Mathematics · Study Guide

How to Solve Quadratic Inequalities (Step by Step)

A quadratic inequality compares a quadratic expression to zero using <, >, \le, or \ge instead of an equals sign — for example $x^2 - 2x - 15 < 0$ . Unlike a quadratic equation, which usually has one or two exact solutions, a quadratic inequality's solution is a *range* of x-values. The method below works for any quadratic inequality, factorable or not.

Step 1 — Move everything to one side

Rewrite the inequality so one side is exactly $0$ . Most textbook problems already start this way, but if you see something like $x^2 - 2x < 15$ , subtract $15$ from both sides first to get $x^2 - 2x - 15 < 0$ .

Step 2 — Find the critical values

The critical values are the x-values where the quadratic equals exactly zero — they're the boundary points between where the expression is positive and where it's negative. Find them by factoring (or using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ , if it doesn't factor nicely) and setting each factor to zero.

Step 3 — Test a value from each interval

The critical values split the number line into intervals. A quadratic's sign can only change AT a critical value, so within each interval the sign stays constant — pick any one test point from each interval, plug it into the ORIGINAL inequality, and check whether it makes the inequality true or false. That tells you whether the whole interval belongs in the answer.

Step 4 — Write the solution as an interval

Collect every interval that tested true. If the inequality uses $<$ or $>$ (strict), the critical values themselves are NOT included in the answer. If it uses $\le$ or $\ge$ , they ARE included.

Worked Example

Problem

Solve

x^2 - 2x - 15 < 0

Factor

Look for two numbers that multiply to

-15

and add to

-2

: those are

-5

and

3

. So

x^2 - 2x - 15 = (x-5)(x+3)

Critical values

Setting each factor to zero:

x - 5 = 0 \Rightarrow x = 5

, and

x + 3 = 0 \Rightarrow x = -3

. These split the number line into three intervals:

x < -3

-3 < x < 5

, and

x > 5

Test x < -3

Try

x = -4

(-4)^2 - 2(-4) - 15 = 16 + 8 - 15 = 9

9

is not less than

0

, so this interval is NOT part of the solution.

Test -3 < x < 5

Try

x = 0

(0)^2 - 2(0) - 15 = -15

-15 < 0

is true, so this interval IS part of the solution.

Test x > 5

Try

x = 6

(6)^2 - 2(6) - 15 = 36 - 12 - 15 = 9

9

is not less than

0

, so this interval is NOT part of the solution.

Final answer

Only the middle interval worked, and the inequality is strict (

<

, not

\le

), so the critical values themselves are excluded:

-3 < x < 5

Common Mistakes

Solving the equation $x^2 - 2x - 15 = 0$ and stopping there — the critical values are the *boundary* of the answer, not the answer itself, for a strict inequality.
Assuming the middle interval is always the solution. Sometimes the OUTER two intervals are correct instead — always test every interval, never guess based on a pattern from a previous problem.
Forgetting to flip the inequality sign when multiplying or dividing both sides by a negative number (this doesn't come up in the example above, but it's the single most common quadratic-inequality mistake overall).
Including the critical values in a strict ( $<$ or $>$ ) inequality's answer — they belong only in a $\le$ / $\ge$ answer.

Practice

Solve

x^2 - x - 6 \le 0

Factor:

(x-3)(x+2) \le 0

. Critical values

x=3

and

x=-2

. Testing shows the middle interval works, and since the inequality is

\le

, the endpoints are included:

-2 \le x \le 3

Solve

x^2 - 9 > 0

Factor:

(x-3)(x+3) > 0

. Critical values

x=3

and

x=-3

. This time the OUTER intervals work:

x < -3

x > 3

Keep studying

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