AQA A-Level Mathematics: Quadratics, Simultaneous Equations and Inequalities — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Quadratics, Simultaneous Equations and Inequalities study guide and revision notes.

Section A

1. Express 2x² + 8x + 3 in the form a(x + p)² + q, and state the minimum value. [4]

2. Use the discriminant to determine the number of real roots of 3x² − 2x + 5. [3]

3. Solve x² + 3x − 10 ≤ 0, giving your answer using “and”/“or” notation. [3]

Section B

4. The line y = x + 4 intersects the curve y = x² − 2x − 1 at two points.

(a) Show that the x-coordinates of the points of intersection satisfy x² − 3x − 5 = 0. [2]

(b) Use the discriminant to confirm that there are two distinct points of intersection, without solving the equation fully. [3]

5. The line y = mx − 3 is a tangent to the curve y = x² + x − 1.

(a) Show that x² + (1 − m)x + 2 = 0. [2]

(b) Find the two possible values of m. [5]

6. The quadratic 4x² − kx + 9 has a repeated root.

(a) State the condition on the discriminant for a repeated root. [1]

(b) Find the possible values of k. [4]

7. Solve x − 2 ≥ 5/x for x > 0, giving your answer as an inequality. [5]

Section C

8. The curve y = x² − 6x + p lies entirely above the x-axis. Find the range of possible values of p. [3]

9. Solve the simultaneous equations y = 2x − 1 and y = x² − 4x + 7, giving your answers as coordinate pairs. [5]

Answers

1. 2(x² + 4x) + 3 [1] = 2[(x + 2)² − 4] + 3 [1] = 2(x + 2)² − 8 + 3 [1] = 2(x + 2)² − 5, minimum value −5 (at x = −2) [1]. Forgetting to multiply the −4 by the leading coefficient 2 is the classic slip here.

2. Discriminant = b² − 4ac = (−2)² − 4(3)(5) = 4 − 60 = −56 [1]. Since the discriminant is negative [1], the quadratic has no real roots [1].

3. Factorise: (x + 5)(x − 2) ≤ 0 [1]. Critical values x = −5 and x = 2 [1]. Upward parabola, so the expression is ≤ 0 between the roots (inclusive, since ≤): −5 ≤ x ≤ 2 [1]. A common error is writing this as “x ≤ −5 or x ≥ 2”, which describes the wrong (outside) region for a “≤ 0” quadratic.

4. (a) x + 4 = x² − 2x − 1 [1] 0 = x² − 2x − x − 1 − 4 = x² − 3x − 5 [1].

(b) Discriminant = (−3)² − 4(1)(−5) = 9 + 20 = 29 [1]. Since 29 > 0 [1], the equation has two distinct real roots, confirming two distinct points of intersection [1].

5. (a) mx − 3 = x² + x − 1 [1] 0 = x² + x − mx − 1 + 3 = x² + (1 − m)x + 2 [1].

(b) Tangent ⟹ discriminant = 0 [1] (1 − m)² − 4(1)(2) = 0 [1] (1 − m)² = 8 [1] 1 − m = ±√8, so m = 1 − √8 or m = 1 + √8 [1] m = 1 − 2√2 or m = 1 + 2√2 [1].

6. (a) A repeated root requires discriminant = 0 [1].

(b) (−k)² − 4(4)(9) = 0 [1] k² − 144 = 0 [1] k² = 144 [1] k = 12 or k = −12 [1].

7. For x > 0, multiply both sides by x without changing the inequality direction: x² − 2x ≥ 5 [1] x² − 2x − 5 ≥ 0 [1] Using the quadratic formula: x = (2 ± √(4 + 20))/2 = (2 ± √24)/2 = 1 ± √6 [1] Since x > 0 is required and 1 − √6 is negative, only the region x ≥ 1 + √6 applies within the given domain [1] (the full quadratic solution “x ≤ 1 − √6 or x ≥ 1 + √6” must be restricted to x > 0) [1].

8. Lying entirely above the x-axis means no real roots, so discriminant < 0 [1] (−6)² − 4(1)(p) < 0, so 36 − 4p < 0 [1] p > 9 [1].

9. 2x − 1 = x² − 4x + 7 [1] 0 = x² − 4x − 2x + 7 + 1 = x² − 6x + 8 [1] (x − 2)(x − 4) = 0, so x = 2 or x = 4 [1] At x = 2: y = 2(2) − 1 = 3, giving (2, 3) [1] At x = 4: y = 2(4) − 1 = 7, giving (2, 3) and (4, 7) [1].

Where marks are usually lost

Forgetting to multiply the completed-square constant term by a leading coefficient other than 1. Writing a bounded (“and”) solution when an unbounded (“or”) solution is required, or vice versa — always sketch the graph first to check which region the inequality describes. Assuming only one method is acceptable for a linear-quadratic simultaneous pair — B4 requires both elimination and substitution, so choose whichever is more convenient. Multiplying both sides of an inequality by an unknown expression such as x without first establishing its sign, which can silently flip the inequality direction or introduce a spurious solution outside the given domain. Stating only the discriminant condition for a tangent (=0) when a question instead asks for two distinct intersection points (>0), or vice versa. Forgetting to substitute a solved x-value back into the original linear equation to find the matching y-coordinate.