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Solving Quadratic Inequalities — Definition, Formula & Examples

Solving a quadratic inequality means finding all values of xx that make an expression like ax2+bx+c>0ax^2 + bx + c > 0 (or ≥\geq, <<, ≤\leq) true. Instead of a single answer, the solution is typically one or two intervals on the number line.

A quadratic inequality is a second-degree polynomial inequality of the form ax2+bx+c  ≶  0ax^2 + bx + c \;\mathord{\lessgtr}\; 0 where a≠0a \neq 0. Its solution set consists of all real numbers xx for which the inequality holds, determined by locating the roots of ax2+bx+c=0ax^2 + bx + c = 0 and analyzing the sign of the quadratic on each resulting interval.

How It Works

First, move all terms to one side so the other side equals zero. Factor the quadratic (or use the quadratic formula) to find its roots — these are the boundary points where the expression equals zero. Plot the roots on a number line, dividing it into intervals. Pick a test value from each interval and substitute it into the factored expression to check whether the result is positive or negative. The intervals where the sign matches your inequality form the solution set. Include the boundary points themselves if the inequality uses ≤\leq or ≥\geq.

Worked Example

Problem: Solve x² − 5x + 6 < 0.
Factor: Factor the left side.
x2−5x+6=(x−2)(x−3)x^2 - 5x + 6 = (x - 2)(x - 3)
Find roots: Set each factor equal to zero to get the boundary points.
x=2andx=3x = 2 \quad \text{and} \quad x = 3
Test intervals: The roots divide the number line into three intervals: (−∞,2)(-\infty, 2), (2,3)(2, 3), and (3,∞)(3, \infty). Pick a test point from each. At x=0x = 0: (0−2)(0−3)=6>0(0-2)(0-3) = 6 > 0. At x=2.5x = 2.5: (0.5)(−0.5)=−0.25<0(0.5)(-0.5) = -0.25 < 0. At x=4x = 4: (2)(1)=2>0(2)(1) = 2 > 0.
Write solution: The expression is negative only on (2,3)(2, 3). Since the inequality is strict (<<), the endpoints are excluded.
2<x<32 < x < 3
Answer: The solution is x∈(2, 3)x \in (2,\, 3).

Why It Matters

Quadratic inequalities appear throughout Algebra 2, Precalculus, and the SAT/ACT. In physics and engineering, they arise when determining the range of values for which a projectile is above a certain height or when a profit function exceeds a threshold.

Common Mistakes

Mistake: Solving (x−2)(x−3)<0(x-2)(x-3) < 0 by setting each factor less than zero separately, concluding x<2x < 2 and x<3x < 3.
Correction: The product of two factors is negative when they have opposite signs. You must test intervals between and beyond the roots rather than solving each factor independently.