Solving Equations and Inequalities

1. 1

   Solve $5x - 3 = 2x + 9$.

   • A

     $x = 2$

   • B

     $x = 4$

   • C

     $x = 6$

   • D

     $x = 12$

   Check answer

   Answer & explanation

   Correct: option B

   Collect $x$ terms: $5x - 2x = 9 + 3 \implies 3x = 12 \implies x = 4$.

2. 2

   Which of the following is the correct solution of $x^2 - 7x + 10 = 0$?

   • A

     $x = 2$ or $x = 5$

   • B

     $x = -2$ or $x = -5$

   • C

     $x = 1$ or $x = 10$

   • D

     $x = 2$ or $x = -5$

   Check answer

   Answer & explanation

   Correct: option A

   Factorise: $(x - 2)(x - 5) = 0 \implies x = 2$ or $x = 5$. Check: $4 - 14 + 10 = 0$ ✓ and $25 - 35 + 10 = 0$ ✓.

3. 3

   What is the discriminant of $2x^2 - 3x + 5$?

   • A

     $31$

   • B

     $-31$

   • C

     $49$

   • D

     $-49$

   Check answer

   Answer & explanation

   Correct: option B

   $b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 - 40 = -31$. Since the discriminant is negative, the equation has no real roots.

4. 4

   Solve the simultaneous equations $x + y = 7$ and $2x - y = 5$.

   • A

     $x = 3, y = 4$

   • B

     $x = 4, y = 3$

   • C

     $x = 6, y = 1$

   • D

     $x = 2, y = 5$

   Check answer

   Answer & explanation

   Correct: option B

   Adding the two equations: $3x = 12 \implies x = 4$. Substituting: $4 + y = 7 \implies y = 3$.

5. 5

   Which solution set correctly represents $3x + 1 > 10$?

   • A

     $x > 3$

   • B

     $x \geq 3$

   • C

     $x > 4$

   • D

     $x < 3$

   Check answer

   Answer & explanation

   Correct: option A

   $3x + 1 > 10 \implies 3x > 9 \implies x > 3$. The inequality is strict ($>$), so $x = 3$ is not included.

6. 6

   The iterative formula $x_{n+1} = \sqrt{3x_n + 1}$ is used to find a root of an equation. Starting with $x_0 = 3$, what is $x_1$?

   • A

     $\sqrt{10}$

   • B

     $\sqrt{7}$

   • C

     $4$

   • D

     $10$

   Check answer

   Answer & explanation

   Correct: option A

   $x_1 = \sqrt{3(3) + 1} = \sqrt{9 + 1} = \sqrt{10} \approx 3.162$.

7. 7

   What are the solution sets for $x^2 - x - 6 < 0$?

   • A

     $x < -2$ or $x > 3$

   • B

     $-2 < x < 3$

   • C

     $x < 2$ or $x > -3$

   • D

     $-3 < x < 2$

   Check answer

   Answer & explanation

   Correct: option B

   Roots of $x^2 - x - 6 = 0$: $(x - 3)(x + 2) = 0 \implies x = -2$ or $x = 3$. The parabola opens upward and is below zero between the roots, so $-2 < x < 3$.

8. 8

   Which method is most efficient for solving $x^2 + 6x + 9 = 0$?

   • A

     Quadratic formula

   • B

     Completing the square

   • C

     Factorising (perfect square)

   • D

     Iterative method

   Check answer

   Answer & explanation

   Correct: option C

   $x^2 + 6x + 9 = (x + 3)^2 = 0 \implies x = -3$ (repeated root). Recognising the perfect square is the quickest approach.

Exam-style multi-part questions. Try each part on paper, then reveal the model answer to compare your working.

1. Q17 marks total

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   Scenario

   A rectangle has length $(3x + 2)$ cm and width $(x - 1)$ cm.

   1. a

      Write an expression for the area of the rectangle and expand it fully.

      [2]
      Show answer

      Model answer · 2 marks

      Area $= (3x + 2)(x - 1) = 3x^2 - 3x + 2x - 2 = 3x^2 - x - 2$ cm²

   2. b

      Given that the area of the rectangle is 12 cm², form a quadratic equation and solve it to find the value of $x$.

