S4 Mathematics Paper: Quadratic Equations, Graphs And Variation (100 Marks)

S4 Mathematics Paper: Quadratic Equations, Graphs and Variation (100 marks)

2026-11-11

S4 Mathematics – quadratic equations and graphs

S4 Mathematics paper on quadratic equations, graphs, variation, logarithms, compound interest, statistics and probability. Answers with working and marking notes.

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Paper details　S4 Mathematics practice paper with six questions on solving quadratic equations, discriminants, quadratic graphs and translations, joint and inverse variation, logarithms, compound interest and simple statistics. Full answers with working and marking notes included.

Section A: Solving Quadratic Equations and the Discriminant (15 marks)

Let a, b and c be real numbers. For the quadratic equation ax² + bx + c = 0 with a not equal to zero, the discriminant is b² – 4ac. A positive discriminant gives two distinct real roots, zero gives a repeated root and a negative value gives no real root. The quadratic formula is x = (-b ± √(b² – 4ac)) / (2a).

#	Question	Answer and marking notes
1	Complete the square to write the left side of x² – 6x + 4 = 0 in the form (x – 3)², then solve the equation to two decimal places.	Answer: (x – 3)² = 5, so x = 3 ± √5, that is x is about 5.24 or x is about 0.76. (Marking: 1 mark for completing the square, 1 mark for the square root, 1 mark for both roots.)
2	Find the discriminant of 2x² + 5x – 3 = 0, state the nature of its roots, then solve the equation.	Answer: the discriminant is 5² – 4(2)(-3) = 25 + 24 = 49, which is positive and a perfect square, so there are two distinct rational roots; x = (-5 ± 7) / 4, that is x = 0.5 or x = -3.0. (Marking: 1 mark discriminant, 1 mark nature of roots, 1 mark both roots.)
3	Given that x² + kx + 9 = 0 has a repeated root, find all possible values of k.	Answer: a repeated root requires the discriminant k² – 36 = 0, so k = 6 or k = -6. (Marking: 2 marks for setting the discriminant to zero, 1 mark for each value.)
4	Let α and β be the roots of 3x² – 7x + 2 = 0. Find α + β, αβ and α² + β², giving the last value as a fraction in lowest terms and as a decimal.	Answer: α + β = 7/3 and αβ = 2/3; α² + β² = (α + β)² – 2αβ = 49/9 – 4/3 = 37/9, about 4.11. (Marking: 2 marks sum and product, 2 marks for the identity, 1 mark answer.)

Section B: Quadratic Graphs and Translations (16 marks)

Consider the function y = -x² + 4x + 5. Its graph is a parabola opening downwards, with a vertical axis of symmetry x = h and vertex (h, k).

#	Question	Answer and marking notes
5	Complete the square to write y = -x² + 4x + 5 as y = -(x – h)² + k and state the coordinates of the vertex.	Answer: y = -(x – 2)² + 9, so the vertex is (2, 9). (Marking: 2 marks for completing the square, 2 marks vertex.)
6	Write down the equation of the axis of symmetry and the coordinates of the y-intercept.	Answer: the axis of symmetry is x = 2; when x = 0, y = 5, so the intercept is (0, 5). (Marking: 2 marks axis, 2 marks intercept.)
7	Find the two x-intercepts of the graph and state the range of x for which y is greater than 0.	Answer: solving -x² + 4x + 5 = 0 gives x = 5 or x = -1, so the intercepts are (5, 0) and (-1, 0); the graph opens downwards, so y is greater than 0 when x lies between -1 and 5. (Marking: 2 marks intercepts, 2 marks range.)
8	The graph is translated 3 units to the right. Write the function of the new graph and the coordinates of its vertex.	Answer: the new function is y = -(x – 5)² + 9 and the new vertex is (5, 9). (Marking: 2 marks for the direction, 2 marks function and vertex.)

Section C: Joint and Inverse Variation (17 marks)

Suppose y varies directly as the square of x and inversely as z, so y = kx² / z where k is a non-zero constant. When x = 3 and z = 2, y = 18.

