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Quadratic and Simultaneous Equations — Paper 2 Challenge MC

Ten harder multiple-choice questions on the same topic. Most carry a distractor that is the right number to the wrong question — an answer that comes from a rule applied where it does not hold. The note under each answer says which option is the trap and why.

10 questions · about 25 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 — new here: the equation is not guaranteed to be quadratic
Question 2 — new here: the cube identity rather than the square one
Question 3 — new here: a sign condition on the roots, with no method named
Question 4 — new here: a leading coefficient that can vanish
Question 5 — new here: a discriminant that is itself a quadratic in the constant
Question 6 — new here: transforming the roots, then rebuilding the equation
Question 7 — new here: an equation reducible to a quadratic
Question 8 — new here: deducing the second root from the coefficients being real
Question 9 — new here: the zero-product rule applied where it does not hold
Question 10 — new here: two equations linked through a shifted constant term

Question 1 · Original

Let $a$ be a real constant. If the equation $a x^{2} + 4 x + 1 = 0$ has exactly one real root, then $a =$

$0$ or $4$
$4$
$0$
there is no such value of $a$

Show answer

A. Option B is the trap: it is what the discriminant alone gives. But $a$ is not said to be non-zero, and when $a = 0$ the equation is linear and still has exactly one root.

Question 2 · Original

If $α$ and $β$ are the roots of $x^{2} - 3 x + 1 = 0$ , then $α^{3} + β^{3} =$

$27$
$18$
$24$
$21$

Show answer

B. Use $(α + β)^{3} - 3 α β (α + β) = 27 - 9$ . Option A is what you get by forgetting the correction term.

Question 3 · Original

Let $k$ be a real constant. If the equation $x^{2} - 2 k x + (k^{2} - 9) = 0$ has two roots of opposite signs, then

$k < - 3$ or $k > 3$
$- 3 \leq k \leq 3$
$- 3 < k < 3$
$k > 3$

Show answer

C. Opposite signs means the product of the roots is negative: $k^{2} - 9 < 0$ . That condition also forces the discriminant to be positive, so nothing further is needed.

Question 4 · Original

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

$k < 6$
$k \leq 6$ and $k \neq = 2$
$k > 6$
$k < 6$ and $k \neq = 2$

Show answer

D. The discriminant condition gives $k < 6$ , but "two roots" also requires the equation to be quadratic, so $k = 2$ must be excluded. Option A is the trap.

Question 5 · Original

Let $m$ be a real constant. If the equation $x^{2} + (m + 1) x + m^{2} = 0$ has real roots, then the greatest value of $m$ is

$\frac{1}{3}$
$1$
$- \frac{1}{3}$
$3$

Show answer

B. The condition $(m + 1)^{2} - 4 m^{2} \geq 0$ factorises to $(1 - m) (3 m + 1) \geq 0$ , giving $- \frac{1}{3} \leq m \leq 1$ .

Question 6 · Original

Let $α$ and $β$ be the roots of $2 x^{2} + 5 x - 1 = 0$ . The quadratic equation with integral coefficients whose roots are $2 α$ and $2 β$ is

$2 x^{2} + 5 x - 1 = 0$
$x^{2} - 5 x - 2 = 0$
$x^{2} + 5 x - 2 = 0$
$2 x^{2} + 10 x - 1 = 0$

Show answer

C. Doubling the roots doubles the sum and multiplies the product by 4: the new sum is −5 and the new product is −2. Option A is the trap — scaling the roots is not the same as scaling the equation.

Question 7 · Original

The number of real roots of the equation $x^{4} - 5 x^{2} + 4 = 0$ is

Show answer

D. Substituting $u = x^{2}$ gives $u = 1$ or $u = 4$ ; each positive value of u returns two values of x.

Question 8 · Original

The equation $x^{2} + p x + q = 0$ has real coefficients, and one of its roots is $1 - 5 i$ . Then $q - p =$

$28$
$24$
$26$
$- 28$

Show answer

A. Real coefficients force the other root to be $1 + 5 i$ , so the sum is 2 and the product is 26. Option C is the trap — it forgets that $p$ is negative.

Question 9 · Original

Solve the equation $(x - 3) (x + 2) = 6$ .

$x = 9$ or $x = 4$
$x = 3$ or $x = - 2$
$x = 4$ or $x = - 3$
$x = - 4$ or $x = 3$

Show answer

C. Option A is the trap: setting each bracket equal to 6 only works when the right-hand side is 0. Expand and rearrange to $x^{2} - x - 12 = 0$ first.

Question 10 · Original

Let $a$ and $b$ be real constants. The equation $x^{2} + a x + b = 0$ has two distinct real roots, and the equation $x^{2} + a x + (b + 4) = 0$ has equal roots. Then $a^{2} - 4 b =$

$16$
$4$
$8$
$- 16$

Show answer

A. The second condition says $a^{2} - 4 (b + 4) = 0$ , so $a^{2} - 4 b = 16$ . The first condition is consistent with it and is not needed for the value.

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

Where this sits in the syllabus

This Challenge set assumes the standard set is already comfortable. The difference is not heavier algebra but less scaffolding: a condition is stated and you choose the tool. Three ideas recur — an equation that need not be quadratic because its leading coefficient can vanish, sign conditions on roots translated into conditions on the coefficients, and claims that must be proved or disproved rather than computed.

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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