Solving Quadratic Inequalities

Quick answer

To solve a quadratic inequality, move everything to one side so that a quadratic is compared with 0, find its zeros, and let them split the number line into intervals. The quadratic keeps one sign on each interval, since it can change sign only at a zero, so one test value per interval decides it. The graph tells the same story: x² − x − 6 > 0 where the parabola is above the x-axis, outside its zeros, and x² − x − 6 < 0 between them.

What you'll learn

Solve quadratic inequalities with zeros and test points
Read the solution of a quadratic inequality from a graph
Write solutions in interval notation, with correct endpoints
Handle quadratics with no real zeros

Where is the parabola above the axis?

The inequality $x^2 - x - 6 > 0$ asks where the graph of $y = x^2 - x - 6$ lies above the $x$ -axis. It crosses the axis at its zeros: $x^2 - x - 6 = (x + 2)(x - 3)$ , so at $x = -2$ and $x = 3$ .

y = x² − x − 6

Where y = x² − x − 6 is positive and negative

The parabola is above the axis to the left of $-2$ and to the right of $3$ , and below it in between. So $x^2 - x - 6 > 0$ when $x < -2$ or $x > 3$ .

The method

Compare with 0: move every term to one side.
Find the zeros by factoring or the quadratic formula.
Test one value in each interval the zeros create.
Write the answer, including the zeros for $\le$ or $\ge$ and excluding them for $<$ or $>$ .

Why one test point is enough

A quadratic changes sign only by passing through $0$ , and it passes through $0$ only at its zeros. Between two neighboring zeros it cannot switch from positive to negative without crossing the axis somewhere in between, and there is no zero there to cross at. The zeros cut the number line into pieces on which the sign never changes, so testing a single number in each piece settles it.

Example 1 — Test points

Solve $x^2 - x - 6 > 0$ .

The zeros $-2$ and $3$ make three intervals. Test a number in each:

Interval	Test value	$x^2 - x - 6$	$> 0$ ?
$x < -2$	$-3$	$6$	yes
$-2 < x < 3$	$0$	$-6$	no
$x > 3$	$4$	$6$	yes

The solution is $x < -2$ or $x > 3$ , in interval notation $(-\infty, -2) \cup (3, \infty)$ .

x < −2 or x > 3

Worked examples

Common mistakes

Practice problems

Solve $x^2 - 9 < 0$ .

Answer

$-3 < x < 3$

Full solution

$(x - 3)(x + 3) < 0$ . The zeros are $\pm 3$ , and the upward parabola is below the axis between them.
Solve $x^2 - 5x + 4 \ge 0$ .

Answer

$x \le 1$ or $x \ge 4$

Full solution

$(x - 1)(x - 4) \ge 0$ . Outside the zeros, and at them.
Solve $x^2 + 2x - 8 > 0$ .

Answer

$x < -4$ or $x > 2$

Full solution

$(x + 4)(x - 2) > 0$ . Test $0$ : $-8$ , negative, so the middle interval is out.
Solve $3x^2 - x - 2 \le 0$ .

Answer

$-\tfrac{2}{3} \le x \le 1$

Full solution

$(3x + 2)(x - 1) \le 0$ , with zeros $-\tfrac{2}{3}$ and $1$ .
Solve $x^2 \le 4x$ .

Answer

$0 \le x \le 4$

Full solution

$x^2 - 4x = x(x - 4) \le 0$ between the zeros $0$ and $4$ .
Solve $x^2 + x + 1 < 0$ .

Answer

No solution

Full solution

The discriminant is $1 - 4 = -3 < 0$ , so there are no real zeros. At $x = 0$ the value is $1$ , so the quadratic is always positive.
Solve $(x - 5)^2 > 0$ .

Answer

Every real number except $5$

Full solution

A square is positive except where it is $0$ , at $x = 5$ .
Solve $-x^2 + 2x + 3 > 0$ .

Answer

$-1 < x < 3$

Full solution

Multiply by $-1$ and flip: $x^2 - 2x - 3 < 0$ , so $(x - 3)(x + 1) < 0$ .
A ball’s height is $h(t) = -5t^2 + 20t + 1$ meters. When is it above $16$ meters?

Answer

For $1 < t < 3$ seconds

Full solution

$-5t^2 + 20t - 15 > 0$ . Divide by $-5$ and flip: $t^2 - 4t + 3 < 0$ , so $(t - 1)(t - 3) < 0$ .
A student solves $x^2 > 4$ by taking square roots and answers $x > 2$ . What went wrong?

Hint

Test $x = -3$ .

Answer

The negative solutions are missing: $x < -2$ or $x > 2$ .

Full solution

$x^2 - 4 = (x - 2)(x + 2) > 0$ outside the zeros $\pm 2$ . For example $(-3)^2 = 9 > 4$ . Taking a square root of both sides gives $\lvert x \rvert > 2$ , not $x > 2$ .

Frequently asked questions

How do you solve a quadratic inequality?

Get 0 on one side, find the zeros of the quadratic, and test one number in each interval they create. Keep the intervals where the inequality holds.

Why is one test point per interval enough?

A quadratic can change sign only where it equals 0. Between two consecutive zeros it never crosses the axis, so its sign is the same across the whole interval.

When are the endpoints included?

For ≤ or ≥ the zeros themselves satisfy the inequality, so they are included, with square brackets. For < or > they are not, with parentheses.

What if the quadratic has no real zeros?

Then it has one sign everywhere. x² + 4 > 0 holds for every real number, and x² + 4 < 0 has no solution.

Can I divide both sides of x² > 5x by x?

No. x might be negative, which would flip the inequality, or zero. Move everything to one side and factor: x(x − 5) > 0.

What to learn next

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

CCSS.MATH.CONTENT.HSA.CED.A.1Creating EquationsCreate equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and simple rational and exponential functions.

For teachersPrint problems onlyTick to print a blank worksheet; leave it clear to print the lesson with the full answer key.