Case a negative, Delta positive

When the concavity is downwards and there are two zeros, the parabola lies below the $x$-axis outside the zeros. Here we also use completing the square to find the vertex.

Example — Case $a<0$, $\Delta>0$ with completing the square

Solve $-2x^2-12x+16 \le 0$.

Concavity: $a=-2 < 0$, parabola opening downwards. We complete the square: $-2(x^2+6x)+16 = -2\bigl[(x+3)^2-9\bigr]+16 = -2(x+3)^2+34.$ Vertex: $V(-3,\,34)$. Zeros: $-2(x+3)^2+34=0 \implies (x+3)^2=17 \implies x=-3\pm\sqrt{17}$. $\Delta=144+128>0$.

We look for where the parabola lies below or on the $x$-axis. For the case $a<0$, $\Delta>0$, this happens outside the zeros.

The solution is made up of the two stretches outside the zeros; the dots are filled because the inequality is $\le 0$.

Solution: $\boxed{x \le -3-\sqrt{17} \;\;\text{or}\;\; x \ge -3+\sqrt{17}}$.

Links

Topics: Inequalities
Concepts: Completing the square · Second-degree inequality · Parabola · Vertex of a parabola
Methods: Second-degree inequality · Parabola vertex
Skills: Interpreting a graph · Solving inequalities