Quadratic Inequalities

Q: How do I remember whether the answer is between or outside the roots?

Sketch the parabola. For $a > 0$ it dips below the axis between the roots, so '$ 0$' is outside.

A quadratic inequality asks for a range of $x$ -values, not just two answers. The method is the same every time: find the critical values, then decide which regions satisfy the inequality using a sketch or a number line.

Solving quadratic inequalities algebraically

Move everything to one side so that the other side is 0.
Factorise to find the critical values.
Decide where the expression is positive or negative.
Write the answer with the correct inequality signs.

Example. $2 x^{2} > 5 x + 3$

$2 x^{2} - 5 x - 3 > 0 \Rightarrow (2 x + 1) (x - 3) > 0$

Critical values: $x = - \frac{1}{2}$ and $x = 3$ .

The expression is positive outside the roots, so $x < - \frac{1}{2}$ or $x > 3$ .

Never divide by a variable or by an expression that could be negative. The direction of the inequality depends on its sign.

Using a sketch or a number line

Sketch method. Sketch $y = a x^{2} + b x + c$ through its roots:

$> 0$ : where the graph is above the $x$ -axis
$< 0$ : where the graph is below the $x$ -axis

For $a > 0$ the parabola opens upwards, so it is negative between the roots and positive outside them.

Number line method. Mark the critical values and test a value in each region.

Example. $x^{2} - x - 6 \leq 0$ : $(x - 3) (x + 2) \leq 0$ , so $- 2 \leq x \leq 3$ .

Special cases. $(x - 2)^{2} < 0$ has no real solution, because a square is never negative. $(x - 2)^{2} \geq 0$ is true for all real $x$ .

Non-quadratic inequalities (such as rational ones) are solved in the context of functions, by reading graphs.

Key terms

Critical values: The values where the expression equals 0; they separate the regions to test.
Interval notation: Writing a solution as an interval, such as $(- 2; 3]$ .
Strict inequality: $<$ or $>$ , where the critical values are not included.

Common mistakes

Writing the answer as $3 < x < - \frac{1}{2}$ instead of 'or' for regions outside the roots.
Dividing by a negative number without reversing the inequality.
Including critical values in a strict inequality.
Reading the regions the wrong way round when $a < 0$ .

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

Solve for $x$ : $x^{2} - x - 6 \leq 0$

Worked solution

Hidden - try the question first

Solution

$(x - 3) (x + 2) \leq 0$
Critical values $x = - 2$ and $x = 3$
$- 2 \leq x \leq 3$

Question 2

4 marks total

routine

Solve for $x$ : $2 x^{2} > 5 x + 3$

Worked solution

Hidden - try the question first

Solution

$(x - 3) (x + 2) \leq 0$
$(2 x + 1) (x - 3) > 0$
$x < - \frac{1}{2}$
or $x > 3$

Question 3

6 marks total

complex

(a) Solve for $x$ : $(x - 1) (x + 2) \geq 4$ (4)

(b) Explain why $x (x - 4) < - 4$ has no real solution. (2)

Worked solution

Hidden - try the question first

Solution

(a) $x^{2} + x - 2 - 4 \geq 0$
$x^{2} + x - 6 \geq 0 \Rightarrow (x + 3) (x - 2) \geq 0$
$x > 3$
or $x \geq 2$
(b) $x^{2} - 4 x + 4 < 0 \Rightarrow (x - 2)^{2} < 0$
A square is never negative, so there is no real solution

Question 4

8 marks total

problem-solving

(a) For which values of $x$ is $x^{2} - 2 x - 15$ real? (3)

(b) For which values of $k$ is $x^{2} + k x + 9 > 0$ for all real values of $x$ ? (5)

Worked solution

Hidden - try the question first

Solution

(a) $x^{2} + x - 2 - 4 \geq 0$
$(x - 5) (x + 3) \geq 0$
$x \leq - 3$ or $x \geq 5$
(b) The parabola opens upwards, so it is always positive if it has no $x$ -intercepts
$Δ < 0$ : $k^{2} - 4 (1) (9) < 0$
$k^{2} - 36 < 0$
$(k - 6) (k + 6) < 0$
$- 6 < k < 6$

FAQs

How do I remember whether the answer is between or outside the roots?

Sketch the parabola. For $a > 0$ it dips below the axis between the roots, so ' $< 0$ ' is between and ' $> 0$ ' is outside.

Can I divide both sides of an inequality by x?

No. You do not know whether $x$ is positive or negative, so you cannot tell whether to reverse the sign.

How do I write 'or' answers?

Write them as two separate inequalities joined by 'or', such as $x < - 3$ or $x > 2$ . Never combine them into one.

Solving quadratic inequalities algebraically

Using a sketch or a number line

Key terms

Common mistakes

Practice questions

FAQs

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