Answer
{$-4\sqrt 2,4\sqrt 2$}
Work Step by Step
Using Property 10.1, which states that for any non-negative real number $a$, $x^{2}=a$ can be written as $x=\pm\sqrt a$, we obtain:
Step 1: $y^{2}=32$
Step 2: $y=\pm \sqrt {32}$
Step 3: $y=\pm \sqrt {16\times 2}$
Step 4: $y=\pm (\sqrt {16}\times \sqrt 2)$
Step 5: $y=\pm (4\times \sqrt 2)$
Step 6: $y=\pm 4\sqrt 2$
The solution set is {$-4\sqrt 2,4\sqrt 2$}.
