Quadratic inequalities

AS · P1 · 10 min

A quadratic inequality asks where a quadratic is positive or negative, for example $x^2 - x - 12 \le 0$ . The answer is a range of values, not a list of numbers. These inequalities appear on their own, and also at the end of almost every discriminant question ("find the set of values of $k$ "), so a reliable method matters more than speed. The reliable method is always the same: find the critical values, sketch, read off.

The idea

A quadratic can only change sign where it equals zero. Its roots, the critical values, split the number line into regions, and in each region the quadratic has a fixed sign.

For $y = (x + 3)(x - 4)$ , the roots are $-3$ and $4$ , and the graph is $\cup$ -shaped:

y = (x + 3)(x - 4) (-3, 0) (4, 0)

Between the roots the curve is below the axis: $(x + 3)(x - 4) < 0$ for $-3 < x < 4$ .
Outside the roots the curve is above the axis: $(x + 3)(x - 4) > 0$ for $x < -3$ or $x > 4$ .

Key result

If $a > 0$ and the roots of $ax^2 + bx + c = 0$ are $\alpha < \beta$ , then

ax^2 + bx + c < 0 \iff \alpha < x < \beta

ax^2 + bx + c > 0 \iff x < \alpha \ \text{ or } \ x > \beta

With $\le$ or $\ge$ the critical values are included: $\alpha \le x \le \beta$ , or $x \le \alpha$ or $x \ge \beta$ .

Notice the shape of the answers. "Between" is a single combined inequality, $\alpha < x < \beta$ . "Outside" is two separate inequalities joined by or, because no number can be both less than $\alpha$ and greater than $\beta$ .

Solving a quadratic inequality

Rearrange so that one side is $0$ . It is easiest to make the $x^2$ coefficient positive (multiply by $-1$ and reverse the inequality sign if needed).
Solve the equation $ax^2 + bx + c = 0$ to find the critical values.
Sketch the parabola, marking the critical values on the $x$ -axis.
Read off where the curve is above the axis (for $>$ , $\ge$ ) or below it (for $<$ , $\le$ ).
Write the answer with the correct inequality signs, including or excluding the critical values.

Writing the answer

All of these are acceptable ways of writing "outside $-3$ and $4$ ":

$x < -3$ or $x > 4$
$\{x : x < -3\} \cup \{x : x > 4\}$

And "between $-3$ and $4$ , inclusive":

$-3 \le x \le 4$
$\{x : -3 \le x \le 4\}$

What is not acceptable: " $-3 > x > 4$ " (no number satisfies it), " $x < -3$ and $x > 4$ " (the word "and" is wrong), or simply " $x = -3$ , $x = 4$ " (those are the critical values, not the solution).

Between the roots

Solve $x^2 - x - 12 \le 0$ .

Solution

Critical values: $x^2 - x - 12 = (x - 4)(x + 3) = 0$ , so $x = -3$ or $x = 4$ .

The graph is $\cup$ -shaped; it is on or below the axis between the roots (see the sketch above). Since the inequality is $\le$ , the critical values are included:

-3 \le x \le 4

Outside the roots

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

Solution

Critical values: $2x^2 - x - 6 = (2x + 3)(x - 2) = 0$ , so $x = -\tfrac{3}{2}$ or $x = 2$ .

y = 2x^2 - x - 6 (-1.5, 0) (2, 0)

The $\cup$ -shaped curve is on or above the axis outside the roots:

x \le -\tfrac{3}{2} \quad \text{or} \quad x \ge 2

A negative coefficient of x squared

Solve $3 + 2x - x^2 > 0$ .

Solution

Multiply both sides by $-1$ to make the $x^2$ coefficient positive, and reverse the inequality:

x^2 - 2x - 3 < 0

Critical values: $(x - 3)(x + 1) = 0$ , so $x = -1$ or $x = 3$ . The $\cup$ -shaped curve $y = x^2 - 2x - 3$ is below the axis between the roots:

-1 < x < 3

Alternatively, sketch $y = 3 + 2x - x^2$ directly. It is $\cap$ -shaped with the same roots, and it is above the axis between them. The answer is the same.

Rearranging first

Solve $x(x + 2) < 15$ .

Solution

It is not valid to say " $x < 15$ or $x + 2 < 15$ ". Expand and bring everything to one side:

x^2 + 2x - 15 < 0 \quad\Rightarrow\quad (x + 5)(x - 3) < 0

Critical values $-5$ and $3$ ; the curve is below the axis between them:

-5 < x < 3

Irrational critical values

Solve $x^2 - 4x - 1 > 0$ , giving your answer in exact form.

Solution

The quadratic does not factorise, so complete the square (or use the formula):

x^2 - 4x - 1 = (x - 2)^2 - 5 = 0 \quad\Rightarrow\quad x = 2 \pm \sqrt{5}

The $\cup$ -shaped curve is above the axis outside the roots:

x < 2 - \sqrt{5} \quad \text{or} \quad x > 2 + \sqrt{5}

Where one graph is above another

Find the set of values of $x$ for which the curve $y = x^2 - 3x + 4$ lies above the line $y = x + 1$ .

Solution

"The curve lies above the line" means

x^2 - 3x + 4 > x + 1 \quad\Rightarrow\quad x^2 - 4x + 3 > 0 \quad\Rightarrow\quad (x - 1)(x - 3) > 0

So $x < 1$ or $x > 3$ .

y = x^2 - 3x + 4 y = x + 1 (1, 2) (3, 4)

The curve and line meet at $(1, 2)$ and $(3, 4)$ ; the curve is above the line to the left of the first point and to the right of the second.

