Question

solve for x and graph the solution. -(x + 3)(x - 3) ≤ 0 plot the endpoints. select an endpoint to change it from closed to open. select the middle of a segment, ray, or line to delete it. submit

Explanation:

Step1: Expand the inequality

$$-(x + 3)(x - 3)\leq0$$
Using the difference - of - squares formula \((a + b)(a - b)=a^{2}-b^{2}\), we get \(-(x^{2}-9)\leq0\), which simplifies to \(-x^{2}+9\leq0\). Then, multiply both sides by \(- 1\) (and reverse the inequality sign) to obtain \(x^{2}-9\geq0\), or \((x + 3)(x - 3)\geq0\).

Step2: Find the critical points

Set \((x + 3)(x - 3)=0\). By the zero - product property, \(x+3 = 0\) gives \(x=-3\), and \(x - 3=0\) gives \(x = 3\).

Step3: Test intervals

• Interval 1: \(x\lt - 3\)

Let \(x=-4\). Then \((-4 + 3)(-4 - 3)=(-1)\times(-7)=7\geq0\).

• Interval 2: \(-3\lt x\lt3\)

Let \(x = 0\). Then \((0 + 3)(0 - 3)=(3)\times(-3)=-9\lt0\).

• Interval 3: \(x\gt3\)

Let \(x = 4\). Then \((4 + 3)(4 - 3)=(7)\times(1)=7\geq0\).

Answer:

The solution of the inequality \(-(x + 3)(x - 3)\leq0\) (or \((x + 3)(x - 3)\geq0\)) is \(x\leq - 3\) or \(x\geq3\). On the number - line, we plot closed circles at \(x=-3\) and \(x = 3\), and shade the regions to the left of \(x=-3\) and to the right of \(x = 3\).