Quadratic Functions Practice

Q: When should I complete the square instead of factorising?

Factorise to find roots quickly. Complete the square when you need the vertex, the minimum or maximum value and where it occurs, because the form (x−p)2+q(x-p)^{2}+q(x−p)2+q shows both at a glance.

Q: What does the discriminant tell me?

For ax2+bx+c=0ax^{2}+bx+c=0ax2+bx+c=0, b2−4ac>0b^{2}-4ac>0b2−4ac>0 gives two distinct real roots, =0=0=0 gives two equal roots, and <0<0<0 gives no real roots. Many questions turn a condition on the roots into a condition on b2−4acb^{2}-4acb2−4ac.

Q: How do I decide which region satisfies a quadratic inequality?

Find the roots, sketch the parabola, and read the sign. An upward parabola is below zero between its roots and above zero outside them. A quick test point confirms your choice.

Q: Do I lose all the marks if my final answer is wrong?

No. Because marking is analytic, a correct method line and correct working still earn marks even if the arithmetic slips at the end. That is why you should always show full working.

How to use these practice questions

The six questions below rise in difficulty across the whole Quadratic Functions chapter, from a single factorisation to reasoning with the discriminant. Give yourself roughly six to nine minutes per question and work on paper first, writing each line as you would in the real exam, a method line, a clear rearrangement, then the final answer.

Do not open the solution until you have committed a full attempt to paper.

When your working differs from ours, stop at the exact line where the two part company; that single step is almost always where the mark was lost. Because Add Math is marked analytically, a clear method line still earns credit even when the final arithmetic slips, so always show the standard form $ax^{2}+bx+c=0$ , the factors, or the discriminant in full.

Treat this page as a rehearsal, the goal is to find weak steps now, while there is still time to fix them.

Six practice questions

Q1[3 marks]

Solve the quadratic equation $x^{2}+2x=15$ .

Show worked solution

First rearrange into the standard form $ax^{2}+bx+c=0$ , so that one side is zero:

x^{2}+2x-15=0

Factorise: find two numbers that multiply to $-15$ and add to $+2$ . Those are $+5$ and $-3$ :

(x+5)(x-3)=0

Set each factor to zero and solve:

x+5=0 \;\Rightarrow\; x=-5 \qquad x-3=0 \;\Rightarrow\; x=3

Answer

$x=-5$ or $x=3$ . Check the first root: $(-5)^{2}+2(-5)=25-10=15$ , as required.

Q2[4 marks]

The quadratic equation $x^{2}-5x+3=0$ has roots $\alpha$ and $\beta$ . Find (a) $\alpha+\beta$ , (b) $\alpha\beta$ , and (c) the value of $\alpha^{2}+\beta^{2}$ .

Show worked solution

For $ax^{2}+bx+c=0$ , the sum of roots is $-\dfrac{b}{a}$ and the product is $\dfrac{c}{a}$ . Here $a=1$ , $b=-5$ , $c=3$ , so read them off directly:

\alpha+\beta=-\frac{b}{a}=-\frac{-5}{1}=5 \qquad \alpha\beta=\frac{c}{a}=\frac{3}{1}=3

(c) Do not solve for the roots. Instead use the identity $\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta$ , which is built from the two values you already have:

\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta=(5)^{2}-2(3)=25-6=19

Answer

$\alpha+\beta=5$ , $\alpha\beta=3$ , and $\alpha^{2}+\beta^{2}=19$ . The identity turns two easy quantities into the answer without ever finding $\alpha$ and $\beta$ separately.

Q3[4 marks]

The quadratic equation $x^{2}+kx+9=0$ has two equal roots. Find the possible values of $k$ .

Show worked solution

Two equal roots means the discriminant is zero. For $ax^{2}+bx+c=0$ the discriminant is $b^{2}-4ac$ .

Here $a=1$ , $b=k$ , $c=9$ , so set it to zero:

b^{2}-4ac=0 \;\Rightarrow\; k^{2}-4(1)(9)=0

Solve for $k$ :

k^{2}=36 \;\Rightarrow\; k=\pm 6

Answer

$k=6$ or $k=-6$ . Check $k=6$ : the equation becomes $x^{2}+6x+9=(x+3)^{2}=0$ , which indeed has the single repeated root $x=-3$ .

Q4[5 marks]

Express $f(x)=x^{2}-6x+11$ in the form $(x-p)^{2}+q$ . Hence state the minimum value of $f(x)$ and the value of $x$ at which it occurs.

Show worked solution

Complete the square. Take half of the coefficient of $x$ , which is $-6$ ; half of it is $-3$ .

