Quadratic Inequalities

To solve x² − x − 6 > 0:

1. Find the zeros: factor to (x − 3)(x + 2) = 0, giving x = 3 and x = −2.
2. Those split the number line into three intervals.
3. Test one value from each: x = −3 gives 6 > 0 ✓; x = 0 gives −6 > 0 ✗; x = 4 gives 6 > 0 ✓.
4. Keep the intervals that worked.

Answer: x < −2 or x > 3.

Thinking graphically is faster once it clicks. The parabola opens up and crosses at −2 and 3, so it is above the axis outside the roots and below between them. "Greater than zero" means above the axis.

The solution set of ax² + bx + c > 0 is determined by the sign of the quadratic, which can only change at its real zeros. Those zeros partition ℝ into intervals of constant sign, so one test point per interval settles the whole interval.

For a > 0 with distinct real roots r₁ < r₂: the expression is positive on (−∞, r₁) ∪ (r₂, ∞) and negative on (r₁, r₂). For a < 0 the pattern inverts. Endpoints are included for ≥ and ≤ and excluded for strict inequalities.

Degenerate cases follow the discriminant. With Δ < 0 the expression never changes sign, so x² + 1 > 0 has solution ℝ while x² + 1 < 0 has none. With Δ = 0 the single root is the only place the expression is zero, so (x − 2)² > 0 excludes just that point.

The same sign-analysis method extends unchanged to higher-degree polynomial and to rational inequalities, where the partition points are the zeros of the numerator together with the zeros of the denominator.