Exercise 9 Page 285

To solve the quadratic inequality algebraically, we will follow three steps.

1. Solve the related quadratic equation.
2. Plot the solutions on a number line.
3. Test a value from each interval to see if it satisfies the original inequality.

Step $1$

We will start by solving the related equation. We see above that $a=1,$ $b=6,$ and $c=\text{-}16.$ Let's substitute these values into the Quadratic Formula to solve the equation.

$x=\frac{\text{-}b\pm \sqrt{b^2-4ac}}{2a}$

SubstituteValues

Substitute values

$x=\frac{\text{-}6\pm \sqrt{(6{)}^2-4(1)(\text{-}16)}}{2(1)}$

▼

Simplify right-hand side

CalcPow

Calculate power

$x=\frac{\text{-}6\pm \sqrt{36-4(1)(\text{-}16)}}{2(1)}$

Multiply

Multiply

$x=\frac{\text{-}6\pm \sqrt{36+64}}{2}$

AddTerms

Add terms

$x=\frac{\text{-}6\pm \sqrt{100}}{2}$

CalcRoot

Calculate root

$x=\frac{\text{-}6\pm 10}{2}$

Now we can calculate the first root using the positive sign and the second root using the negative sign.

$x=\frac{\text{-}6\pm 10}{2}$
$x=\frac{\text{-}6+10}{2}$	$x=\frac{\text{-}6-10}{2}$
$x=\text{-}\frac{6}{2}+\frac{10}{2}$	$x=\text{-}\frac{6}{2}-\frac{10}{2}$
$x=2$	$x=\text{-}8$

Step $2$

The solutions of the related equation are $\text{-}8$ and $2.$ Let's plot them on a number line. Since the original is a strict inequality, the points will be open.

Step $3$

Finally, we must test a value from each interval to see if it satisfies the original inequality. Let's choose a value from the first interval, $x<\text{-}8.$ For simplicity, we will choose $x=\text{-}10.$

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

Substitute

$x=\text{-}10$

$(\text{-}10{)}^2+6(\text{-}10)-16\stackrel{?}{<}0$

▼

Simplify left-hand side

CalcPow

Calculate power

$100+6(\text{-}10)-16\stackrel{?}{<}0$

Multiply

Multiply

$100-60-16\stackrel{?}{<}0$

SubTerms

Subtract terms

$24\nless 0$

Since $x=\text{-}10$ did not produce a true statement, the interval $x<\text{-}8$ is not part of the solution. Similarly, we can test the other two intervals.

Interval	Test Value	Statement	Is It Part of the Solution?
$\text{-}8<x<2$	$0$	$\text{-}16<0$ $✓$	Yes
$x>2$	$3$	$11\nless 0$ $\times$	No

We can now write the solution set and show it on a number line.

$$\begin{array}{c}\{x|\text{-}8<x<2\}\text{or}(\text{-}8,2)\end{array}$$