Practice · HKDSE Mathematics · Paper 2 multiple-choice

Quadratic and simultaneous equations — Paper 2 multiple-choice

Twelve multiple-choice questions on quadratic and simultaneous equations, matched to HKDSE Paper 2 style from 2014 to 2023. The real papers return to three ideas again and again: the discriminant, substituting a root back into an expression, and the sum and product of roots. All three are here, along with the variants the papers have not used recently.

12 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.

Question 1 · Modelled on 2016 Paper 2 Q8 — equal roots with the constant appearing twice

Let $k$ be a constant. If the quadratic equation $x^{2}+kx+6k-27=0$ has equal roots, then $k=$

$6$
$6$ or $18$
$18$
$-6$ or $-18$

Show answer

B. $\Delta=0$ gives $k^{2}=4(6k-27)$ , so $k^{2}-24k+108=0$ and $k=6$ or $k=18$ . Both satisfy the original condition, so neither can be dropped.

Question 2 · Modelled on 2015 Paper 2 Q7 — substituting a root back into an expression

If $\beta$ is a root of the equation $3x^{2}-7x-2=0$ , then $6\beta^{2}-14\beta+5=$

$1$
$5$
$11$
$9$

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D. From $3\beta^{2}-7\beta-2=0$ we get $3\beta^{2}-7\beta=2$ , so $6\beta^{2}-14\beta=4$ and the expression equals $9$ . Solving for $\beta$ first is slower and introduces surds.

Question 3 · Modelled on 2015 Paper 2 Q34 — a different symmetric combination of the roots

Let $k$ be a constant. If the roots of the quadratic equation $x^{2}+kx-5=0$ are $\alpha$ and $\beta$ , then $\dfrac{1}{\alpha}+\dfrac{1}{\beta}=$

$\dfrac{k}{5}$
$-\dfrac{k}{5}$
$-5k$
$\dfrac{k^{2}}{5}$

Show answer

A. $\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{\alpha+\beta}{\alpha\beta}=\dfrac{-k}{-5}=\dfrac{k}{5}$ . The two minus signs cancel — that is the whole trap.

Question 4 · Modelled on 2014 Paper 2 Q4 — inverted to ask for a range rather than a value

If the quadratic equation $2x^{2}+5x+c=0$ has two distinct real roots, then the range of values of $c$ is

$c>\dfrac{25}{8}$
$c<\dfrac{8}{25}$
$c<\dfrac{25}{8}$
$c>\dfrac{8}{25}$

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C. $\Delta>0$ gives $25-8c>0$ , so $c<\dfrac{25}{8}$ . Dividing the wrong way round produces option B.

Question 5 · Modelled on 2022 Paper 2 Q4 — a common factor hidden on both sides

Let $a$ be a constant. Solve the equation $(x-2a)(x+a)=(x-2a)(3a-2x)$ .

$x=2a$
$x=\dfrac{2a}{3}$
$x=-a$ or $x=2a$
$x=2a$ or $x=\dfrac{2a}{3}$

Show answer

D. Move everything to one side: $(x-2a)\big[(x+a)-(3a-2x)\big]=(x-2a)(3x-2a)=0$ . Cancelling $(x-2a)$ instead of factorising loses the root $x=2a$ .

Question 6 · Original — non-real roots given as the condition, testing that sum and product still hold

If the equation $x^{2}-6x+c=0$ has two non-real roots $\alpha$ and $\beta$ , which of the following must be true?
I. $c>9$
II. $\alpha+\beta=6$
III. $\alpha\beta<0$

II and III only
I and II only
I and III only
I, II and III

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B. $\Delta<0$ gives $36-4c<0$ , so $c>9$ , and the sum of roots is $6$ whether or not the roots are real. But $\alpha\beta=c>9$ , so III is false.

Question 7 · Original — forming a new equation from transformed roots, a sub-type absent from the 2014–2023 MC papers

If $\alpha$ and $\beta$ are the roots of $x^{2}-4x+1=0$ , then the quadratic equation whose roots are $2\alpha$ and $2\beta$ is

$x^{2}-4x+4=0$
$x^{2}-2x+4=0$
$x^{2}-8x+4=0$
$x^{2}-8x+2=0$

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C. The new sum is $2(\alpha+\beta)=8$ and the new product is $4\alpha\beta=4$ . Doubling the roots doubles the sum but multiplies the product by four.

Question 8 · Original — reducible to quadratic, counting real roots

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

Show answer

A. Putting $t=x^{2}$ gives $t=1$ or $t=4$ . Both are positive, so each yields two real values of $x$ : $\pm 1$ and $\pm 2$ . Stopping at $t$ and answering $2$ is the usual error.

Question 9 · Original — tangency as a discriminant condition, the MC form of the long-question line-meets-curve work

If the straight line $y=x+k$ is a tangent to the curve $y=x^{2}+3x+7$ , then $k=$

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D. Eliminating $y$ gives $x^{2}+2x+(7-k)=0$ . Tangency means one repeated root, so $4-4(7-k)=0$ and $k=6$ .

Question 10 · Original — a simultaneous linear and quadratic system, with a maximum asked at the end

If $(x,\,y)$ satisfies both $y=x-1$ and $x^{2}+y^{2}=25$ , then the greatest possible value of $x+y$ is

$-7$
$1$
$7$
$5$

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C. Substituting gives $2x^{2}-2x-24=0$ , so $x=4$ or $x=-3$ , giving the points $(4,3)$ and $(-3,-4)$ . Their values of $x+y$ are $7$ and $-7$ .

Question 11 · Modelled on 2014 Paper 2 Q4 — the unknown constant now sits on $x^{2}$

Let $k$ be a non-zero constant. If the equation $kx^{2}-4x+1=0$ has two equal real roots, then $k=$

$\dfrac{1}{4}$
$4$
$2$
$16$

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B. $\Delta=16-4k=0$ gives $k=4$ . Note the condition $k\neq 0$ is needed for the equation to be quadratic at all.

Question 12 · Original — roots in a given ratio, the MC counterpart of the long-question $q,\,3q$ technique

Let $c$ be a constant. If one root of the equation $x^{2}+cx+18=0$ is twice the other, then $c=$

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

Show answer

D. Let the roots be $q$ and $2q$ . Then $2q^{2}=18$ , so $q=\pm 3$ , and $3q=-c$ gives $c=-9$ or $c=9$ . Both cases are real, so both count.

Where this sits in the syllabus

Paper 2 rewards the short route. Most of these questions have one, and it usually runs through the sum and product of roots rather than through solving the equation. If you found yourself reaching for the quadratic formula more than twice, that is the habit to work on.

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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