## Elementary Algebra

Recall, $i=\sqrt{-1}$. We know that if $x^{2}=a$, then $x=\pm \sqrt{a}$. Step 1: $(x)^{2}=-8$ Step 2: $x=\pm \sqrt {-8}$ Step 3: $x=\pm \sqrt {-1\times4\times2}$ Step 4: $x=\pm (\sqrt {-1}\times\sqrt {4}\times\sqrt 2)$ Step 5: $x=\pm (i\times2\times\sqrt {2})$ [as $i=\sqrt {-1}$] Step 6: $x=\pm (2i\sqrt {2})$ Step 7: $x=+2i\sqrt {2}$ or $x=-2i\sqrt {2}$ Therefore, the solution set is {$-2i\sqrt {2},2i\sqrt {2}$}. This means that the statement in the question is true.