Case Delta zero — tangent parabola

With $\Delta=0$ the parabola is tangent to the $x$-axis at a single point: there the value is exactly zero. The sense of the inequality (strict or weak) then makes all the difference at the point of tangency.

Example — Case $\Delta = 0$, tangent parabola

Solve $x^2-6x+9 \ge 0$.

Notice that $x^2-6x+9 = (x-3)^2$. $\Delta = 36-36 = 0$: the parabola is tangent to the $x$-axis at $x_0=3$.

The parabola touches the $x$-axis only at $x=3$, where it equals exactly $0$.

A square is always $\ge 0$: the inequality is satisfied for all $x$. Solution: $\boxed{\forall x\in\mathbb{R}}$.

Variation: the same parabola, with the inequality $(x-3)^2 > 0$ (strict!), has solution $\boxed{x\neq 3}$ — at $x=3$ the value is exactly zero, not strictly positive.

Another variation: $(x-3)^2 \le 0$ has as its only solution $\boxed{x=3}$ (the value is $\le 0$ only at the point of tangency). And $(x-3)^2 < 0$ has solution $\boxed{\emptyset}$.

Links

Topics: Inequalities
Concepts: Discriminant · Second-degree inequality · Parabola
Skills: Reasoning by cases · Solving inequalities