Being radical part 2/5

$\sqrt { { x }^{ 4 }-8{ x }^{ 2 }+16 } =0$
Let $x^2$ $=$ $y$ .
$\sqrt { { y }^{ 2 }-8y+16 } =0$
$\sqrt{y^2-4y-4y+16}=0$
$\sqrt{(y-4)(y-4)}=0$
$(y-4)=0$
$y=4$
$x^2=4$
$\therefore x=\pm2$
Now, their sum becomes, $2+(-2)=\boxed{0}$