Practice › Number and Algebra › Quadratic Inequalities

Quadratic Inequalities: Paper 2 Challenge MC

This is the Challenge set. Each question has a tidy-looking wrong answer built in, such as a missed edge case or a strict end-point counted as included. Work the standard Paper 2 set first if the basic method is still new.

10 questions · about 20 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

The solution of

(x - 1) (x + 3) \geq 5

$x \geq 6$
$- 4 \leq x \leq 2$
$x \leq - 4$ or $x \geq 2$
$x \geq 2$

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C. Expand: $x^{2} + 2 x - 3 \geq 5$ , so $x^{2} + 2 x - 8 \geq 0$ , that is $(x + 4) (x - 2) \geq 0$ , giving $x \leq - 4$ or $x \geq 2$ . Option D is what you get by splitting into $x - 1 \geq 5$ or $x + 3 \geq 5$ : it is correct on one branch only, and shows a product compared with 5 cannot be split into factors.

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

The solution of

a x^{2} + b x + c > 0

- 2 < x < 3

, where

a

b

c

are real constants. The solution of

c x^{2} + b x + a > 0

$x < - \frac{1}{2}$ or $x > \frac{1}{3}$
$- \frac{1}{2} < x < \frac{1}{3}$
$x < - \frac{1}{3}$ or $x > \frac{1}{2}$
$- \frac{1}{3} < x < \frac{1}{2}$

Show answer

A. The solution is between the roots, so $a < 0$ and $a x^{2} + b x + c = a (x + 2) (x - 3) = a (x^{2} - x - 6)$ , giving $b = - a$ and $c = - 6 a$ . Then $c x^{2} + b x + a = - a (6 x^{2} + x - 1) > 0$ , and since $- a > 0$ this is $(3 x - 1) (2 x + 1) > 0$ : $x < - \frac{1}{2}$ or $x > \frac{1}{3}$ . Option B is what you get if you forget that $a < 0$ .

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

The equation

x^{2} + 2 k x + (k + 6) = 0

, where

k

is an integer with

- 6 \leq k \leq 6

, has two distinct real roots. How many values of

k

are possible?

Show answer

C. $Δ = 4 k^{2} - 4 (k + 6) > 0$ gives $k^{2} - k - 6 > 0$ , so $(k - 3) (k + 2) > 0$ : $k < - 2$ or $k > 3$ . In the window: $- 6, - 5, - 4, - 3$ and $4, 5, 6$ , which is 7. Option D (9) wrongly keeps $k = - 2$ and $k = 3$ , which give $Δ = 0$ .

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

The set of all

x

satisfying both

x^{2} - 2 x - 3 \leq 0

and

x^{2} - 5 x + 4 \geq 0

$- 1 \leq x \leq 3$
$- 1 \leq x \leq 1$
$1 \leq x \leq 3$
$x \leq 1$ or $x \geq 4$

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B. The first gives $- 1 \leq x \leq 3$ ; the second gives $x \leq 1$ or $x \geq 4$ . The overlap is $- 1 \leq x \leq 1$ . Option C is the part of the first interval that the second inequality rules out (apart from the single point $x = 1$ ).

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

The solution of

x^{4} - 10 x^{2} + 9 < 0

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

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D. Factorise as $(x^{2} - 1) (x^{2} - 9) < 0$ , so $1 < x^{2} < 9$ . This means $1 < ∣ x ∣ < 3$ : $- 3 < x < - 1$ or $1 < x < 3$ . Option A forgets the lower bound $x^{2} > 1$ .

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

The inequality

k x^{2} - k x + 1 > 0

holds for every real number

x

. The range of values of

k

$0 \leq k < 4$
$0 < k < 4$
$0 \leq k \leq 4$
$k > 0$

Show answer

A. If $k = 0$ the expression is 1, which is always positive. If $k \neq = 0$ we need $k > 0$ and $Δ = k^{2} - 4 k < 0$ , so $0 < k < 4$ . Combining gives $0 \leq k < 4$ . Option B forgets $k = 0$ ; option C wrongly includes $k = 4$ , where the expression is $(2 x - 1)^{2}$ , equal to 0 at $x = \frac{1}{2}$ .

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

Let

f (x) = x^{2} - 4 x + 3

. The solution of

f (x) \leq f (5)

$0 \leq x \leq 5$
$2 \leq x \leq 5$
$- 5 \leq x \leq 5$
$- 1 \leq x \leq 5$

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D. $f (5) = 8$ , so $x^{2} - 4 x + 3 \leq 8$ gives $x^{2} - 4 x - 5 \leq 0$ , that is $(x - 5) (x + 1) \leq 0$ : $- 1 \leq x \leq 5$ . Equivalently, the axis of symmetry is $x = 2$ , and 5 is 3 units to its right, so the other end-point is 3 units to its left, at $- 1$ .

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

How many integers satisfy

(x - 2)^{2} < 4

Show answer

B. $(x - 2)^{2} < 4$ gives $- 2 < x - 2 < 2$ , so $0 < x < 4$ . The integers are 1, 2, 3: three. Option D (5) is the answer to the non-strict $(x - 2)^{2} \leq 4$ , where 0 and 4 also count.

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

Which of the following is true about

- x^{2} + 4 x - 5

for real

x

It is positive for every real $x$ .
It is positive for some $x$ and negative for others.
It is zero for exactly one value of $x$ .
It is negative for every real $x$ .

Show answer

D. The graph opens downwards ( $- 1 < 0$ ). The discriminant is $16 - 20 = - 4 < 0$ , so it never meets the x-axis. Hence it lies below the axis everywhere. Equivalently, $- x^{2} + 4 x - 5 = - (x - 2)^{2} - 1 \leq - 1$ .

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

A rectangle has width

x

cm and length

(x + 6)

cm. Its area is less than

55 cm^{2}

. The possible values of

x

are

$0 < x < 5$
$- 11 < x < 5$
$x < 5$
$0 < x < 11$

Show answer

A. $x (x + 6) < 55$ gives $x^{2} + 6 x - 55 < 0$ , that is $(x + 11) (x - 5) < 0$ , so $- 11 < x < 5$ . A width must be positive, so $0 < x < 5$ . Option B is the algebraic answer without the physical condition.

Continue practising this topic: Go back to the Paper 2 multiple-choice if you want the standard version first, or move to Paper 1 Challenge — also available as Paper 1 long questions.

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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