S4 Mathematics Practice Paper: Quadratic Equations and Quadratic Functions (12 Questions, 100 Marks, 75 Minutes, with Answers)
- 2026-11-09
12 questions, 100 marks and 75 minutes covering quadratic equations, the discriminant, completing the square, vertices and extreme values, with applied problems and marking points.
Instructions: This paper has three sections, A, B and C. Answer all questions. Total marks: 100. Time allowed: 75 minutes.
Show all working. Answers without working earn no method marks. Write down a formula before substituting numbers.
Give numerical answers to three significant figures unless stated otherwise, and state the correct units.
Spend about 35 minutes on Section A, then use the remaining time for Section B graphs and Section C applications.
Section A: Quadratic Equations and the Discriminant
Factorisation, the quadratic formula and the nature of roots
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | Solve the equation x² − 7x + 12 = 0 by factorisation and state the product of its two roots. | (a) x² − 7x + 12 = (x − 3)(x − 4), so x = 3 or x = 4 (6 marks). (b) The product of the roots is 3 × 4 = 12 (2 marks). |
| 2 | The equation kx² + 4x + 2 = 0 has two distinct real roots. Find the range of values of k. | (a) Discriminant = 4² − 4(k)(2) = 16 − 8k (3 marks). (b) Two distinct real roots require the discriminant to be greater than 0, so 16 − 8k is greater than 0 and k is less than 2 (4 marks). (c) The coefficient k of x² cannot be 0, so k is less than 2 and k is not 0 (1 mark). |
| 3 | If α and β are the roots of x² − 6x + 5 = 0, find the value of α² + β². | (a) α + β = 6 (2 marks). (b) αβ = 5 (2 marks). (c) α² + β² = (α + β)² − 2αβ = 36 − 10 = 26 (4 marks). |
| 4 | Solve 2x² + 5x − 3 = 0 using the quadratic formula and state the sum of the two roots. | (a) Discriminant = 5² − 4(2)(−3) = 25 + 24 = 49 (3 marks). (b) x = (−5 ± 7) / 4 (3 marks). (c) x = 1/2 or x = −3 (2 marks). (d) Sum of roots = −5/2 (1 mark). |
Section B: Graphs and Extreme Values of Quadratic Functions
Completing the square, vertex and axis of symmetry
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | Rewrite y = x² − 8x + 13 in the form y = (x − h)² + k, and state the vertex, the axis of symmetry and the minimum value. | (a) y = (x − 4)² − 3 (4 marks). (b) Vertex (4, −3) (2 marks). (c) Axis of symmetry x = 4 (1 mark). (d) Minimum value −3 (1 mark). |
| 6 | For y = −2x² + 12x − 13, find the maximum value and the value of x at which it occurs. | (a) y = −2(x² − 6x) − 13 = −2(x − 3)² + 18 − 13 = −2(x − 3)² + 5 (4 marks). (b) The maximum occurs at x = 3 (2 marks). (c) Maximum value = 5 (2 marks). |
| 7 | The graph of y = x² + 2x + k touches the x-axis at exactly one point. Find k, and describe the intersections with the x-axis when k is less than 1. | (a) Discriminant = 4 − 4k = 0, so k = 1 (4 marks). (b) When k is less than 1 the discriminant is greater than 0, so the graph cuts the x-axis at two points (4 marks). |
| 8 | The line y = mx + 1 intersects the curve y = x² − 4x + 5 at two points. Find the range of m. | (a) x² − 4x + 5 = mx + 1 gives x² − (4 + m)x + 4 = 0 (3 marks). (b) Discriminant = (4 + m)² − 16 must be greater than 0 (3 marks). (c) m² + 8m is greater than 0, so m is greater than 0 or m is less than −8 (3 marks). |
Section C: Applied and Integrated Problems
Maximum area, projectile motion and number problems
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A rectangular garden is enclosed by 40 m of fencing. Find its maximum possible area. | (a) Let the width be x, so the length is 20 − x (2 marks). (b) Area A = x(20 − x) = −(x − 10)² + 100 (4 marks). (c) Maximum area = 100 m² when x = 10 (2 marks). |
| 10 | A ball is thrown and its height is h = −5t² + 20t metres after t seconds. Find the greatest height and the time when the ball returns to the ground. | (a) h = −5(t − 2)² + 20 (3 marks). (b) Greatest height 20 m at t = 2 s (3 marks). (c) −5t² + 20t = 0 gives t(t − 4) = 0, so t = 4 s, rejecting t = 0 (2 marks). |
| 11 | Form the quadratic equation whose roots are 2 + √3 and 2 − √3. | (a) Sum of roots = 4 (3 marks). (b) Product of roots = 4 − 3 = 1 (3 marks). (c) The equation is x² − 4x + 1 = 0 (3 marks). |
| 12 | The product of two consecutive positive integers is 156. Find the two integers. | (a) Let the integers be n and n + 1, so n(n + 1) = 156 and n² + n − 156 = 0 (3 marks). (b) Discriminant = 1 + 624 = 625 and its square root is 25 (3 marks). (c) (n − 12)(n + 13) = 0, so n = 12, rejecting n = −13 (2 marks). (d) The integers are 12 and 13 (1 mark). |
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