      [4]
      Show answer

      Model answer · 4 marks

      $3x^2 - x - 2 = 12 \implies 3x^2 - x - 14 = 0$. Using the quadratic formula or factorising: $(3x + 7)(x - 2) = 0 \implies x = 2$ (taking the positive value since $x - 1 > 0$). Check: length $= 8$ cm, width $= 1$ cm, area $= 8$ cm² — wait, area must be 12 cm²: length $= 8$, width $= 1$ gives 8, not 12. Recheck: $3(4) - 2 - 2 = 8 \neq 12$. Correct factorisation: $3x^2 - x - 14 = (3x - 7)(x + 2) = 0 \implies x = \frac{7}{3}$ or $x = -2$. Since $x - 1 > 0$, $x > 1$, so $x = \frac{7}{3}$. Length $= 3(\frac{7}{3}) + 2 = 9$ cm, width $= \frac{7}{3} - 1 = \frac{4}{3}$ cm, area $= 9 \times \frac{4}{3} = 12$ cm² ✓

   3. c

      Explain why the other solution for $x$ cannot be used.

      [1]
      Show answer

      Model answer · 1 mark

      $x = -2$ gives width $= -2 - 1 = -3$ cm, which is negative and therefore not a valid length.

   Examiner note

   Part (b) requires setting up the equation from context, factorising or applying the formula, and rejecting the negative solution with a reason.

2. Q26 marks total

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   Scenario

   The equation $x^3 - 4x + 1 = 0$ has a root in the interval $(1, 2)$.

   1. a

      Show that there is a root in the interval $(1, 2)$ by evaluating $f(1)$ and $f(2)$.

      [2]
      Show answer

      Model answer · 2 marks

      $f(1) = 1 - 4 + 1 = -2 < 0$ and $f(2) = 8 - 8 + 1 = 1 > 0$. There is a change of sign, so there is a root in the interval $(1, 2)$.

   2. b

      Use the iterative formula $x_{n+1} = \dfrac{x_n^3 + 1}{4}$ with $x_0 = 1.5$ to find the root correct to 2 decimal places. Show your iterations.

      [3]
      Show answer

      Model answer · 3 marks

      $x_1 = \frac{(1.5)^3 + 1}{4} = \frac{4.375}{4} = 1.09375$; $x_2 = \frac{(1.09375)^3 + 1}{4} = \frac{2.3097...}{4} \approx 0.5774$; continue iterating until the sequence converges. The root is $x \approx 0.25$ (the iteration shown converges to a different root near 0.254 — if the starting value and formula lead to the root near $x = 1.86$, use $x_0 = 1.8$ instead). Root $\approx 1.86$ (2 d.p.) for the root in $(1, 2)$.

   3. c

      How can you verify that your answer is correct to 2 decimal places?

      [1]
      Show answer

      Model answer · 1 mark

      Evaluate $f(1.855)$ and $f(1.865)$; if one is negative and one positive, the root lies between them and rounds to 1.86 to 2 d.p.

   Examiner note

   The change-of-sign check in part (a) must include numerical values and the statement about the sign change.

3. Q36 marks total

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   Scenario

   Solve these simultaneous equations: $y = x + 3$ $x^2 + y^2 = 29$

   1. a

      Solve the simultaneous equations algebraically.

      [5]
      Show answer

      Model answer · 5 marks

      Substitute $y = x + 3$ into $x^2 + y^2 = 29$: $x^2 + (x+3)^2 = 29 \implies x^2 + x^2 + 6x + 9 = 29 \implies 2x^2 + 6x - 20 = 0 \implies x^2 + 3x - 10 = 0 \implies (x+5)(x-2) = 0$. So $x = -5, y = -2$ or $x = 2, y = 5$.

   2. b

      Interpret your solutions geometrically.

      [1]
      Show answer

      Model answer · 1 mark

      The solutions are the two points of intersection of the line $y = x + 3$ with the circle $x^2 + y^2 = 29$ (radius $\sqrt{29}$, centred at the origin).

   Examiner note

   Full marks in part (a) require showing the substitution step, expanding and simplifying to a standard quadratic, factorising, and giving both solution pairs correctly matched.