# How do you solve the inequality 2<=abs(x^2-9)<9 and write your answer in interval notation?

Jun 25, 2017

$\left\{x | - 3 \sqrt{2} < x \le 11 , - \sqrt{7} \le x < 0 , 0 < x \le \sqrt{7} , \sqrt{11} \le x < 3 \sqrt{2}\right\}$

#### Explanation:

We can split this into two separate equations:

$2 \le \left\mid {x}^{2} - 9 \right\mid < 9$

$2 \le {x}^{2} - 9 < 9 \text{ " or " } - 9 < {x}^{2} - 9 \le - 2$

Solving both of these separately will then give us our solution set.

$2 \le {x}^{2} - 9 < 9$

$11 \le {x}^{2} < 18$

$\sqrt{11} \le x < 3 \sqrt{2} \text{ " or " } - 3 \sqrt{2} < x \le - \sqrt{11}$

$- 9 < {x}^{2} - 9 \le - 2$

$0 < {x}^{2} \le 7$

$0 < x \le \sqrt{7} \text{ " or " } - \sqrt{7} \le x < 0$

Therefore, our set of numbers that $x$ could be is:

$\left\{x | - 3 \sqrt{2} < x \le 11 , - \sqrt{7} \le x < 0 , 0 < x \le \sqrt{7} , \sqrt{11} \le x < 3 \sqrt{2}\right\}$

On the number line, the solution set looks like this: