What if the quadratic has no real roots?

The expression never crosses the x-axis, meaning it is either always positive or always negative. Test any single point, such as $x=0$, to determine if the solution is all real numbers or the empty set.

What if there is exactly one real root?

The graph touches the x-axis at exactly one point without crossing it. For a strict inequality, the solution is either the empty set or all real numbers except that root. Test intervals on both sides to confirm.

Can I solve by just taking the square root of both sides?

Only if the inequality is in the form $x^2 < k$. If so, the solution involves absolute values, $|x| < \sqrt{k}$, which expands to $-\sqrt{k} < x < \sqrt{k}$. The interval testing method is less prone to sign errors.

How to solve a quadratic inequality

A quadratic inequality takes the form $ax^2 + bx + c > 0$ , or similar expressions using $<, \ge, \le$ . The method finds the roots of the corresponding quadratic equation to divide the real number line into intervals, then tests each interval to determine where the inequality holds true.

The setup

Write the inequality in standard form $ax^2 + bx + c < 0$ (or $>$ , $\le$ , $\ge$ ). Ensure one side of the inequality is zero and combine all like terms. Identify the inequality type as strict ( $<, >$ ) or non-strict ( $\le, \ge$ ).

The steps

Replace the inequality sign with an equals sign to form $ax^2 + bx + c = 0$ . 2. Solve for $x$ using factoring, completing the square, or the quadratic formula to find the critical values. 3. Plot these critical values on a number line to create distinct test intervals. 4. Select one test point from each interval and evaluate the original quadratic expression. 5. Keep the intervals that satisfy the original inequality. Use open bounds for strict inequalities and closed bounds for non-strict inequalities.

Checking the result

Select a random value strictly inside your final solution set and verify it satisfies the original inequality. If the inequality is non-strict, verify that the endpoints equal zero when evaluated in the quadratic expression.

Common errors

A frequent error is dividing by a variable expression, which eliminates roots and obscures the sign changes. Another common mistake is including the critical values in the solution of a strict inequality, or forgetting to flip the inequality sign if dividing the entire inequality by a negative constant.

Worked example

Solve $x^2 - 3x - 10 < 0$ .

Standard form is $x^2 - 3x - 10 < 0$ . The corresponding equation is $x^2 - 3x - 10 = 0$ . Factoring yields $(x - 5)(x + 2) = 0$ . Critical values are $x = -2$ and $x = 5$ . The intervals are $(-\infty, -2)$ , $(-2, 5)$ , and $(5, \infty)$ . Test $x = -3$ : $(-3)^2 - 3(-3) - 10 = 9 + 9 - 10 = 8$ . Since $8 < 0$ is false, $(-\infty, -2)$ is not in the solution. Test $x = 0$ : $0^2 - 3(0) - 10 = -10$ . Since $-10 < 0$ is true, $(-2, 5)$ is in the solution. Test $x = 6$ : $6^2 - 3(6) - 10 = 36 - 18 - 10 = 8$ . Since $8 < 0$ is false, $(5, \infty)$ is not in the solution. The inequality is strict, so endpoints are excluded. The final solution is $(-2, 5)$ .

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References: OpenStax College Algebra, Chapter 2: Equations and Inequalities · Khan Academy, Quadratic inequalities