Q. 18 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 18–19, solve each quadratic inequality.

$x^2+6x-16<0$

Step-by-Step Solution

Answer

The solution of inequality is $\left\{x|-8<x<2\right\}$ or interval notation $\left(-8,2\right)$.

Step 1. Given information.

Solve the given quadratic inequality.

$x^2+6x-16<0$ 

Step 2. Graph the function.

Graph the function $f\left(x\right)=x^2+6x-16$.

Step 3. Find intercept.

y- intercept: $f\left(0\right)=0^2+6\left(0\right)-16=-16$

x-intercept : Solve $f\left(x\right)=0$

$$\begin{gathered} x^2+6x-16=0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ {x}_{1,2}=\frac{-6\pm \sqrt{6^2-4·1·\left(-16\right)}}{2·1} \\ {x}_{1,2}=\frac{-6\pm \sqrt{36+64}}{2} \\ {x}_{1,2}=\frac{-6\pm \sqrt{100}}{2} \\ {x}_{1,2}=\frac{-6\pm 10}{2} \\ {x}_1=\frac{-6+10}{2}, x_2=\frac{-6-10}{2} \\ {x}_1=2, x_2=-8 \end{gathered}$$

So, the x-intercept are $x_1=2 \mathrm{and} x_2=-8$

The graph is below the x-axis $\left(f\left(x\right)<0\right)$ between the $x=-8 \mathrm{and} x=2$. Since the inequality is strict the solution set is$\left\{\overline{)x}-8<x<2\right\}$ or using interval notation $\left(-8,2\right)$.

Step 4. Conclusion.

The solution set of inequality is $\left\{\overline{)x}-8<x<2\right\}$or interval notation$\left(-8,2\right)$.