Answer
No solution
Work Step by Step
Given
\begin{equation}
\sqrt{x-4}=\sqrt{2 x+8}.
\end{equation} Start by squaring both sides to eliminate the radical sign on both sides and then solve for $x$.
\begin{equation}
\begin{aligned}
\sqrt{x-4}&=\sqrt{2 x+8}\\
(\sqrt{x-4})^2 & =(\sqrt{2 x+8})^2 \\
x-4 & =2x+8 \\
x-2x& =8+4 \\
-x & = 12 \\
x& =-12.
\end{aligned}
\end{equation} Check:
\begin{equation}
\begin{aligned}
\sqrt{-12-4} & \stackrel{?}{=} \sqrt{2 (-12)+8} \\
\sqrt{-16} & \stackrel{?}{=} \sqrt{-16} \quad \text { False} \\
\end{aligned}
\end{equation} This solution does not exist as the radicals are not defined for $x=-12$.