#	Question	Answer and marking notes
9	Find the value of the constant k and write the equation connecting y, x and z.	Answer: 18 = k(3²)/2 = 4.5k, so k = 4 and the equation is y = 4x² / z. (Marking: 2 marks substitution, 2 marks k and equation.)
10	Find y when x = 5 and z = 4.	Answer: y = 4(5²)/4 = 4 x 25 / 4 = 25. (Marking: 2 marks substitution, 2 marks answer.)
11	Find the positive value of x when y = 50 and z = 8.	Answer: 50 = 4x²/8, so x² = 100 and x = 10. (Marking: 2 marks for simplifying, 2 marks for the square root, 1 mark for taking the positive value.)
12	If x increases by 20 per cent and z decreases by 20 per cent, by what percentage does y change?	Answer: x² becomes 1.44 times and z becomes 0.8 times the original, so y becomes 1.44 / 0.8 = 1.8 times, an increase of 80 per cent. (Marking: 1 mark for each multiplier, 2 marks for the conclusion.)

Section D: Applications of Quadratic Equations (18 marks)

A rectangular garden has a perimeter of 34 m and an area of 60 m². Let the width be w metres and the length be (17 – w) metres.

#	Question	Answer and marking notes
13	Express the length in terms of w and show that w satisfies w² – 17w + 60 = 0.	Answer: the length is (17 – w) metres; from w(17 – w) = 60 we get 17w – w² = 60, so w² – 17w + 60 = 0. (Marking: 2 marks for the length, 2 marks for the derivation.)
14	Solve the equation and state the length and width of the garden.	Answer: w² – 17w + 60 = 0 factorises as (w – 5)(w – 12) = 0, so w = 5 or w = 12; the width is 5 m and the length is 12 m. (Marking: 2 marks factorising, 2 marks roots, 1 mark dimensions.)
15	Fencing costs 45 dollars per metre. Find the cost of fencing the whole perimeter once.	Answer: the cost is 34 x 45 = 1530 dollars. (Marking: 2 marks for using the perimeter, 2 marks answer.)
16	If the perimeter stays the same but the area becomes 72 m², find the new length and width.	Answer: w² – 17w + 72 = 0 has discriminant 289 – 288 = 1, so w = (17 ± 1) / 2 = 9 or 8; the new size is 8 m by 9 m. (Marking: 2 marks equation, 2 marks discriminant, 1 mark dimensions.)

Section E: Logarithms and Compound Interest (17 marks)

The laws of logarithms include log(ab) = log a + log b and log(a / b) = log a – log b. Compound interest is given by A = P(1 + r)^n, where P is the principal, r the interest rate per period and n the number of periods.

#	Question	Answer and marking notes
17	Evaluate log₂ 16 + log₃ (1/9).	Answer: log₂ 16 = 4 and log₃ (1/9) = -2, so the total is 2. (Marking: 2 marks for each term, 1 mark for the total.)
18	Solve 2^(x + 1) = 32.	Answer: 32 = 2⁵, so x + 1 = 5 and x = 4. (Marking: 2 marks for the same base, 2 marks for equating indices, 1 mark answer.)
19	A person deposits 20000 dollars at 5 per cent per year compounded yearly. Using logarithms, find the least whole number of years after which the balance exceeds 30000 dollars, and the balance at that time, correct to two decimal places.	Answer: we need 20000(1.05)^n to exceed 30000, that is (1.05)^n above 1.5; n is above log 1.5 / log 1.05 which is about 8.31, so the least whole number of years is 9; the balance is 20000(1.05)^9, about 31026.56 dollars, while the eighth year gives only 29549.11 dollars. (Marking: 2 marks formulation, 2 marks logarithms, 2 marks n, 1 mark balance.)

Section F: Statistical Measures and Probability (17 marks)

The marks of eight students in a mathematics test are 12, 15, 15, 18, 20, 22, 23 and 25.

#	Question	Answer and marking notes
20	Find the mean and the median of the marks.	Answer: the total is 150, so the mean is 150 / 8 = 18.75; the median is (18 + 20) / 2 = 19. (Marking: 1 mark total, 2 marks mean, 2 marks median.)
21	One student is chosen at random. Find the probability that the mark is above the mean.	Answer: the marks above 18.75 are 20, 22, 23 and 25, that is 4 marks, so the probability is 4 / 8 = 0.5. (Marking: 3 marks for the four marks, 3 marks probability.)
22	If the lowest mark 12 is changed to 20, find the new mean and the new median.	Answer: the new total is 150 – 12 + 20 = 158, so the new mean is 158 / 8 = 19.75; the two middle values are 20 and 20, giving a median of 20. (Marking: 2 marks total, 2 marks mean, 2 marks median.)

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