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I need some help here...

Solve the inequality [tex]7 \leq \left|9-x^2\right|[/tex]

The attempt at a solution

So, you've got two cases: [tex]\left|9-x^2\right| =[/tex]

1. [tex]9 - x^2[/tex]

2. [tex]x^2 - 9[/tex]

For the first case [tex]7 \leq 9-x^2[/tex]

[tex]-2 \leq -x^2[/tex]

[tex]\Rightarrow \sqrt{2}\geq x[/tex]

For the second case [tex]7 \leq x^2 - 9[/tex]

[tex]\sqrt{16} \leq x[/tex]

[tex]4 \leq x[/tex]

Anyway, my answer is only partially correct. The actual answer has to be;

[tex]x \in (-\infty, -4] \cup [-\sqrt{2}, \sqrt{2}] \cup [-4, \infty)[/tex]

I'm very curious to know how you to get that answer. Thanks.

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# Homework Help: Solving inequality 7 <= |9 - x^2|

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