 Solution

Isolate the square root: $\sqrt{x+16}=x-4$. A real solution must have $x\ge 4$, because a real principal square root is nonnegative. Squaring gives $x+16=(x-4{)}^2$, hence $x(x-9)=0$. The candidate zero violates $x\ge 4$; substitution verifies the other candidate. ThusThe sign restriction prevents the extraneous solution introduced by squaring.