Solving $2\left(\sqrt{2s-16}-\sqrt{s}\right)-8=0$ I am trying to solve the equation 
$$2\left(\sqrt{2s-16}-\sqrt{s}\right)-8=0$$ 
Using regular method I found two roots of this namely $32(2+\sqrt{3})$ and $32(2-\sqrt{3})$. But when I tried to confirm them only $32(2+\sqrt{3})$ worked as a root whereas $32(2-\sqrt{3})$ gave me negative value. 
Can someone help with this?
 A: Whenever you square an equation you get additional roots. For example $x=1$ has a unique solution but $x^{2}=1^{2}$ has two solutions $x=1$ and $x=-1$. Your 'regular method' involves squaring so you got an extra root. After getting the two values for $s$ you have to go back to the given equation and keep only the one that really satisfies that equation. 
A: The domain is $s\geq8$, but $32(2-\sqrt3)<8$, which says that $32(2-\sqrt3)$ is not a root of the equation.
A: First, rewrite your expression as: $2\sqrt{2s-16}=8+2\sqrt{s}$; in other words: $\sqrt{2s-16}=4+\sqrt{s}$. Let $P(s)=\sqrt{2s-16}$ and $Q(s)=4+\sqrt{s}$.
Now as a general rule, when you have to find the solution of an equation with root, you have to solve a system:
$$\left\{\begin{matrix}
P(s)\geq0
\\ Q(s)\geq0
\\ P(s)^2=Q(s)^2
\end{matrix}\right.$$
If you substitute:
$$\left\{\begin{matrix}
\sqrt{2s-16}\geq0
\\ 4+\sqrt{s}\geq0
\\ (\sqrt{2s-16})^2=(4+\sqrt{s})^2
\end{matrix}\right.$$
Solving, I obtain only one solution: $32(2+\sqrt{3})$.
