S4 Mathematics Practice Paper: Quadratic Equations and Functions

Solve the quadratic equation 2x² − 7x + 3 = 0.

(a) Factorise the left-hand side by the cross method. (3 marks)
(b) Hence find the two roots of the equation. (2 marks)
(c) Verify that the sum and the product of the roots equal −b/a and c/a. (2 marks)

(a) 2x² − 7x + 3 = (2x − 1)(x − 3): the two numbers with product 2 × 3 = 6 and sum −7 are −1 and −6, so splitting the middle term and grouping gives (2x − 1)(x − 3). (3 marks)
(b) 2x − 1 = 0 or x − 3 = 0, so x = 0.5 or x = 3. (2 marks)
(c) Sum of roots = 0.5 + 3 = 3.5 = −b/a = 7/2; product = 0.5 × 3 = 1.5 = c/a = 3/2. Both check out. (2 marks)

The two roots of a quadratic equation are 4 and −3.

(a) Find the sum and the product of the roots. (3 marks)
(b) Hence write down the quadratic equation with leading coefficient 1. (2 marks)
(c) If the roots are α and β, find α² + β². (2 marks)

(a) Sum = 4 + (−3) = 1; product = 4 × (−3) = -12. (3 marks)
(b) The equation is x² − (sum)x + (product) = 0, i.e. x² − x − 12 = 0. (2 marks)
(c) α² + β² = (α + β)² − 2αβ = 1² − 2(-12) = 1 + 24 = 25. (2 marks)

The quadratic equation x² + kx + 9 = 0 has a repeated root.

(a) Find all possible values of k. (3 marks)
(b) When k = −6, find the repeated root. (2 marks)
(c) When k = 6, find the repeated root, and state how many times the graph meets the x-axis in each case. (2 marks)

(a) A repeated root means the discriminant is zero: k² − 4(1)(9) = 0, k² = 36, so k = 6 or k = -6. (3 marks)
(b) When k = −6 the equation is x² − 6x + 9 = 0, i.e. (x − 3)² = 0, so the repeated root is x = 3. (2 marks)
(c) When k = 6 the equation is x² + 6x + 9 = 0, i.e. (x + 3)² = 0, so the repeated root is x = -3. In both cases the discriminant is zero, so each graph meets the x-axis at exactly one point (it is tangent to the axis). (2 marks)

Consider the quadratic function y = x² + 8x + 5.

(a) By completing the square, write the function as y = (x + h)² + k. (3 marks)
(b) Hence write down the coordinates of the vertex and the minimum value. (2 marks)
(c) Solve x² + 8x + 5 = 0, giving answers correct to 3 significant figures. (2 marks)

(a) x² + 8x + 5 = (x + 4)² − 16 + 5 = (x + 4)² − 11. (3 marks)
(b) The vertex is (-4, -11); since (x + 4)² ≥ 0, the minimum value is -11. (2 marks)
(c) (x + 4)² = 11, x + 4 = ±√11, so x = -0.683 or x = -7.317. (2 marks)

The sum of two numbers is 14 and their product is 45.

(a) Let one number be x, and form a quadratic equation in x. (3 marks)
(b) Solve the equation and find the two numbers. (3 marks)
(c) If the sum stays 14 but the product becomes 60, does the equation have real roots? Explain using the discriminant. (2 marks)

(a) The other number is 14 − x; since the product is 45, x(14 − x) = 45, i.e. x² − 14x + 45 = 0. (3 marks)
(b) Discriminant = 196 − 180 = 16, x = (14 ± 4) / 2, giving x = 9 or 5. The numbers are 5 and 9. (3 marks)
(c) The new equation is x² − 14x + 60 = 0 with discriminant 196 − 240 = −44 < 0, so it has no real roots: with a fixed sum of 14 the product cannot reach 60 (the maximum is 7 × 7 = 49). (2 marks)

Consider the graph of y = x² − 4x + 1.

(a) Write the function in vertex form by completing the square. (2 marks)
(b) State the vertex, the axis of symmetry and the minimum value. (2 marks)
(c) Find the y-intercept of the graph. (2 marks)
(d) Find the x-intercepts, correct to 3 significant figures. (2 marks)

(a) y = x² − 4x + 1 = (x − 2)² − 4 + 1 = (x − 2)² − 3. (2 marks)
(b) Vertex (2, -3), axis of symmetry x = 2, minimum value -3. (2 marks)
(c) Put x = 0: y = 1, so the y-intercept is 1. (2 marks)
(d) (x − 2)² = 3, x = 2 ± √3, so the x-intercepts are 3.732 and 0.268. (2 marks)

Let α and β be the roots of x² − 3x − 5 = 0.

