Solving $2\left(\sqrt{2s-16}-\sqrt{s}\right)-8=0$

Question

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?

Answer 1

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.

Answer 2

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: $32(2+\sqrt{3})$.

Answer 3

The domain is $s\geq8$, but $32(2-\sqrt3)<8$, which says that $32(2-\sqrt3)$ is not a root of the equation.