Linear and Quadratic Equations and Inequalities

Lessons

Senior 4 · Mathematics

Linear and Quadratic Equations and Inequalities

5 lessons

Lesson 4 of 5

Quadratic inequalities

25 min

When is the ball above 15 metres?

A ball is thrown straight up at $20\,\mathrm{m\,s^{-1}}$ . Taking $g \approx 10\,\mathrm{m\,s^{-2}}$ and ignoring air resistance, its height after $t$ seconds is $h = 20 t - 5 t^{2}$ metres. Asking "for how long is it higher than 15 m?" means solving $20t - 5t^2 \gt 15$ , a quadratic inequality. Every such inequality reduces to one question: where is a quadratic expression positive, and where is it negative?

Definition

Quadratic inequality in $x$ : an inequality that can be written $ax^2 + bx + c \gt 0$ (or $\ge 0$ , $\lt 0$ , $\le 0$ ) with $a \ne 0$ . The expression $a x^{2} + b x + c$ is called a quadratic trinomial.

Example: $20t - 5t^2 \gt 15$ is the quadratic inequality $-5t^2 + 20t - 15 \gt 0$ .

The sign of $a x^{2} + b x + c$

If $\Delta \gt 0$ the trinomial factorises as $a (x - x_{1}) (x - x_{2})$ with $x_1 \lt x_2$ . For $x \gt x_2$ both brackets are positive; for $x \lt x_1$ both are negative, so their product is again positive; between the roots exactly one bracket is negative. If $\Delta \le 0$ , completing the square gives

ax^2 + bx + c = a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{\Delta}{4a^2}\right],

and the square bracket is never negative, so the trinomial has the sign of $a$ .

Important

Sign rule for $a x^{2} + b x + c$

Discriminant	Sign of $a x^{2} + b x + c$
$\Delta \gt 0$ , roots $x_1 \lt x_2$	sign of $a$ outside the roots, opposite sign between them, 0 at the roots
$\Delta = 0$ , double root $x_{0}$	sign of $a$ everywhere except $x_{0}$ , where it is 0
$\Delta \lt 0$	sign of $a$ for every real $x$

Memory hook: "same sign as $a$ outside the roots".

For $y = 2 x^{2} - x - 6$ ( $a = 2 \gt 0$ ) the graph is below the $x$ -axis only between the roots $-\frac{3}{2}$ and $2$ : the trinomial is negative there and positive outside.

Worked example 1: two roots

Solve $2x^2 - x - 6 \le 0$ .

Solution $\Delta = 1 + 48 = 49$ , so $x = \frac{1 \pm 7}{4}$ : $x_1 = -\frac{3}{2}$ , $x_{2} = 2$ . Since $a = 2 \gt 0$ , the trinomial is negative between the roots and zero at them. Solution set $\left[-\frac{3}{2}, 2\right]$ .

Worked example 2: $\Delta = 0$ and $\Delta \lt 0$

Solve (a) $-x^2 + 4x - 4 \lt 0$ (b) $x^2 - x + 1 \gt 0$ (c) $x^2 - x + 1 \le 0$ .

Solution

$- x^{2} + 4 x - 4 = - (x - 2)^{2}$ . It is negative for every $x \ne 2$ and zero at 2. Solution set $\mathbb{R} \setminus \{2\}$ .
$\Delta = 1 - 4 = -3 \lt 0$ and $a = 1 \gt 0$ , so $x^2 - x + 1 \gt 0$ for every real $x$ . Solution set $\mathbb{R}$ .
By (b), never true. Solution set $\varnothing$ .

Worked example 3: a quotient of quadratics

Solve $\frac{x^2 - 4}{x^2 - 4x + 3} \ge 0$ .

Solution Factorise: $\frac{(x - 2)(x + 2)}{(x - 1)(x - 3)} \ge 0$ . Critical values: $- 2, 2$ (zeros) and $1, 3$ (excluded).

Each trinomial is positive outside its roots and negative between them (a = 1 > 0).
$x$		$- 2$		$1$		$2$		$3$
$x^{2} - 4$	+	0	−		−	0	+		+
$x^{2} - 4 x + 3$	+		+	0	−		−	0	+
quotient	+	0	−	‖	+	0	−	‖	+

Solution set $(-\infty, -2] \cup (1, 2] \cup (3, +\infty)$ . Note the round brackets at 1 and 3, where the quotient is undefined.

Inequalities with a parameter, and systems

The trinomial $a x^{2} + b x + c$ is positive for every real $x$ exactly when $a \gt 0$ and $\Delta \lt 0$ , and negative for every $x$ exactly when $a \lt 0$ and $\Delta \lt 0$ . The condition $\Delta \lt 0$ is itself an inequality in the parameter.

Worked example 4: true for every $x$

Find the values of $m$ for which $x^2 + (m - 1)x + m + 2 \gt 0$ for every real $x$ .

Solution Here $a = 1 \gt 0$ , so we need $\Delta \lt 0$ :

(m - 1)^2 - 4(m + 2) \lt 0 \iff m^2 - 6m - 7 \lt 0 \iff (m - 7)(m + 1) \lt 0 .

This trinomial in $m$ is negative between its roots: $-1 \lt m \lt 7$ .

