A Level Mathematics: Quadratics — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Quadratics revision notes

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Section A

1. Express 3x² − 12x + 7 in the form a(x + p)² + q, and state the minimum value. [4]

2. Solve x² − 5x − 14 > 0. [3]

3. Solve x⁴ − 13x² + 36 = 0. [4]

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Section B

4. The line y = mx + 2 is a tangent to the curve y = x² + 3x + 6.

(a) Show that x² + (3 − m)x + 4 = 0. [2]

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

5. The quadratic 2x² + kx + 8 is positive for all real values of x.

(a) State the two conditions that must be satisfied. [2]

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

6. Solve x − 8√x + 15 = 0. [4]

7. Solve 3^(2x) − 10(3^x) + 9 = 0. [4]

8. The line y = kx − 1 intersects the curve y = x² + 2x + 3 at two distinct points. Find the range of possible values of k. [5]

9. The curve y = x² + 4x + p does not meet the x-axis at any point. Find the range of possible values of p. [3]

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Answers

1. 3(x² − 4x) + 7 [1] = 3[(x − 2)² − 4] + 7 [1] = 3(x − 2)² − 12 + 7 [1] = 3(x − 2)² − 5, minimum value −5 (at x = 2) [1]. Forgetting to multiply the −4 by 3 is the classic slip.

2. Factorise: (x − 7)(x + 2) > 0 [1]. Critical values x = 7 and x = −2 [1]. Upward parabola, so positive outside the roots: x < −2 or x > 7 [1]. Writing 7 < x < −2 is impossible — it must be two separate intervals.

3. Let u = x² [1]: u² − 13u + 36 = 0 → (u − 4)(u − 9) = 0 [1] u = 4 or u = 9 [1] x² = 4 → x = ±2; x² = 9 → x = ±3, so x = ±2, ±3 [1]. All four solutions are required.

4. (a) mx + 2 = x² + 3x + 6 [1] 0 = x² + 3x − mx + 4 = x² + (3 − m)x + 4 [1].

(b) Tangent ⟹ discriminant = 0 [1] (3 − m)² − 4(1)(4) = 0 [1] (3 − m)² = 16, so 3 − m = ±4 [1] m = 3 − 4 = −1 or m = 3 + 4 = 7 [1].

5. (a) The coefficient of x² must be positive (a > 0, satisfied here as a = 2) [1], and the discriminant must be negative [1].

(b) b² − 4ac < 0 [1] k² − 4(2)(8) < 0 [1] k² < 64 [1] −8 < k < 8 [1].

6. Let u = √x [1]: u² − 8u + 15 = 0 → (u − 3)(u − 5) = 0 [1] u = 3 or u = 5, both valid since √x ≥ 0 [1] x = 9 or x = 25 [1]. Substituting back is essential — stopping at u loses the final mark.

7. Let u = 3^x [1]: u² − 10u + 9 = 0 → (u − 1)(u − 9) = 0 [1] u = 1 or u = 9, both valid since 3^x is always positive [1] 3^x = 1 → x = 0; 3^x = 9 → x = 2 [1].

8. kx − 1 = x² + 2x + 3 [1] 0 = x² + 2x − kx + 4 = x² + (2 − k)x + 4 [1] Two distinct points ⟹ discriminant > 0 [1] (2 − k)² − 4(1)(4) > 0 → (2 − k)² > 16 [1] 2 − k > 4 or 2 − k < −4, so k < −2 or k > 6 [1].

9. Not meeting the x-axis at any point is the third of the three discriminant cases, so it requires discriminant < 0 [1] 4² − 4(1)(p) < 0 → 16 − 4p < 0 [1] p > 4 [1].

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Where marks are usually lost

• Not multiplying q by a when completing the square with a ≠ 1.
• Writing an impossible double inequality instead of two intervals.
• Giving only two of the four roots in a quartic.
• Giving only the discriminant condition for “always positive”.
• Failing to substitute back after a substitution.
• Forgetting that a term such as 3^x or 2^x is always positive, so a negative value of u must be rejected.
• Using discriminant = 0 (tangent) instead of discriminant > 0 (two distinct intersection points) when the question asks for two points, not one.

Work through the Quadratics revision notes alongside these questions: the notes summarise the completed-square method, the three discriminant cases and the substitution technique in condensed form, while these questions test whether you can recognise which case applies — tangent, two intersections, or no intersection — from the wording of an unfamiliar problem, rather than just recall the rule.