Practice › Number and Algebra › Quadratic Inequalities

Quadratic Inequalities: Paper 1 Challenge

This is the Challenge set. The questions state a condition and leave the method to you, and several ask you to explain or disprove a claim. Work the standard Paper 1 set first if the basic method is still new.

10 questions · 71 marks · about 75 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 set of the inequality

x^{2} + b x + c < 0

- 3 < x < 5

, where

b

and

c

are constants.

(a) Find $b$ and $c$ .(3)

(b) Hence solve $c x^{2} + b x + 1 > 0$ .(4)

Show answer

(a) $b = - 2$ , $c = - 15$ . The end-points $- 3$ and $5$ are the roots, so $x^{2} + b x + c = (x + 3) (x - 5)$ .

(b) $- \frac{1}{3} < x < \frac{1}{5}$ . The inequality becomes $- 15 x^{2} - 2 x + 1 > 0$ ; multiplying by $- 1$ flips the sign, giving $(5 x - 1) (3 x + 1) < 0$ .

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

Find all values of

k

for which

(k - 1) x^{2} - 2 (k - 1) x + 3 > 0

for every real number

x

(7)

Show answer

$1 \leq k < 4$ . When $k = 1$ the expression is the constant 3, which is positive. When $k \neq = 1$ we need $k - 1 > 0$ and $Δ < 0$ , where $Δ = 4 (k - 1) (k - 4)$ , giving $1 < k < 4$ .

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

The curve

C

has equation

y = x^{2} - 2 x + 3

and the line

L

has equation

y = k x

, where

k

is a constant.

(a) $L$ meets $C$ at two distinct points. Find the range of values of $k$ .(5)

(b) Find the values of $k$ for which $L$ meets $C$ at exactly one point.(2)

Show answer

(a) $k < - 2 - 23$ or $k > - 2 + 23$ . The intersection equation is $x^{2} - (k + 2) x + 3 = 0$ and two points need $Δ > 0$ , so $(k + 2)^{2} > 12$ .

(b) $k = - 2 \pm 23$ , from $Δ = 0$ .

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

Peter solves

(x - 1) (x - 4) \geq 6

as follows:
“ $x - 1 \geq 6$ or $x - 4 \geq 6$ , so $x \geq 7$ or $x \geq 10$ , so $x \geq 7$ .”

(a) Show that $x = - 1$ satisfies the inequality $(x - 1) (x - 4) \geq 6$ , but is not in Peter's answer.(2)

(b) Explain why Peter's method is not valid.(1)

(c) Solve $(x - 1) (x - 4) \geq 6$ correctly.(4)

Show answer

(a) At $x = - 1$ : $(- 2) (- 5) = 10 \geq 6$ , true. But $- 1 \geq 7$ is false.

(b) Splitting into factors works only when the product is compared with 0. A product of 6 does not force either factor to be at least 6; for example $2 \times 3 = 6$ .

(c) $x \leq \frac{5 - 33}{2}$ or $x \geq \frac{5 + 33}{2}$ . The inequality becomes $x^{2} - 5 x - 2 \geq 0$ , which does not factorise, so use the quadratic formula.

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

Find all real numbers

x

such that

x^{4} - 5 x^{2} + 4 < 0

(6)

Show answer

$- 2 < x < - 1$ or $1 < x < 2$ . Factorise as $(x^{2} - 1) (x^{2} - 4) < 0$ , so $1 < x^{2} < 4$ .

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

(a) Solve the simultaneous inequalities $x^{2} - x - 6 > 0$ and $x^{2} - 3 x - 4 < 0$ .(5)

(b) Ada says: “3 and 4 are the integers that satisfy both inequalities.” Is Ada correct? Explain your answer.(2)

Show answer

(a) $3 < x < 4$ . The first gives $x < - 2$ or $x > 3$ ; the second gives $- 1 < x < 4$ ; the overlap is $3 < x < 4$ .

(b) No. At $x = 3$ the first expression is 0, not $> 0$ ; at $x = 4$ the second is 0, not $< 0$ . In fact no integer lies strictly between 3 and 4.

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

A rectangle has perimeter

40

cm and its area is not less than

75 cm^{2}

. Find the range of values of the length of its shorter side.

(6)

Show answer

$5 \leq s \leq 10$ . The area condition gives $5 \leq s \leq 15$ , but the shorter side cannot exceed 10 (a square), so the upper end is 10.

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

Consider the equation

x^{2} - (k + 1) x + (k - 1) = 0

, where

k

is a real constant.

(a) Prove that the equation has two distinct real roots for every real value of $k$ .(3)

(b) Find the range of values of $k$ for which both roots are positive.(4)

Show answer

(a) $Δ = (k - 1)^{2} + 4 \geq 4 > 0$ for every real $k$ .

(b) $k > 1$ . With real roots guaranteed, both are positive exactly when the sum $k + 1 > 0$ and the product $k - 1 > 0$ .

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

The quadratic function

f (x) = - x^{2} + 6 x + c

has greatest value

14

(a) Find $c$ .(2)

(b) Solve $f (x) > f (0)$ .(3)

(c) Find the set of values of $k$ for which the equation $f (x) = k$ has no real roots. Explain your answer.(3)

Show answer

(a) $c = 5$ . $f (x) = - (x - 3)^{2} + 9 + c$ , so the greatest value is $9 + c = 14$ .

(b) $0 < x < 6$ . With $f (0) = 5$ the inequality reduces to $x^{2} - 6 x < 0$ .

(c) $k > 14$ . The function never takes a value above its greatest value 14, so $f (x) = k$ has no real solution when $k > 14$ .

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

A ball is thrown upwards. Its height

h

metres above the ground after

t

seconds is given by

h = 20 t - 5 t^{2}

, for

0 \leq t \leq 4

(a) For how long is the ball more than 15 m above the ground?(3)

(b) A student claims: “The ball never reaches a height of 20 m, because the inequality $h \geq 20$ has no solution.” Is the student correct? Explain your answer.(3)

(c) For how long is the ball at least 19 m above the ground? Give your answer correct to 2 decimal places.(3)

Show answer

(a) 2 seconds. $20 t - 5 t^{2} > 15$ gives $t^{2} - 4 t + 3 < 0$ , so $1 < t < 3$ .

(b) No. $h = 20 - 5 (t - 2)^{2}$ , so $h \geq 20$ means $(t - 2)^{2} \leq 0$ , whose only solution is $t = 2$ . The ball reaches exactly 20 m at its highest point.

(c) 0.89 seconds. $(t - 2)^{2} \leq \frac{1}{5}$ gives $∣ t - 2∣ \leq \frac{1}{5}$ , so the duration is $\frac{2}{5}$ .

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

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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