AS & A Level · Practice questions · The discriminant

The sign that decides.

Six original Cambridge-style questions on the discriminant: classifying roots, finding constants for equal, distinct or no real roots, proving roots are always real and a context where the discriminant gives the maximum.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Rearrange, read \(a, b, c\), then decide.

Rearrange to \(ax^{2} + bx + c = 0\) and read \(a\), \(b\), \(c\) with their signs.
Translate the words: equal \(= 0\); distinct \(> 0\); real \(\ge 0\); no real \(< 0\).
Substitute with brackets: \((b)^{2} - 4(a)(c)\).
Solve the resulting equation or inequality in \(k\); a quadratic inequality has a two-ended answer.
Always positive needs \(a > 0\) as well as \(b^{2} - 4ac < 0\).

Recall

[2 marks]

Find the discriminant of \(5x^{2} - 2x + 3\) and hence state the number of real roots of \(5x^{2} - 2x + 3 = 0\).

\((-2)^{2} - 4(5)(3) = 4 - 60 = -56\). ✓
\(-56 < 0\), so there are no real roots. ✓

Calculation

[3 marks]

Find the values of \(k\) for which \(4x^{2} + kx + 9 = 0\) has a repeated root. For the positive value of \(k\), find the root.

\(k^{2} - 4(4)(9) = 0 \Rightarrow k^{2} = 144 \Rightarrow k = \pm 12\). ✓
\(k = 12\): \(4x^{2} + 12x + 9 = (2x + 3)^{2} = 0\). ✓
The repeated root is \(x = -\dfrac{3}{2}\). ✓

Calculation

[3 marks]

Find the set of values of \(k\) for which the equation \(x^{2} + 6x + 2k - 1 = 0\) has two distinct real roots.

\(36 - 4(1)(2k - 1) > 0\). ✓
\(36 - 8k + 4 > 0 \Rightarrow 40 > 8k\). ✓
\(k < 5\). ✓

Analysis

[4 marks]

Find the set of values of \(k\) for which \(kx^{2} - 4x + k > 0\) for all real values of \(x\).

The parabola must open upwards: \(k > 0\). ✓
It must not meet the \(x\)-axis: \(16 - 4k^{2} < 0 \Rightarrow k^{2} > 4\). ✓
\(k^{2} > 4 \Rightarrow k < -2\) or \(k > 2\). ✓
Combined with \(k > 0\): \(k > 2\). ✓

Proof

[3 marks]

Show that the equation \(2x^{2} + (p - 4)x - p = 0\) has two distinct real roots for all real values of \(p\).

\(b^{2} - 4ac = (p - 4)^{2} - 4(2)(-p) = p^{2} - 8p + 16 + 8p\). ✓
\(= p^{2} + 16\). ✓
\(p^{2} \ge 0\), so \(p^{2} + 16 \ge 16 > 0\): two distinct real roots for every \(p\). ✓

Analysis

[5 marks]

A rectangular garden is to be fenced with exactly 40 m of fencing. One side has length \(x\) m and the area is \(A\) m².

(a) Show that \(x^{2} - 20x + A = 0\).

(b) Use the discriminant to find the greatest possible area, and the dimensions of the garden in that case.

(a) The other side is \(\dfrac{40 - 2x}{2} = 20 - x\), so \(A = x(20 - x) = 20x - x^{2}\). ✓
Rearranging: \(x^{2} - 20x + A = 0\). ✓
(b) For a real length \(x\), \(b^{2} - 4ac \ge 0\): \(400 - 4A \ge 0 \Rightarrow A \le 100\). ✓
Greatest area 100 m²; then \(x^{2} - 20x + 100 = (x - 10)^{2} = 0\), so the garden is a 10 m by 10 m square. ✓
(c) \(A = 105\) gives \(400 - 420 = -20 < 0\): no real value of \(x\), so this area cannot be made. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.