Question

at least one scored part is incorrect. jump to first changable incorrect part. next question get a similar question you can retry this question below solve the following inequality and graph the solution: $(x - 6)(x + 6) > 0$ choose test values and indicate whether the inequality is true or false in each region. enter the test values from smallest to largest. draw the solution above. write the solution as a compound inequality $x < -6$ or $x > 6$

Explanation:

Step1: Find the critical points

Set \((x - 6)(x + 6)=0\). Using the zero - product property \(a\times b = 0\) implies \(a = 0\) or \(b = 0\). So \(x-6=0\) gives \(x = 6\), and \(x + 6=0\) gives \(x=-6\). These are the critical points that divide the number line into three intervals: \((-\infty,-6)\), \((-6,6)\), and \((6,\infty)\).

Step2: Test the intervals

• For the interval \((-\infty,-6)\):

Choose a test value \(x=-7\). Substitute into \((x - 6)(x + 6)\): \((-7-6)(-7 + 6)=(-13)\times(-1)=13>0\). So the inequality is True for \(x\in(-\infty,-6)\).

• For the interval \((-6,6)\):

Choose a test value \(x = 0\). Substitute into \((x - 6)(x + 6)\): \((0-6)(0 + 6)=(-6)\times6=-36<0\). So the inequality is False for \(x\in(-6,6)\).

• For the interval \((6,\infty)\):

Choose a test value \(x = 7\). Substitute into \((x - 6)(x + 6)\): \((7-6)(7 + 6)=(1)\times(13)=13>0\). So the inequality is True for \(x\in(6,\infty)\).

Step3: Write the solution

Since the inequality \((x - 6)(x + 6)>0\) is True for \(x<-6\) and \(x>6\), the solution in compound - inequality form is \(x<-6\) or \(x>6\).

Answer:

The solution of the inequality \((x - 6)(x + 6)>0\) is \(x<-6\) or \(x>6\). On the number - line, we use open circles at \(x=-6\) and \(x = 6\) (because the inequality is strict, i.e., \(>\) not \(\geq\)) and shade the regions to the left of \(x=-6\) and to the right of \(x = 6\).