(a) Without solving, write down α + β and αβ. (3 marks)
(b) Hence find α² + β². (3 marks)
(c) Find 1/α + 1/β. (2 marks)

(a) By the roots-coefficients relations: α + β = 3, αβ = -5. (3 marks)
(b) α² + β² = (α + β)² − 2αβ = 3² − 2(-5) = 9 + 10 = 19. (3 marks)
(c) 1/α + 1/β = (α + β) / (αβ) = 3 / (-5) = -0.6. (2 marks)

Solve the following quadratic inequalities and give the answers in inequality or interval form.

(a) x² − 5x + 6 < 0 (4 marks)
(b) x² − 2x − 8 ≥ 0 (4 marks)

(a) x² − 5x + 6 = (x − 2)(x − 3), roots 2 and 3. The parabola opens upwards, so it is negative between the roots: 2 < x < 3, i.e. (2, 3). (4 marks)
(b) x² − 2x − 8 = (x − 4)(x + 2), roots 4 and −2. The parabola opens upwards, so it is non-negative outside the roots: x ≤ −2 or x ≥ 4. (4 marks)

A rectangular garden has perimeter 40 m and area 96 m².

(a) Let the width be w m, form an equation and find the length and width. (4 marks)
(b) If the perimeter stays 40 m, find the greatest possible area. (3 marks)
(c) Use completing the square or the graph to explain why the greatest area occurs for a square. (3 marks)

(a) Length + width = 20, so length = (20 − w) m. Then w(20 − w) = 96, i.e. w² − 20w + 96 = 0. Discriminant = 400 − 384 = 16, w = (20 ± 4) / 2 = 8 or 12. The garden is 8 m by 12 m. (4 marks)
(b) With width w, area A = w(20 − w) = −(w − 10)² + 100, so the greatest area is 100 m². (3 marks)
(c) After completing the square A = 100 − (w − 10)²; since (w − 10)² ≥ 0, A is greatest only when w = 10, where length = width = 10 m, i.e. a square. So for a fixed perimeter the square has the greatest area. (3 marks)

A small shop sells x items per month. Its monthly profit P, in hundreds of dollars, satisfies P = −2x² + 40x − 150, where x is a positive integer.

(a) Find the break-even values of x (where P = 0). (3 marks)
(b) Find the maximum monthly profit and the corresponding sales. (3 marks)
(c) If the shop can sell at most 8 items per month, find the maximum profit then, and explain why the answer in (b) cannot be reached. (4 marks)

(a) Put P = 0: −2x² + 40x − 150 = 0; dividing by −2 gives x² − 20x + 75 = 0, i.e. (x − 5)(x − 15) = 0, so x = 15 or x = 5. (3 marks)
(b) Completing the square: P = −2(x² − 20x) − 150 = −2[(x − 10)² − 100] − 150 = −2(x − 10)² + 50. So the profit is greatest when x = 10, at 50 hundred dollars (HK$5000). (3 marks)
(c) The restriction x ≤ 8 lies to the left of the vertex, where P is increasing, so the maximum under the cap is at x = 8: P(8) = −2(64) + 320 − 150 = 42 hundred dollars (HK$4200). The vertex value of 50 hundred dollars cannot be reached because at most 8 items can be sold. (4 marks)

The curve y = x² + (k − 2)x + 9 touches the x-axis (it has exactly one intersection point).

(a) Find all possible values of k. (4 marks)
(b) When k = 8, find the coordinates of the point of contact. (3 marks)
(c) When k = −4, find the point of contact and compare the axes of symmetry of the two curves. (3 marks)

(a) Tangency means the discriminant is zero: (k − 2)² − 4(1)(9) = 0, (k − 2)² = 36, k − 2 = ±6, so k = 8 or k = -4. (4 marks)
(b) When k = 8, y = x² + 6x + 9 = (x + 3)², so the point of contact is (-3, 0). (3 marks)
(c) When k = −4, y = x² − 6x + 9 = (x − 3)², so the point of contact is (3, 0). The axes of symmetry are x = -3 and x = 3, mirror images about the y-axis. (3 marks)

Solve the simultaneous equations y = x² − 2x and y = 2x − 3.

(a) Use substitution to find the values of x. (4 marks)
(b) Find the coordinates of the intersection points. (3 marks)
(c) Hence describe the geometric relationship between the line y = 2x − 3 and the parabola y = x² − 2x. (3 marks)

(a) Substituting: x² − 2x = 2x − 3, i.e. x² − 4x + 3 = 0, (x − 1)(x − 3) = 0, so x = 1 or x = 3. (4 marks)
(b) When x = 1, y = -1; when x = 3, y = 3. The points are (1, -1) and (3, 3). (3 marks)
(c) Two distinct real roots mean the line cuts the parabola at two points: it is a secant (a chord) of the parabola, not a tangent. (3 marks)