Question

Solve each inequality. Graph the solution set.

$\frac{x^2-8x+12}{x^2-16}>0$

Short answer

The real zeroes of the function are 6, 2 , 4 and $-4$and the graph is

Step-by-step solution

Step 1. Given Information 

The given inequality is$\frac{x^2-8x+12}{x^2-16}>0$

Step 2. Calculation 

The inequality is already in a form with zero on the right side.

Solve the given inequality by finding the real zeroes of the numerator and denominator.

$$\begin{gathered} x^2-8x+12=0 \\ (x-6)(x-2)=0 \\ x=6,2 \end{gathered}$$

And

$$\begin{gathered} x^2-16=0 \\ {x}^2=16 \\ x=\pm 4 \end{gathered}$$

Thus, real zeroes of the function are 6, 2, 4 and$-4$

Step 3. Graph 

Construct the table by using a number from each interval to check whether the function is positive or negative.

Thus, the solution set is $\left\{x|x<-4or2<x<4orx>6\right\}$

And the solution set can be graphed as