S4 Mathematics Practice Paper: Quadratic Equations, Graphs and Applications

Solve the following quadratic equations, showing your steps.

(a) x² – 7x + 12 = 0

(b) 2x² + 5x – 3 = 0

(a) (x – 3)(x – 4) = 0, so x = 3 or x = 4.

(b) (2x – 1)(x + 3) = 0, so x = 1/2 or x = -3.

Marking: 1 mark for each correct factorisation, 1 mark for each pair of roots.

It is given that x² + kx + 9 = 0, where k is a constant.

(a) If the equation has a repeated root, find all possible values of k.

(b) If k = -10, find the roots and state the nature of the roots.

(a) A repeated root requires discriminant zero: k² – 36 = 0, so k = 6 or k = -6.

(b) When k = -10, x² – 10x + 9 = 0; discriminant = 100 – 36 = 64 > 0, so there are two distinct real roots; (x – 1)(x – 9) = 0, x = 1 or x = 9.

Marking: 2 marks for the discriminant condition, 2 for both values of k, 2 for the discriminant, 2 for the roots.

By completing the square, express y = x² – 6x + 5 in the form y = (x – h)² + k, and state the vertex, the axis of symmetry and the minimum value.

y = x² – 6x + 5 = (x – 3)² – 4.

Vertex (3, -4); axis of symmetry x = 3; minimum value -4.

Marking: 4 marks for completing the square, 2 for the vertex, 1 for the axis, 2 for the minimum.

Let α and β be the roots of 3x² – 7x + 2 = 0.

(a) Find α + β and αβ.

(b) Find the value of 1/α + 1/β.

(c) Hence find α and β.

(a) α + β = 7/3 and αβ = 2/3.

(b) 1/α + 1/β = (α + β)/(αβ) = (7/3) ÷ (2/3) = 7/2.

(c) 3x² – 7x + 2 = (3x – 1)(x – 2), so the roots are 1/3 and 2.

Marking: 1 mark each for sum and product, 2 for the substitution, 4 for factorising and both roots.

A ball is thrown upwards from a height of 1 m with a speed of 20 m/s. Its height h (in metres) after t seconds is h = -5t² + 20t + 1.

(a) Find the time taken to reach the maximum height and the maximum height.

(b) Find the time when the ball hits the ground, correct to 3 significant figures.

(a) h = -5(t – 2)² + 21, so the ball reaches the maximum height of 21 m at t = 2 s.

(b) Set h = 0: -5t² + 20t + 1 = 0, t = (20 + √420)/10 ≈ 4.05 s (reject the negative root).

Marking: 3 marks for completing the square, 1 for the maximum height, 4 for the equation and its solution, 1 for the positive root with units.

Solve the following quadratic inequalities.

(a) x² – 5x + 6 < 0

(b) x² – 2x – 8 ≥ 0

(a) x² – 5x + 6 = (x – 2)(x – 3) < 0, so 2 < x < 3.

(b) x² – 2x – 8 = (x – 4)(x + 2) ≥ 0, so x ≤ -2 or x ≥ 4.

Marking: 1 mark for each factorisation, 3 marks for each correct inequality answer.

Consider the function y = 2x² – 8x + 3.

(a) Find the axis of symmetry and the vertex of its graph.

(b) State the minimum value and the range of the function.

(c) Describe the transformation of the graph y = 2x² that gives this graph.

(a) y = 2(x – 2)² – 5, so the axis of symmetry is x = 2 and the vertex is (2, -5).

(b) Since a = 2 > 0, the minimum value is -5 and the range is y ≥ -5.

(c) The graph of y = 2x² is translated 2 units to the right and 5 units down.

Marking: 3 marks for completing the square, 2 for the axis and vertex, 2 for the minimum and range, 1 for the description.

A shop sells a stationery item at $(80 – x) each and sells (20 + 2x) items per month, where 0 ≤ x ≤ 60.

(a) Show that the monthly revenue R (in dollars) is R = -2x² + 140x + 1600.

(b) Find the value of x that maximises the revenue, and the maximum revenue.

(a) R = (80 – x)(20 + 2x) = 1600 + 160x – 20x – 2x² = -2x² + 140x + 1600.

(b) R = -2(x – 35)² + 4050, so the maximum revenue is $4050 when x = 35.

Marking: 3 marks for expanding, 3 for completing the square, 2 for the value of x, 1 for the maximum revenue.

Solve the simultaneous equations y = x² – 2x and y = x + 4.

x² – 2x = x + 4 gives x² – 3x – 4 = 0, (x – 4)(x + 1) = 0, so x = 4 or x = -1.

When x = 4, y = 8; when x = -1, y = 3. Solutions: (4, 8) and (-1, 3).

Marking: 3 marks for the quadratic equation, 2 for factorising, 1 mark for each solution, 2 for the y values.

The roots of a quadratic equation are 2 + √3 and 2 – √3.

(a) Find the sum and the product of the roots.

(b) Hence find the quadratic equation with leading coefficient 1.

(c) Find the discriminant of the equation and explain the result.

(a) Sum = 4 and product = (2 + √3)(2 – √3) = 1.

(b) The equation is x² – 4x + 1 = 0.

(c) Discriminant = 16 – 4 = 12 > 0, so the equation has two distinct real roots, consistent with the given roots.

Marking: 1 mark each for sum and product, 4 for the equation, 2 for the discriminant and the explanation.

Solve x⁴ – 5x² + 4 = 0.

Let u = x². Then u² – 5u + 4 = 0, (u – 1)(u – 4) = 0, so u = 1 or u = 4.

If x² = 1, x = ±1; if x² = 4, x = ±2. Hence x = ±1 or x = ±2.

Marking: 2 marks for the substitution, 2 for solving in u, 1 mark for each of the four roots.

A rectangular garden is fenced with 40 m of fencing. One side of the garden is against a wall and needs no fencing.

(a) Let the length of each side perpendicular to the wall be x m. Express the area A (in m²) in terms of x.

(b) Find the maximum area and the corresponding dimensions.

(a) The two sides perpendicular to the wall are x m each and the side parallel to the wall is (40 – 2x) m, so A = x(40 – 2x) = -2x² + 40x, where 0 < x < 20.

(b) A = -2(x – 10)² + 200, so the maximum area is 200 m² when x = 10, giving dimensions 10 m by 20 m.

Marking: 3 marks for the expression, 3 for completing the square, 2 for the value of x, 1 for the area and dimensions.