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Answer:
[tex]x < -3\quad \mathrm{or}\quad \:x > 3[/tex]
Step-by-step explanation:
Given:
[tex]9-x^2 < \:0[/tex]
Solve:
[tex]\mathrm{Subtract\:}9\mathrm{\:from\:both\:sides}[/tex]
[tex]9-x^2-9 < 0-9[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]-x^2 < -9[/tex]
[tex]Multiply\:both\:sides\:by\:-1\:[/tex]
[tex]\left(-x^2\right)\left(-1\right) > \left(-9\right)\left(-1\right)[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]x^2 > 9[/tex]
[tex]\mathrm{For\:}u^n\: > \:a\mathrm{,\:if\:}n\:\mathrm{is\:even}\mathrm{\:then\:}u\: < \:-\sqrt[n]{a}\:or\:u\: > \:\sqrt[n]{a}[/tex]
[tex]x < -\sqrt{9}\quad \mathrm{or}\quad \:x > \sqrt{9}[/tex]
[tex]\sqrt{9}=3[/tex]
[tex]x < -3\quad \mathrm{or}\quad \:x > 3[/tex]
~lenvy~