \(12x - 3y = 30\) — M1
    • Adds or subtracts correctly, \(14x = 42\) — M1
    • \(x = 3\) — A1
    • \(y = 2\) — A1
  5. 5 Work out [4 marks]

    3 adults and 2 children go to the cinema. The total cost is \(\pounds 29\). 2 adults and 4 children go to the cinema. The total cost is \(\pounds 30\). Work out the cost of an adult ticket and the cost of a child ticket.

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    Model answer

    \(3a + 2c = 29\) and \(2a + 4c = 30\), so \(a + 2c = 15\). Subtracting this from the first equation gives \(2a = 14\), so \(a = 7\). Then \(2c = 8\) and \(c = 4\). An adult ticket costs \(\pounds 7\) and a child ticket \(\pounds 4\).

    Mark scheme

    • Two correct equations — M1
    • A correct elimination step — M1
    • Adult ticket \(\pounds 7\) — A1
    • Child ticket \(\pounds 4\) — A1
  6. 6 Solve [5 marks]

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

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    Model answer

    Setting them equal gives \(x^2 - 4 = 2x - 1\), so \(x^2 - 2x - 3 = 0\) and \((x - 3)(x + 1) = 0\). So \(x = 3\) or \(x = -1\). Then \(y = 5\) when \(x = 3\), and \(y = -3\) when \(x = -1\).

    Mark scheme

    • \(x^2 - 4 = 2x - 1\) — M1
    • \(x^2 - 2x - 3 = 0\) — M1
    • \(x = 3\) and \(x = -1\) — A1
    • One correct \(y\)-value — M1
    • \((3, 5)\) and \((-1, -3)\) — 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 integer values of \(n\) that satisfy \(-3 \leq n < 2\).

    Show answerHide answer

    Model answer

    The integers are \(-3, -2, -1, 0, 1\).

    Mark scheme

    • At least four correct and no more than one wrong — M1
    • \(-3, -2, -1, 0, 1\) — A1
  2. 2 Write down [5 marks]

    The diagram shows a shaded region \(R\). (a) Write down the three inequalities that define the region \(R\). (3 marks) (b) Write down the number of points with integer coordinates inside or on the boundary of \(R\). (2 marks)

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

    Model answer

    (a) \(x \geq 2\), \(y \geq 1\) and \(x + y \leq 7\). (b) For \(x = 2, 3, 4, 5, 6\) there are \(5, 4, 3, 2, 1\) points, so the total is \(5 + 4 + 3 + 2 + 1 = 15\).

    Mark scheme

    • (a) \(x \geq 2\) — B1
    • (a) \(y \geq 1\) — B1
    • (a) \(x + y \leq 7\) — B1
    • (b) Counts by columns, such as 5, 4, 3, 2, 1 — M1
    • (b) 15 — A1
  3. 3 Solve [2 marks]

    Solve the inequality \(3x - 2 > 10\).

    Show answerHide answer

    Model answer

    Add 2: \(3x > 12\). Divide by 3: \(x > 4\).

    Mark scheme

    • \(3x > 12\) — M1
    • \(x > 4\) — A1
  4. 4 Solve [3 marks]

    Solve the inequality \(x^2 - 2x - 8 < 0\).

    Show answerHide answer

    Model answer

    \(x^2 - 2x - 8 = (x - 4)(x + 2)\), with roots \(-2\) and 4. The curve is below the \(x\)-axis between the roots, so \(-2 < x < 4\).

    Mark scheme

    • \((x - 4)(x + 2)\) or the roots \(-2\) and 4 — M1
    • A sketch or a statement that the quadratic is negative between the roots — M1
    • \(-2 < x < 4\) — A1
  5. 5 Solve [4 marks]

    (a) Solve \(x^2 \geq 16\). (2 marks) (b) Solve \(x^2 - 5x + 6 > 0\). (2 marks)

    Show answerHide answer

    Model answer

    (a) \(x^2 = 16\) gives \(x = \pm 4\), and the curve is above the axis outside the roots, so \(x \leq -4\) or \(x \geq 4\). (b) \((x - 2)(x - 3) > 0\), so \(x < 2\) or \(x > 3\).

    Mark scheme

    • (a) Roots \(\pm 4\) — M1
    • (a) \(x \leq -4\) or \(x \geq 4\) — A1
    • (b) Roots 2 and 3 — M1
    • (b) \(x < 2\) or \(x > 3\) — A1
  6. 6 Work out [3 marks]

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

    Show answerHide answer

    Model answer

    Subtract 1: \(-3 < 2n \leq 8\). Divide by 2: \(-1.5 < n \leq 4\). The integers are \(-1, 0, 1, 2, 3, 4\).

    Mark scheme

    • \(-3 < 2n \leq 8\) or \(-1.5 < n \leq 4\) — M1
    • At least four correct integers — M1
    • \(-1, 0, 1, 2, 3, 4\) — 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 \(\sqrt{75}\).

    Show answerHide answer

    Model answer

    \(\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}\).

    Mark scheme

    • \(\sqrt{25 \times 3}\) — M1
    • \(5\sqrt{3}\) — A1
  2. 2 Show that [3 marks]

    Show that \(\sqrt{18} + \sqrt{8} = 5\sqrt{2}\).

    Show answerHide answer

    Model answer

    \(\sqrt{18} = 3\sqrt{2}\) and \(\sqrt{8} = 2\sqrt{2}\), so the sum is \(3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2}\).