Two inequalities at once

Find the set of values of $x$ that satisfy both $x^2 - x - 6 > 0$ and $2x^2 - 13x + 15 \le 0$ .

Solution

First inequality: $(x - 3)(x + 2) > 0$ , so $x < -2$ or $x > 3$ .

Second inequality: $(2x - 3)(x - 5) \le 0$ , so $\tfrac{3}{2} \le x \le 5$ .

Both must hold. Put them on a number line:

the first allows everything left of $-2$ and everything right of $3$ ;
the second allows only $\tfrac{3}{2}$ to $5$ .

The overlap is the part of $\left[\tfrac{3}{2}, 5\right]$ to the right of $3$ :

3 < x \le 5

The end $x = 3$ is excluded because the first inequality is strict; $x = 5$ is included because the second is not.

Watch out

Dividing by $x$ . From $x^2 > 4x$ you cannot divide by $x$ to get $x > 4$ , because $x$ might be negative (which reverses the sign) or zero. Write $x^2 - 4x > 0$ , so $x(x - 4) > 0$ , giving $x < 0$ or $x > 4$ . The answer $x < 0$ is lost by dividing.

Square rooting an inequality. $x^2 > 9$ does not give $x > 3$ ; it gives $x < -3$ or $x > 3$ . $x^2 < 9$ gives $-3 < x < 3$ .

Forgetting to reverse the sign when multiplying by a negative number.

Stopping at the critical values. $x = -3$ and $x = 4$ are not the answer to an inequality.

Getting "between" and "outside" the wrong way round. Never decide from memory; always look at a sketch.

Exam tip

A quick sketch of the parabola with the critical values marked is the single best habit for inequalities. It is acceptable working and it prevents the most common error.
The critical values usually earn a method mark and the final inequality an accuracy mark. If your final inequality uses the wrong sign (for example $<$ instead of $\le$ ), you lose that accuracy mark.
When an inequality comes from a discriminant (in terms of $k$ ), it is solved in exactly the same way. Treat $k$ as the variable.
In context questions (lengths, areas), remember physical restrictions such as lengths being positive; they may cut the solution set down further.

Summary

Rearrange to $ax^2 + bx + c \ (\text{sign}) \ 0$ , ideally with $a > 0$ .
Find the critical values by solving $= 0$ .
Sketch: for $a > 0$ , the quadratic is negative between the roots and positive outside them.
"Between" is one inequality, $\alpha < x < \beta$ ; "outside" is two, $x < \alpha$ or $x > \beta$ .
$\le$ and $\ge$ include the critical values; $<$ and $>$ exclude them.
Never divide by $x$ or take square roots of an inequality.
For two inequalities together, find the overlap on a number line.

Practice questions

Question

Solve $x^2 - 7x + 10 < 0$ .
Solve $x^2 + 3x - 18 \ge 0$ .
Solve $6 + x - 2x^2 > 0$ .
Solve $x^2 > 9$ .
Solve $(2x - 1)(x + 3) \le 4x$ .
Solve $x^2 - 6x + 4 < 0$ , giving your answer in exact form.
Find the set of values of $x$ for which the curve $y = 2x^2 - 3x$ lies below the line $y = x + 6$ .
Find all the integers $x$ that satisfy both $x^2 - 2x - 15 < 0$ and $x^2 > 4$ .
Find the set of values of $x$ that satisfy both $x^2 - x - 6 > 0$ and $2x^2 - 11x + 5 < 0$ .
A rectangle has length $(x + 3)$ cm and width $(x - 1)$ cm. Its area is less than $21\ \text{cm}^2$ and its perimeter is greater than $12$ cm. Find the set of possible values of $x$ .

Answers

$(x - 2)(x - 5) < 0$ , so $2 < x < 5$ .
$(x + 6)(x - 3) \ge 0$ , so $x \le -6$ or $x \ge 3$ .
Multiply by $-1$ : $2x^2 - x - 6 < 0$ , i.e. $(2x + 3)(x - 2) < 0$ . So $-\tfrac{3}{2} < x < 2$ .
$x^2 - 9 > 0$ , i.e. $(x - 3)(x + 3) > 0$ . So $x < -3$ or $x > 3$ .
$2x^2 + 5x - 3 \le 4x$ , so $2x^2 + x - 3 \le 0$ , i.e. $(2x + 3)(x - 1) \le 0$ . So $-\tfrac{3}{2} \le x \le 1$ .
$(x - 3)^2 - 5 = 0$ gives $x = 3 \pm \sqrt{5}$ . Between the roots: $3 - \sqrt{5} < x < 3 + \sqrt{5}$ .
$2x^2 - 3x < x + 6$ , so $2x^2 - 4x - 6 < 0$ , i.e. $x^2 - 2x - 3 < 0$ , $(x - 3)(x + 1) < 0$ . So $-1 < x < 3$ .
First: $(x - 5)(x + 3) < 0$ , so $-3 < x < 5$ . Second: $x < -2$ or $x > 2$ . Overlap: $-3 < x < -2$ or $2 < x < 5$ . There are no integers strictly between $-3$ and $-2$ , so the integers are $3$ and $4$ .
First: $x < -2$ or $x > 3$ . Second: $(2x - 1)(x - 5) < 0$ , so $\tfrac{1}{2} < x < 5$ . Overlap: $3 < x < 5$ .
The width must be positive: $x > 1$ . Area: $(x + 3)(x - 1) < 21$ , so $x^2 + 2x - 24 < 0$ , $(x + 6)(x - 4) < 0$ , giving $-6 < x < 4$ . Perimeter: $2(x + 3) + 2(x - 1) > 12$ , so $4x + 4 > 12$ , $x > 2$ . All three together: $2 < x < 4$ .

The idea

Writing the answer

Practice questions

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