# Square root confusion: Why am I getting an answer if it doesn’t work?

Alright, so I have $\sqrt{x-15} = 3-\sqrt{x}$.
I first square both sides to get $x-15 = (3-\sqrt{x})(3-\sqrt{x})$ which simplifies to $x-15 = 9 -6\sqrt{x} + x$.

I solved for $x$ and got $x = 16$, however, when I plug it in, the equation doesn’t work. Why does this happen?

Whenever you square both sides of an equation, you potentially introduce new solutions that may or may not be solutions of the original equation. Consider the simple example $x=1$. Obviously $x=1$ is the solution, but squaring both sides gives $x^2 = 1$, which has solutions $x=\pm 1$. Over $\mathbb{R}$ it is most definitely not true that $-1=1$, so $x=-1$ cannot be a solution here.