Chapter 1.9, Problem 2E

Solution

To Explain: The solving procedure of $x^2+6x-8<0$ using a table, by graphing and algebraically.

The values of $x$ are $\left(-3-\sqrt{17},-3+\sqrt{17}\right)$ .

Given information:

Given expression, $x^2+6x-8<0$

Calculation:

Solve $x^2+6x-8<0$

by using the table:

Rewrite the equation as:

  $y=x^2+6x-8$

Make a table of $x$ and $y$ values that includes points where $y=x^2+6x-8=0$ .

There will be intervals separated by where $y=0$ . Negative intervals of $y$ are the solution intervals.

Substitute $-2$ for $x$ and find the result for $y$ :

  $\begin{array}{l}y={\left(-2\right)}^2+6\left(-2\right)-8\\ y=4-12-8\\ y=-16\end{array}$

Substitute $-1$ for $x$ and find the result for $y$ :

  $\begin{array}{l}y={\left(-1\right)}^2+6\left(-1\right)-8\\ y=1-6-8\\ y=-13\end{array}$

Substitute $0$ for $x$ and find the result for $y$ :

  $\begin{array}{l}y={\left(0\right)}^2+6\left(0\right)-8\\ y=-8\end{array}$

Substitute $1$ for $x$ and find the result for $y$ :

  $\begin{array}{l}y={\left(1\right)}^2+6\left(1\right)-8\\ y=7-8\\ y=-1\end{array}$

Substitute $2$ for $x$ and find the result for $y$ :

  $\begin{array}{l}y={\left(2\right)}^2+6\left(2\right)-8\\ y=4+12-8\\ y=8\end{array}$

$x$	$y$
$-2$	$-16$
$-1$	$-13$
$0$	$-8$
$1$	$-1$
$2$	$8$

The solution is where the $y$ is negative, so the graph will be:

  

Thus, the value of $x$ in $\left(-3,-\sqrt{17}\right),\left(3,-\sqrt{17}\right)$

Solve by algebraically:

Convert the inequality to an equation:

  $x^2+6x-8=0$

Use the quadratic formula to find the solutions.

  $x=\frac{-b\pm \sqrt{b^2-4\left(ac\right)}}{2a}$

Substitute the values $a=1,b=6,$ and $c=-8$ into the quadratic formula and solve for $x$ .

  $\begin{array}{l}x=\frac{-6\pm \sqrt{6^2-4\cdot (1\cdot -8)}}{2\cdot 1}\\ x=\frac{-6\pm 2\sqrt{17}}{2\cdot 1}\\ x=-3\pm \sqrt{17}\end{array}$

The solutions are: $x=-3+\sqrt{17},-3-\sqrt{17}$

Choose a test value from each interval and plug this value into the original inequality to

determine which intervals satisfy the inequality.

Test a value on the interval $x<-3-\sqrt{17}$ to see if it makes the inequality true:

Choose a value on the interval $x<-3-\sqrt{17}$ and see if this value makes the original inequality true.

  $x=-10$

Replace $x$ with $-10$ in the original inequality.

  ${\left(-10\right)}^2+6\left(-10\right)-8<0$

The left side $32$ is not less than the right side $0$ , which means that the given statement is false.

Test a value on the interval $-3-\sqrt{17}<x<-3+\sqrt{17}$ to see if it makes the inequality true.

Choose a value on the interval $-3-\sqrt{17}<x<-3+\sqrt{17}$ and see if this value makes the original inequality true:

  $x=0$

Replace $x$ with $0$ in the original inequality.

  ${\left(0\right)}^2+6\left(0\right)-8<0$

The left side $-8$ is less than the right side $0$ , which means that the given statement is always true.

Test a value on the interval $x>-3+\sqrt{17}$ to see if it makes the inequality true.

Choose a value on the interval $x>-3+\sqrt{17}$ and see if this value makes the original inequality true.

  $x=4$

Replace $x$ with $4$ in the original inequality.

  ${\left(4\right)}^2+6\left(4\right)-8<0$

The left side $32$ is not less than the right side $0$ , which means that the given statement is false.

Hence, the solution consists of all of the true intervals is $-3-\sqrt{17}<x<-3+\sqrt{17}$ .