Write the square $(x-3)^{2}$ , which expands to $x^{2}-6x+9$ , so subtract the extra $9$ and keep the $+11$ :

x^{2}-6x+11=(x-3)^{2}-9+11=(x-3)^{2}+2

So $p=3$ and $q=2$ . A squared term $(x-3)^{2}$ is never negative, and its smallest value is $0$ , reached when $x-3=0$ .

At that point $f(x)$ takes its minimum:

f(x)_{\min}=0+2=2 \quad\text{at}\quad x=3

Answer

$f(x)=(x-3)^{2}+2$ ; the minimum value is $2$ , occurring at $x=3$ . Check by expanding: $(x-3)^{2}+2=x^{2}-6x+9+2=x^{2}-6x+11$ , which matches the original.

Q5[5 marks]

Find the range of values of $x$ for which $x^{2}-x-6<0$ .

Show worked solution

First find the roots of the matching equation $x^{2}-x-6=0$ by factorising. Two numbers that multiply to $-6$ and add to $-1$ are $-3$ and $+2$ :

(x-3)(x+2)=0 \;\Rightarrow\; x=3 \;\text{or}\; x=-2

The graph of $y=x^{2}-x-6$ is a parabola opening upward, so it dips below the $x$ -axis only between its two roots. The inequality asks where $y<0$ , which is that inside region:

-2<x<3

Answer

$-2<x<3$ . Check with a test point inside, $x=0$ : $0-0-6=-6<0$ , true; and outside, $x=4$ : $16-4-6=6>0$ , so $x=4$ is correctly excluded.

Q6[6 marks]

The quadratic equation $x^{2}+px+(p+3)=0$ has no real roots. Find the range of values of $p$ .

Show worked solution

No real roots means the discriminant is negative: $b^{2}-4ac<0$ . Here $a=1$ , $b=p$ , $c=p+3$ .

Substitute and expand carefully:

p^{2}-4(1)(p+3)<0 \;\Rightarrow\; p^{2}-4p-12<0

This is itself a quadratic inequality in $p$ . Factorise the left side: two numbers multiplying to $-12$ and adding to $-4$ are $-6$ and $+2$ :

(p-6)(p+2)<0 \;\Rightarrow\; p=6 \;\text{or}\; p=-2 \;\text{are the boundaries}

The expression $(p-6)(p+2)$ is a parabola opening upward, so it is negative only between its roots:

-2<p<6

Answer

$-2<p<6$ . Check $p=0$ , which lies inside: the discriminant is $0-0-12=-12<0$ , so there are no real roots, as required.

At the boundary $p=6$ the discriminant is $36-24-12=0$ , giving equal roots, so the boundary is correctly excluded by the strict inequality.

How to mark yourself like an examiner

Add Math is marked analytically, which means marks are attached to steps, not only to the final number. When you check your own script, award yourself credit the way a marker would: look for the correct method being started, the right substitution, and a clean final statement.

Method mark: did you write the correct move, standard form $ax^{2}+bx+c=0$ , the sum/product formulas, or $b^{2}-4ac$ for the discriminant?
Working mark: are the factors, the completed square, or the substituted discriminant all correct?
Answer mark: is the final value or range stated clearly, and does a test point or substitution confirm it?
For an inequality, the correct boundary roots earn marks even before you decide which region satisfies the sign.
If your final number is wrong but the method line is right, give yourself the method mark, that is exactly what a real marker does.

How a teacher helps

Marking yourself is powerful, but it is hard to see your own blind spots. In a one-to-one lesson our teacher watches the exact line where a mark slips away, a sign error while completing the square, or the wrong region chosen in an inequality, and corrects the habit on the spot.

Because our teachers are experienced, you work with someone who explains the why behind each step. Lessons are taught in English, while SPM papers are set in both Malay and English, so we make sure the notation reads the same to you either way.

Get 1-to-1 help.

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Frequently asked questions

When should I complete the square instead of factorising?

Factorise to find roots quickly. Complete the square when you need the vertex, the minimum or maximum value and where it occurs, because the form $(x-p)^{2}+q$ shows both at a glance.

What does the discriminant tell me?

For $ax^{2}+bx+c=0$ , $b^{2}-4ac>0$ gives two distinct real roots, $=0$ gives two equal roots, and $<0$ gives no real roots. Many questions turn a condition on the roots into a condition on $b^{2}-4ac$ .

How do I decide which region satisfies a quadratic inequality?

Find the roots, sketch the parabola, and read the sign. An upward parabola is below zero between its roots and above zero outside them.

A quick test point confirms your choice.

Do I lose all the marks if my final answer is wrong?

No. Because marking is analytic, a correct method line and correct working still earn marks even if the arithmetic slips at the end.

That is why you should always show full working.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

Quadratic Functions, Practice Questions

How to use these practice questions

Six practice questions

How to mark yourself like an examiner

How a teacher helps

Frequently asked questions

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