    Mark scheme

    • \(\sqrt{18} = 3\sqrt{2}\) or \(\sqrt{8} = 2\sqrt{2}\) — M1
    • Both simplified correctly — M1
    • \(3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2}\) with a conclusion — C1
  3. 3 Expand [3 marks]

    Expand and simplify \((2 + \sqrt{3})(4 - \sqrt{3})\).

    Show answerHide answer

    Model answer

    \(8 - 2\sqrt{3} + 4\sqrt{3} - 3 = 5 + 2\sqrt{3}\).

    Mark scheme

    • At least three of the four terms correct — M1
    • \(8 - 2\sqrt{3} + 4\sqrt{3} - 3\) — M1
    • \(5 + 2\sqrt{3}\) — A1
  4. 4 Rationalise [2 marks]

    Rationalise the denominator of \(\dfrac{14}{\sqrt{7}}\). Give your answer in its simplest form.

    Show answerHide answer

    Model answer

    \(\dfrac{14}{\sqrt{7}} \times \dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac{14\sqrt{7}}{7} = 2\sqrt{7}\).

    Mark scheme

    • Multiplies the top and bottom by \(\sqrt{7}\) — M1
    • \(2\sqrt{7}\) — A1
  5. 5 Rationalise [3 marks]

    Write \(\dfrac{6}{3 + \sqrt{3}}\) in the form \(a - \sqrt{b}\), where \(a\) and \(b\) are integers.

    Show answerHide answer

    Model answer

    Multiply the top and bottom by \(3 - \sqrt{3}\): \(\dfrac{6(3 - \sqrt{3})}{9 - 3} = \dfrac{6(3 - \sqrt{3})}{6} = 3 - \sqrt{3}\).

    Mark scheme

    • Multiplies by \(3 - \sqrt{3}\) — M1
    • Denominator \(9 - 3 = 6\) — M1
    • \(3 - \sqrt{3}\) — A1
  6. 6 Work out [4 marks]

    A right-angled triangle has shorter sides of length \(\sqrt{2}\) cm and \(\sqrt{18}\) cm. (a) Work out the area of the triangle. (2 marks) (b) Work out the length of the hypotenuse. Give your answer in the form \(a\sqrt{b}\). (2 marks)

    Show answerHide answer

    Model answer

    (a) \(\dfrac{1}{2} \times \sqrt{2} \times \sqrt{18} = \dfrac{1}{2} \times \sqrt{36} = 3\) cm\(^2\). (b) \(\sqrt{2 + 18} = \sqrt{20} = 2\sqrt{5}\) cm.

    Mark scheme

    • (a) \(\dfrac{1}{2} \times \sqrt{2} \times \sqrt{18}\) — M1
    • (a) 3 cm\(^2\) — A1
    • (b) \(\sqrt{20}\) — M1
    • (b) \(2\sqrt{5}\) — 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\(2\sqrt{2}\)
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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\(\dfrac{2 - \sqrt{3}}{7}\)
    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 two consecutive odd numbers is a multiple of 4.

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    Model answer

    Let the odd numbers be \(2n + 1\) and \(2n + 3\). Their sum is \(4n + 4 = 4(n + 1)\), which is a multiple of 4.

    Mark scheme

    • \(2n + 1\) and \(2n + 3\) — M1
    • \(4n + 4\) — M1
    • \(4(n + 1)\) with a conclusion — C1
  2. 2 Simplify [2 marks]

    Simplify \(\dfrac{x^2 - 16}{x - 4}\).

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    Model answer

    \(\dfrac{(x - 4)(x + 4)}{x - 4} = x + 4\).

    Mark scheme

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

    Solve \(\dfrac{x + 3}{2} + \dfrac{x - 1}{4} = 5\).

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    Model answer

    Multiply every term by 4: \(2(x + 3) + (x - 1) = 20\). Then \(3x + 5 = 20\), so \(x = 5\).

    Mark scheme

    • \(2(x + 3) + (x - 1) = 20\) — M1
    • \(3x + 5 = 20\) — M1
    • \(x = 5\) — A1
  4. 4 Simplify [3 marks]

    Simplify \(\dfrac{x^2 + x - 6}{x^2 - 9}\).

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    Model answer

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

    Mark scheme

    • \((x + 3)(x - 2)\) or \((x - 3)(x + 3)\) — M1
    • Both factorised — M1
    • \(\dfrac{x - 2}{x - 3}\) — A1
  5. 5 Show that [3 marks]

    Show that \((n + 3)^2 - (n - 3)^2\) is a multiple of 12 for every integer \(n\).

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    Model answer

    \((n + 3)^2 = n^2 + 6n + 9\) and \((n - 3)^2 = n^2 - 6n + 9\). The difference is \(12n\), which is a multiple of 12.

    Mark scheme

    • \(n^2 + 6n + 9\) or \(n^2 - 6n + 9\) — M1
    • \(12n\) — M1
    • States that \(12n\) is a multiple of 12 — C1
  6. 6 Write [3 marks]

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

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    Model answer

    Use the common denominator \((x + 2)(x - 1)\): \(\dfrac{(x - 1) + (x + 2)}{(x + 2)(x - 1)} = \dfrac{2x + 1}{(x + 2)(x - 1)}\).

    Mark scheme

    • Common denominator \((x + 2)(x - 1)\) — M1
    • Numerator \((x - 1) + (x + 2)\) — M1
    • \(\dfrac{2x + 1}{(x + 2)(x - 1)}\) — 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\(n^2\)
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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\(\dfrac{5x + 6}{3x + 2}\)
    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: \(4n\)

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