Chapter 2.8, Problem 36E

Solution

To solve: The given inequality using sign chart.

The solution set is equal to $\left(-\infty ,-2\right)\cup \left(2,\infty \right)$ .

Given information:

Consider the given inequality.

  $\frac{x^2-4}{x^2+4}>0$

Calculation:

Rewrite the given equality below:

  $\begin{array}{c}\frac{x^2-2^2}{x^2+4}>0\\ \frac{\left(x-2\right)\left(x+2\right)}{x^2+4}>0\end{array}$

To set numerator and the denominator equal to zero. These are the values that make the quotient zero or undefined and solve for $x$ :

  $\begin{array}{c}\left(x-2\right)\left(x+2\right)=0\\ x=-2,2\end{array}$

And

  $\begin{array}{c}{x}^2+4=0\\ x^2=-4\\ x=\pm 2i\end{array}$

Since $x=-2i,2i$ are the imaginary points. So, the boundary points are $-2,2$ .

Now, locate these boundary points on the number line (or sign chart) and dividing the number line into intervals and check the function $f\left(x\right)$ is positive or negative at each interval below:

  

The boundary points divide the number line into three test intervals. The intervals are below:

  $\left(-\infty ,-2\right),\left(-2,2\right),\left(2,\infty \right)$

Now, take one representative number within each test interval and substitute that number into the inequality below:

  $\begin{array}{l}\left(-\infty ,-2\right),\left(-2,2\right),\left(2,\infty \right)\\ \frac{{\left(-3\right)}^2-4}{{\left(-3\right)}^2+4}>0\frac{{\left(0\right)}^2-4}{{\left(0\right)}^2+4}>0\frac{{\left(3\right)}^2-4}{{\left(3\right)}^2+4}>0\\ \frac{5}{13}>0\text{True }-1>0\text{False}\frac{5}{13}>0\text{True}\end{array}$

Therefore, the intervals $\left(-\infty ,-2\right)$ and $\left(2,\infty \right)$ belongs to the solution set and the intervals $\left(-2,2\right)$ does not belong to the solution set.

Hence, the solution set are intervals $\left(-\infty ,-2\right)\cup \left(2,\infty \right)$ .