Answer
{$\pm4\sqrt 3$}
Work Step by Step
$48-x^{2}=0$ can be rearranged as $x^{2}=48$
According to the square root property,
if $x^{2}=48$
then $x=\sqrt (48)=\sqrt (16\times3)=\sqrt 16\sqrt 3=4\sqrt 3$
or $x=-\sqrt (48)=-\sqrt (16\times3)=-\sqrt 16\sqrt 3=-4\sqrt 3$
Therefore, the solution set is {$-4\sqrt 3,+4\sqrt 3$} or {$\pm4\sqrt 3$}
