Question

solve the following inequality and graph the solution:

(x + 7)(x - 7) leq 0

write the solution in interval notation

and draw the solution

12 - 11 - 10 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1 0 1 2 3 4 5 6 7 8 9 10 11

clear all draw:

Explanation:

Step1: Find the critical points

Set \((x + 7)(x - 7)=0\).
Using the zero - product property \(a\times b = 0\) implies \(a = 0\) or \(b = 0\).
If \(x+7=0\), then \(x=-7\). If \(x - 7=0\), then \(x = 7\).

Step2: Test intervals

The critical points \(x=-7\) and \(x = 7\) divide the real number line into three intervals: \((-\infty,-7)\), \((-7,7)\), and \((7,\infty)\).

• For the interval \((-\infty,-7)\), let \(x=-8\). Then \((-8 + 7)(-8 - 7)=(-1)\times(-15)=15>0\).
• For the interval \((-7,7)\), let \(x = 0\). Then \((0 + 7)(0 - 7)=(7)\times(-7)=-49<0\).
• For the interval \((7,\infty)\), let \(x = 8\). Then \((8 + 7)(8 - 7)=(15)\times(1)=15>0\).

Step3: Consider the inequality \((x + 7)(x - 7)\leq0\)

The inequality is satisfied when \(x\in[-7,7]\) (since the inequality includes the equal sign, we include the critical points \(x=-7\) and \(x = 7\)).

Answer:

The solution in interval notation is \([-7,7]\). To graph the solution, place a closed (filled - in) circle at \(x=-7\) and \(x = 7\) and draw a line segment connecting them on the number line.