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Quadratic and Simultaneous Equations — Paper 1 Challenge

Six harder Paper 1 questions on the same topic. Each one states a condition without naming the method, and several ask you to justify or disprove a claim rather than compute a number. Work the standard Paper 1 set first — this one assumes it.

6 questions · about 55 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 assumed quadratic — the linear case must be tested separately
Question 2 — new here: a justification that has to cite the discriminant and the product of roots together
Question 3 — new here: translating sign conditions on roots into conditions on the coefficients
Question 4 — new here: a claim that must be settled by locating the point, not by a condition on m
Question 5 — new here: proving a property for all k by completing the square on the discriminant
Question 6 — new here: relating two equations by a shift of the roots, then disproving a claim by counter-example

Question 1 · Original — a stated condition with a hidden second case

Let $a$ be a real constant. It is given that the equation $a x^{2} + (a + 4) x + 4 = 0$ has exactly one real root.

(a) Find all possible values of $a$ . (5 marks)

(b) For each value of $a$ found in (a), write down that root. (2 marks)

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(a) $a = 0$ or $a = 4$ . (b) In both cases the root is $x = - 1$ . The phrase is "exactly one real root", not "equal roots": when $a = 0$ the equation is linear and has a single root, and that case is lost if you go straight to the discriminant. For $a \neq = 0$ the discriminant is $(a - 4)^{2}$ , so only $a = 4$ works.

Question 2 · Original — building a new equation, then justifying a claim about its roots

Let $α$ and $β$ be the roots of the equation $x^{2} - 5 x + 2 = 0$ .

(a) Find the quadratic equation with integral coefficients whose roots are $\frac{α}{β}$ and $\frac{β}{α}$ . (5 marks)

(b) Someone claims that $\frac{α}{β}$ and $\frac{β}{α}$ are both real and positive. Do you agree? Explain your answer. (3 marks)

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(a) $2 x^{2} - 21 x + 2 = 0$ . (b) Agree. The original discriminant is 17, which is positive, so $α$ and $β$ are real; and $α β = 2 > 0$ , so they have the same sign, which makes each ratio positive. The point of (b) is that both facts are needed — real, and same sign.

Question 3 · Original — sign conditions on roots, stated as conditions rather than as formulas

Let $k$ be a real constant. The equation $x^{2} - 2 (k + 1) x + (k^{2} + 3 k - 4) = 0$ has two real roots $α$ and $β$ .

(a) Find the range of values of $k$ . (3 marks)

(b) Suppose further that $α$ and $β$ are both positive. Find the range of values of $k$ . (4 marks)

(c) Is it possible for $α$ and $β$ to have opposite signs? Explain your answer. (2 marks)

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(a) $k \leq 5$ . (b) $1 < k \leq 5$ . (c) Yes, whenever $- 4 < k < 1$ . Nothing in the question names a method: "both positive" has to be translated into a positive sum and a positive product, and "opposite signs" into a negative product alone.

Question 4 · Original — tangency, with a claim that looks plausible and is not true

Let $m$ be a real constant. Consider the straight line $L : y = m x + 2$ and the curve $C : y = x^{2} - 2 x + 6$ .

(a) Find the range of values of $m$ such that $L$ and $C$ intersect at two distinct points. (4 marks)

(b) Find the values of $m$ such that $L$ and $C$ intersect at exactly one point. (2 marks)

(c) A student claims that for one of the values of $m$ found in (b), the single point of intersection lies on the $y$ -axis. Determine whether the claim is correct. (3 marks)

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(a) $m < - 6$ or $m > 2$ . (b) $m = 2$ or $m = - 6$ . (c) The claim is incorrect: the two points are $(2, 6)$ and $(- 2, 14)$ , and neither has $x = 0$ . Part (c) cannot be answered from the discriminant alone — the point itself has to be found.

Question 5 · Original — a coefficient that can vanish, and a proof that holds for every other value

Let $k$ be a real constant. Consider the equation $(k + 1) x^{2} - (k + 3) x + 1 = 0 \dots\dots (*)$

(a) Find the value of $k$ for which $(*)$ is a linear equation, and solve $(*)$ in that case. (3 marks)

(b) Show that for every k ≠ −1, $(*)$ has two distinct real roots. (4 marks)

(c) Find the value of $k$ for which the two roots of $(*)$ are reciprocals of each other. (3 marks)

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(a) $k = - 1$ , and then $x = \frac{1}{2}$ . (b) The discriminant is $k^{2} + 2 k + 5 = (k + 1)^{2} + 4$ , which is positive for every real $k$ . (c) $k = 0$ . Part (b) is a "show that" — completing the square on the discriminant is what makes it a proof rather than a check of a few values.

Question 6 · Original — an unstated relationship between two equations

Let $p$ and $q$ be real constants. The roots of the equation $x^{2} - p x + q = 0$ are $α$ and $β$ . It is given that $α - 3$ and $β - 3$ are the roots of the equation $x^{2} + p x + q = 0$ .

(a) Find $p$ . (4 marks)

(b) Suppose further that $x^{2} - p x + q = 0$ has two distinct real roots. Find the range of values of $q$ . (3 marks)

(c) Someone claims that under the conditions above, $α - 3$ and $β - 3$ must both be negative. Do you agree? Explain your answer. (4 marks)

Show answer

(a) $p = 3$ . (b) $q < \frac{9}{4}$ . (c) Disagree. A counter-example settles it: when $q = - 4$ the shifted roots are 1 and −4, so one of them is positive. The product of the shifted roots is $q$ , and $q$ is allowed to be negative.

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

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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