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Quadratic and Simultaneous Equations — Paper 2 Multiple-Choice

Twenty multiple-choice questions in HKDSE Paper 2 style. Paper 2 rewards recognising the form of an equation rather than grinding through it, so several of these can be answered without solving anything.

20 questions · about 30 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.

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Question 1 — factorising a simple quadratic
Question 2 — square-root method; keeping both signs
Question 3 — sum of roots from the coefficients
Question 4 — combining the two root relations
Question 5 — discriminant equal to zero
Question 6 — discriminant negative; strict versus non-strict
Question 7 — substitution; both solution pairs
Question 8 — solving the discriminant inequality
Question 9 — using one known root with the root relations
Question 10 — factorising by grouping with a literal coefficient
Question 11 — symmetric function of the roots
Question 12 — reading the discriminant's sign across four equations
Question 13 — forming an equation from words; rejecting a root
Question 14 — recognising a factorisation instead of solving
Question 15 — a discriminant that is quadratic in the unknown constant
Question 16 — building an equation from its roots
Question 17 — fractional equation reducing to a quadratic
Question 18 — discriminant of the eliminated equation
Question 19 — difference of roots; spotting equal roots
Question 20 — root relations with a conjugate pair

Question 1 · Modelled on 2013 P2 Q6

Solve the equation $x^{2} - x - 12 = 0$ .

$x = - 3$ or $x = 4$
$x = 3$ or $x = - 4$
$x = - 2$ or $x = 6$
$x = 2$ or $x = - 6$

Show answer

A. The factors are $(x + 3) (x - 4)$ , so the roots carry the opposite signs to the numbers inside the brackets.

Question 2 · Original

Solve the equation $(x - 5)^{2} = 49$ .

$x = 12$
$x = - 2$ or $x = 12$
$x = 2$ or $x = - 12$
$x = - 7$ or $x = 7$

Show answer

B. Taking the square root gives $x - 5 = \pm 7$ . Taking only the positive root loses one answer.

Question 3 · Original

If $α$ and $β$ are the roots of $3 x^{2} + 7 x - 6 = 0$ , then $α + β =$

$\frac{7}{3}$
$- 2$
$- \frac{7}{3}$
$2$

Show answer

C. The sum of the roots is $- \frac{b}{a}$ , and the minus sign is what most answers drop.

Question 4 · Original

If $α$ and $β$ are the roots of $2 x^{2} - 9 x + 4 = 0$ , then $\frac{1}{α} + \frac{1}{β} =$

$\frac{4}{9}$
$- \frac{9}{4}$
$\frac{9}{2}$
$\frac{9}{4}$

Show answer

D. Combine the fractions first: the expression equals $\frac{α + β}{α β}$ , which is $\frac{9/2}{2}$ .

Question 5 · Modelled on 2016 P2 Q8

Let $k$ be a constant. If the equation $x^{2} - 6 x + k = 0$ has equal roots, then $k =$

$- 9$
$9$
$3$
$- 3$

Show answer

B. Equal roots means the discriminant is zero: $36 - 4 k = 0$ .

Question 6 · Original

Let $c$ be a constant. If the equation $x^{2} + 5 x + c = 0$ has no real roots, then

$c < \frac{25}{4}$
$c \leq \frac{25}{4}$
$c > \frac{25}{4}$
$c \geq \frac{25}{4}$

Show answer

C. No real roots means $25 - 4 c < 0$ . The inequality is strict, so the boundary value is excluded.

Question 7 · Modelled on 2015 P2 Q34

Solve the simultaneous equations ${y = x + 1 x^{2} + y^{2} = 25$

$x = 3, y = 4$ only
$x = 4, y = 3$ or $x = - 3, y = - 4$
$x = - 4, y = - 3$ only
$x = 3, y = 4$ or $x = - 4, y = - 3$

Show answer

D. Substitution gives $x^{2} + x - 12 = 0$ , so $x = 3$ or $x = - 4$ ; each x has its own y from the linear equation.

Question 8 · Modelled on 2022 P1 Q10(b)(ii)

Let $k$ be a constant. If the equation $x^{2} + k x + (k + 3) = 0$ has two distinct real roots, then

$k < - 2$ or $k > 6$
$- 2 < k < 6$
$k < - 6$ or $k > 2$
$- 6 < k < 2$

Show answer

A. The condition is $k^{2} - 4 k - 12 > 0$ , which factorises as $(k - 6) (k + 2) > 0$ — an "outside" solution set.

Question 9 · Original

If one of the roots of the equation $x^{2} + b x - 20 = 0$ is 5, then $b =$

$1$
$4$
$- 1$
$- 4$

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C. The product of the roots is −20, so the other root is −4; the sum is then 1, and the sum equals $- b$ .

