Solve 9x^2+24x-16geq0

$$9 { x }^{ 2 } +24x-16 \geq 0$$

Solve for x

$$x\in (-\infty,\frac{-4\sqrt{2}-4}{3}]\cup [\frac{4\sqrt{2}-4}{3},\infty)$$

Steps Using the Quadratic Formula

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$9x^{2}+24x-16=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 9 for a, 24 for b, and -16 for c in the quadratic formula.

$$x=\frac{-24±\sqrt{24^{2}-4\times 9\left(-16\right)}}{2\times 9}$$

Do the calculations.

$$x=\frac{-24±24\sqrt{2}}{18}$$

Solve the equation x=\frac{-24±24\sqrt{2}}{18} when ± is plus and when ± is minus.

$$x=\frac{4\sqrt{2}-4}{3} x=\frac{-4\sqrt{2}-4}{3}$$

Rewrite the inequality by using the obtained solutions.

$$9\left(x-\frac{4\sqrt{2}-4}{3}\right)\left(x-\frac{-4\sqrt{2}-4}{3}\right)\geq 0$$

For the product to be ≥0, x-\frac{4\sqrt{2}-4}{3} and x-\frac{-4\sqrt{2}-4}{3} have to be both ≤0 or both ≥0. Consider the case when x-\frac{4\sqrt{2}-4}{3} and x-\frac{-4\sqrt{2}-4}{3} are both ≤0.

$$x-\frac{4\sqrt{2}-4}{3}\leq 0 x-\frac{-4\sqrt{2}-4}{3}\leq 0$$

The solution satisfying both inequalities is x\leq \frac{-4\sqrt{2}-4}{3}.

$$x\leq \frac{-4\sqrt{2}-4}{3}$$

Consider the case when x-\frac{4\sqrt{2}-4}{3} and x-\frac{-4\sqrt{2}-4}{3} are both ≥0.

$$x-\frac{-4\sqrt{2}-4}{3}\geq 0 x-\frac{4\sqrt{2}-4}{3}\geq 0$$

The solution satisfying both inequalities is x\geq \frac{4\sqrt{2}-4}{3}.

$$x\geq \frac{4\sqrt{2}-4}{3}$$

The final solution is the union of the obtained solutions.

$$x\leq \frac{-4\sqrt{2}-4}{3}\text{; }x\geq \frac{4\sqrt{2}-4}{3}$$