Quadratic Inequalities

By the end of this lesson, you will be able to solve quadratic inequalities algebraically and graphically, express solutions in interval notation, and apply quadratic inequalities to real-world constraints.

The Big Idea

A quadratic inequality asks: for what $x$-values is $ax^2 + bx + c$ positive, negative, or zero? The key insight is to find the zeros (where the parabola crosses the $x$-axis) and use the shape of the parabola to determine where the inequality holds. Test points in each interval to confirm.

Strategy

1. Move all terms to one side: $ax^2 + bx + c , \square , 0$
2. Find the zeros (set $= 0$ and solve)
3. The zeros divide the number line into regions
4. Test a point in each region
5. Write the solution using interval notation

Example 1 — Parabola Opens Up ($a > 0$)

Solve: $x^2 - 5x + 6 > 0$

Zeros: $x^2 - 5x + 6 = 0 \implies (x-2)(x-3) = 0 \implies x = 2, 3$

Test regions:

• $x < 2$: test $x = 0$: $6 > 0$ ✓
• $2 < x < 3$: test $x = 2.5$: $6.25 - 12.5 + 6 = -0.25 < 0$ ✗
• $x > 3$: test $x = 4$: $16 - 20 + 6 = 2 > 0$ ✓

Solution: $x < 2$ or $x > 3$ → $(-\infty, 2) \cup (3, \infty)$

Graphical: parabola opens up; above $x$-axis outside the zeros.

Example 2 — Non-Strict Inequality

Solve: $x^2 - 5x + 6 \leq 0$

Same zeros: $x = 2, 3$.

The parabola is ≤ 0 between the zeros (where it's below the $x$-axis) and equals 0 at the zeros.

Solution: $[2, 3]$

Example 3 — Parabola Opens Down ($a < 0$)

Solve: $-x^2 + 4x - 3 > 0$

Zeros: $-x^2 + 4x - 3 = 0 \implies x^2 - 4x + 3 = 0 \implies (x-1)(x-3) = 0 \implies x = 1, 3$

Since $a < 0$, parabola opens down — positive BETWEEN the zeros.

Solution: $(1, 3)$

Example 4 — No Real Zeros (Discriminant < 0)

Solve: $x^2 + 2x + 5 > 0$

$D = 4 - 20 = -16 < 0$ → no real zeros.

Since $a > 0$, the parabola opens up and is always above the $x$-axis.

Solution: All real numbers → $(-\infty, \infty)$

Example 5 — With Quadratic Formula

Solve: $2x^2 - 3x - 2 < 0$