Practice › Number and Algebra › Quadratic Inequalities

Quadratic Inequalities: Paper 2 Multiple-Choice

Ten multiple-choice questions on quadratic inequalities. Each wrong option is a mistake students actually make, such as taking the wrong region or dividing by x, so read the reason under each answer.

10 questions · about 15 minutes · answers below each question

All questions are original, written for Math and AI Academy and modelled on past-paper style. They are not reproductions of HKEAA examination questions.

Question 1 · Original — extension topic, no past-paper model

Tests: reading the right region off a factorised quadratic.

The solution of

x^{2} - 4 x - 5 > 0

$- 1 < x < 5$
$x < - 1$ or $x > 5$
$- 5 < x < 1$
$x < - 5$ or $x > 1$

Show answer

B. $x^{2} - 4 x - 5 = (x - 5) (x + 1)$ . The graph opens upwards, so it is above the x-axis outside the roots $- 1$ and $5$ . Option A is the answer to $< 0$ .

Question 2 · Original — extension topic, no past-paper model

Tests: counting integers in a quadratic solution interval.

How many integers satisfy

x^{2} + 2 x - 8 < 0

Show answer

C. $x^{2} + 2 x - 8 = (x + 4) (x - 2) < 0$ gives $- 4 < x < 2$ . The integers are $- 3, - 2, - 1, 0, 1$ : five of them.

Question 3 · Original — extension topic, no past-paper model

Tests: non-monic quadratic, closed inequality, outside region.

The solution of

2 x^{2} - x - 1 \geq 0

$- \frac{1}{2} \leq x \leq 1$
$x \leq - 1$ or $x \geq \frac{1}{2}$
$- 1 \leq x \leq \frac{1}{2}$
$x \leq - \frac{1}{2}$ or $x \geq 1$

Show answer

D. $2 x^{2} - x - 1 = (2 x + 1) (x - 1)$ , with roots $- \frac{1}{2}$ and $1$ . The graph opens upwards, so it is on or above the axis outside the roots.

Question 4 · Original — extension topic, no past-paper model

Tests: x squared at least a multiple of x: do not divide by x.

The solution of

x^{2} \geq 4 x

$x \leq 0$ or $x \geq 4$
$0 \leq x \leq 4$
$x \geq 4$
$x \leq 4$

Show answer

A. $x^{2} - 4 x \geq 0$ , so $x (x - 4) \geq 0$ , giving $x \leq 0$ or $x \geq 4$ . Option C comes from dividing by $x$ , which loses the case $x \leq 0$ .

Question 5 · Original — extension topic, no past-paper model

Tests: square less than a number: a between-roots interval.

The solution of

(x - 2)^{2} < 9

$x < - 1$ or $x > 5$
$- 5 < x < 1$
$- 1 < x < 5$
$x < 5$

Show answer

C. $(x - 2)^{2} < 9$ means $- 3 < x - 2 < 3$ , so $- 1 < x < 5$ .

Question 6 · Original — extension topic, no past-paper model

Tests: a quadratic that is at most zero: the single-point solution.

The solution of

x^{2} - 6 x + 9 \leq 0

there is no real solution
$x = 3$ only
$x \leq 3$
all real numbers

Show answer

B. $x^{2} - 6 x + 9 = (x - 3)^{2}$ . A square cannot be negative, so $(x - 3)^{2} \leq 0$ only when $x = 3$ .

Question 7 · Original — extension topic, no past-paper model

Tests: an inequality true for every real x gives a discriminant inequality in k.

The inequality

x^{2} + k x + 9 > 0

is true for every real number

x

. The range of values of

k

$- 6 < k < 6$
$k < - 6$ or $k > 6$
$- 3 < k < 3$
$k > - 6$

Show answer

A. The graph opens upwards, so it stays above the axis exactly when there are no real roots: $Δ = k^{2} - 36 < 0$ , giving $- 6 < k < 6$ .

Question 8 · Original — extension topic, no past-paper model

Tests: reading an inequality off a graph.

The figure shows the graph of

y = f (x)

, where

f (x)

is a quadratic function. The graph cuts the x-axis at

x = - 1

and

x = 3

. The solution of

f (x) < 0

Graph of y = f(x), a parabola opening downwards, cutting the x-axis at x = −1 and x = 3.

$- 1 < x < 3$
$x > 3$
$x < - 1$
$x < - 1$ or $x > 3$

Show answer

D. $f (x) < 0$ where the curve is below the x-axis. The graph opens downwards, so that is outside the roots: $x < - 1$ or $x > 3$ .

Question 9 · Original — extension topic, no past-paper model

Tests: counting integers when the quadratic needs rearranging first.

How many integers satisfy

x^{2} - 3 x \leq 4

Show answer

C. $x^{2} - 3 x - 4 \leq 0$ factorises as $(x - 4) (x + 1) \leq 0$ , giving $- 1 \leq x \leq 4$ . The integers are $- 1, 0, 1, 2, 3, 4$ : six, with both end-points included.

Question 10 · Original — extension topic, no past-paper model

Tests: compound inequality: intersecting a quadratic and a linear solution set.

The solution of the compound inequality

x^{2} - 9 \geq 0

and

x + 1 < 5

$x \leq - 3$ or $3 \leq x < 4$
$x \leq - 3$ or $x \geq 3$
$3 \leq x < 4$
$- 3 \leq x < 4$

Show answer

A. The first gives $x \leq - 3$ or $x \geq 3$ ; the second gives $x < 4$ . Keeping both: $x \leq - 3$ or $3 \leq x < 4$ .

Continue practising this topic: Try the Paper 2 Challenge MC for harder variants, or move to Paper 1 long questions — also available as Paper 1 Challenge.

Where this sits in the syllabus

We found no HKDSE Compulsory Part past-paper question on quadratic inequalities between 2012 and 2026, so these sets are an extension of the exam-style practice on this site rather than a set modelled on a particular paper. The skills underneath are exactly the ones the exam does test: factorising, completing the square, the discriminant and reading a graph.

Full coverage of this topic, and the rest of the course, is on the Mathematics page. More sets are listed on the practice index.

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