Question 10 · Modelled on 2024 P2 Q7

Let $m$ be a constant. Solve the equation $x^{2} - 2 m x = 3 x - 6 m$ .

$x = - 2 m$ or $x = 3$
$x = 2 m$ or $x = - 3$
$x = m$ or $x = 3$
$x = 2 m$ or $x = 3$

Show answer

D. Group instead of expanding: $x (x - 2 m) - 3 (x - 2 m) = 0$ , so $(x - 2 m) (x - 3) = 0$ .

Question 11 · Original

If $α$ and $β$ are the roots of $x^{2} - 4 x + 7 = 0$ , then $α^{2} + β^{2} =$

$2$
$30$
$- 2$
$16$

Show answer

A. Use $(α + β)^{2} - 2 α β = 16 - 14$ . The roots themselves are not real, but this value is.

Question 12 · Modelled on 2012 P2 Q34

Which of the following equations has two non-real roots?

$x^{2} - 4 x + 4 = 0$
$x^{2} + 2 x + 5 = 0$
$x^{2} + 4 x - 4 = 0$
$2 x^{2} - 3 x - 1 = 0$

Show answer

B. Only $x^{2} + 2 x + 5 = 0$ has a negative discriminant $(4 - 20 = - 16)$ . The first has a zero discriminant, which means equal real roots, not non-real ones.

Question 13 · Original

The sum of a positive number and its square is 72. The number is

$6$
$- 9$
$9$
$8$

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D. The equation $n^{2} + n - 72 = 0$ gives $n = 8$ or $n = - 9$ , and the number is stated to be positive.

Question 14 · Original

If $2 x^{2} + 5 x - 3 = 0$ , then $(2 x - 1) (x + 3) =$

$0$
$1$
$3$
$- 3$

Show answer

A. The product is exactly the factorised form of the left-hand side, which is given as 0. No solving is needed.

Question 15 · Original

Let $a$ be a non-zero constant. If the equation $a x^{2} + 6 x + a = 0$ has equal roots, then $a =$

$3$
$3$ or $- 3$
$- 3$
$9$ or $- 9$

Show answer

B. The discriminant is $36 - 4 a^{2}$ , so $a^{2} = 9$ and both signs are admissible.

Question 16 · Original

The quadratic equation whose roots are 2 and $- 7$ is

$x^{2} - 5 x - 14 = 0$
$x^{2} + 5 x + 14 = 0$
$x^{2} + 5 x - 14 = 0$
$x^{2} - 5 x + 14 = 0$

Show answer

C. The sum is −5 and the product is −14, and the equation is $x^{2} - (sum) x + (product) = 0$ .

Question 17 · Original

Solve the equation $\frac{x}{x - 2} + \frac{3}{x} = 2$ .

$x = 1$ or $x = 6$
$x = - 1$ or $x = - 6$
$x = 2$ or $x = 3$
$x = 6$

Show answer

A. Clearing the denominators gives $x^{2} - 7 x + 6 = 0$ . Both roots are admissible because neither is 0 or 2.

Question 18 · Original

Let $k$ be a constant. If the simultaneous equations ${y = 3 x + k y = x^{2} + 5$ have only one solution, then $k =$

$- \frac{11}{4}$
$\frac{11}{4}$
$\frac{4}{11}$
$\frac{29}{4}$

Show answer

B. Eliminating y gives $x^{2} - 3 x + (5 - k) = 0$ ; one solution means its discriminant $9 - 4 (5 - k)$ is zero.

Question 19 · Original

If $α$ and $β$ are the roots of $4 x^{2} - 12 x + 9 = 0$ , then $(α - β)^{2} =$

$18$
$9$
$0$
$3$

Show answer

C. $(α - β)^{2} = (α + β)^{2} - 4 α β = 9 - 9$ . The equation is a perfect square, so the two roots coincide.

Question 20 · Original

In the equation $x^{2} + p x + q = 0$ , $p$ and $q$ are real constants. If the two roots are $3 + 2 i$ and $3 - 2 i$ , then $p + q =$

$19$
$5$
$- 7$
$7$

Show answer

D. The sum of the roots is 6, so $p = - 6$ ; the product is $9 + 4 = 13$ , so $q = 13$ .

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

Where this sits in the syllabus

Quadratic and simultaneous equations sit in the Number and Algebra strand of the HKDSE Compulsory Part. Paper 1 tends to ask for the equation to be formed first, from a variation, a rate or a geometric setting, and only then solved. The discriminant carries most of the marks that are actually lost: it decides whether roots are two, one or non-real, and it turns questions about an unknown constant into inequalities. Everything here also feeds the next chapter, Functions and Graphs of Quadratic Functions, where the same discriminant tells you how the curve meets the x-axis.

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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