Check $m = 3$ : $x^2 + 2x + 5 = (x + 1)^2 + 4 \gt 0$ ✓. $m = 7$ : $x^{2} + 6 x + 9 = (x + 3)^{2}$ , which is 0 at $x = - 3$ , so 7 is rightly excluded.

Worked example 5: a system of two inequalities

Solve the system $x^2 - 5x + 4 \lt 0$ and $x^2 - 9 \ge 0$ .

Solution First: roots 1 and 4, negative between them: $(1, 4)$ . Second: roots $\pm 3$ , non-negative outside: $(-\infty, -3] \cup [3, +\infty)$ . The system needs both, so take the intersection: $[3, 4)$ .

The solution of a system is the part covered by every row: here $[3, 4)$ , closed at 3 and open at 4.

Worked example 6: back to the ball

For how long is the ball with $h = 20 t - 5 t^{2}$ at least 15 m above the ground?

Solution $20t - 5t^2 \ge 15 \iff 5t^2 - 20t + 15 \le 0 \iff t^2 - 4t + 3 \le 0 \iff (t - 1)(t - 3) \le 0$ , so $1 \le t \le 3$ . The ball is at or above 15 m for $3 - 1 = 2$ seconds.

Warning

Dividing by a negative number reverses the inequality: $-5t^2 + 20t - 15 \ge 0$ becomes $t^2 - 4t + 3 \le 0$ after dividing by $- 5$ . And never "take the square root of both sides" of $x^2 \gt 9$ to get $x \gt 3$ : the full answer is $|x| \gt 3$ , that is $x \lt -3$ or $x \gt 3$ .

Tip

Draw a tiny sketch of the parabola: which way it opens (sign of $a$ ) and where it cuts the axis (the roots). The answer to " $\gt 0$ " is where the sketch is above the axis, and to " $\lt 0$ " where it is below. It takes five seconds and prevents most sign errors.

Khan Academy: Quadratic inequalities (about 10 min). Watch how the zeros of the quadratic split the number line, and how a sketch of the parabola tells which pieces satisfy the inequality.

✦

Practice

Give each answer as an interval or a union of intervals.

Solve $x^2 - x - 12 \gt 0$ .
Solve $-3x^2 + 5x + 2 \ge 0$ .
Solve $4x^2 - 4x + 1 \gt 0$ .
Solve $x(x + 2) \lt 2x + 9$ .
Solve $\frac{x^2 - 1}{x + 2} \le 0$ .
Find $m$ such that $mx^2 + 4x + m \lt 0$ for every real $x$ .
Solve the system $x^2 + x - 6 \le 0$ and $2x^2 - x - 1 \gt 0$ .
A farmers' cooperative models its daily profit, in thousands of francs, as $P = - 2 x^{2} + 120 x - 1000$ , where $x$ is the number of crates sold (illustrative figures). For which $x$ is the profit at least 600 thousand francs?

Show answersHide answers

$(x - 4)(x + 3) \gt 0$ : $(-\infty, -3) \cup (4, +\infty)$ .
$\Delta = 25 + 24 = 49$ , roots $2$ and $-\frac{1}{3}$ ; $a \lt 0$ , so non-negative between: $\left[-\frac{1}{3}, 2\right]$ .
$(2x - 1)^2 \gt 0$ for all $x \ne \frac{1}{2}$ : $\mathbb{R} \setminus \{\frac{1}{2}\}$ .
$x^2 - 9 \lt 0$ : $(- 3, 3)$ .
Sign table with $- 2$ (excluded), $- 1$ , $1$ : $(-\infty, -2) \cup [-1, 1]$ .
Need $m \lt 0$ and $\Delta = 16 - 4m^2 \lt 0$ , that is $m^2 \gt 4$ . With $m \lt 0$ : $m \lt -2$ .
First: $[- 3, 2]$ . Second: roots $-\frac{1}{2}$ and $1$ , so $(-\infty, -\frac{1}{2}) \cup (1, +\infty)$ . Both: $[-3, -\frac{1}{2}) \cup (1, 2]$ .
$-2x^2 + 120x - 1000 \ge 600 \iff x^2 - 60x + 800 \le 0 \iff (x - 20)(x - 40) \le 0$ : from 20 to 40 crates.

Quick check

The trinomial −x² + 2x + 8 has roots −2 and 4. Where is it positive?

Key points

$a x^{2} + b x + c$ has the sign of $a$ outside its roots and the opposite sign between them; with $\Delta \le 0$ it has the sign of $a$ (except at a double root).
For products and quotients, factorise and use a sign table; exclude the zeros of the denominator.
$ax^2 + bx + c \gt 0$ for all $x \iff a \gt 0$ and $\Delta \lt 0$ .
A system of inequalities: solve each one, then intersect the solution sets.

Sources and credits

Written by Ganzaa: the explanations, the diagrams, the worked examples (the cooperative's profit figures are illustrative), the practice items and the quiz questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced.
Photo: Parabolic water trajectory, GuidoB, Wikimedia Commons, CC BY-SA 3.0.
Video: Quadratic inequalities, Khan Academy, embedded from YouTube.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1When is the ball above 15 metres?

2The sign of ax2+bx+c ax^2 + bx + c ax2+bx+c

3Inequalities with a parameter, and systems

When is the ball above 15 metres?

The sign of $a x^{2} + b x + c$

Inequalities with a parameter, and systems