[tex]\sqrt{x^{2}-8}=1\\Ca\ radicalii\ sa\ fie\ de.finiti\\
x^{2}-8\geq0\\x^{2}\geq0+8\\x^{2}\geq8\\x\ apartine\ intervalului\ [2\sqrt{2},\infty)\ reunit\ cu\ (-\infty,-2\sqrt{2}]\\
Acum\ putem\ ridica\ la\ patrat\\
x^{2}-8=1\\x^{2}=1+8\\x^{2}=9\\\sqrt{x^{2}}=\sqrt{9}\\|x|=3\\x_1=3\\x_2=-3\\
Solutia\ x=3\ se\ incadreaza\ in\ [2\sqrt{2},\infty)\\
Solutia x=-3\ se\ incadreaza\ in\ (-\infty,-2\sqrt{2}][/tex]
