Answer
$z= -\frac{1}{9}$
Work Step by Step
Given \begin{equation}
\sqrt{4+2 z}=\sqrt{3-7 z}.
\end{equation} Start by squaring both sides of the equation to eliminate the radical signs and solve for $z$.
\begin{equation}
\begin{aligned}
\sqrt{4+2 z}&=\sqrt{3-7 z} \\
\left( \sqrt{4+2 z}\right)^2&= \left(\sqrt{3-7 z}\right)^2\\
4+2 z& = 3-7 z\\
2 z+7z& = 3-4\\
9z& =-1\\
z&=-\frac{1}{9}.
\end{aligned}
\end{equation} Check. \begin{equation}
\begin{aligned}
\sqrt{4+ 2\cdot (-1/9)}& \stackrel{?}{=} \sqrt{3-7 \cdot (-1/9)} \\
\sqrt{\frac{34}{9}}& =\sqrt{\frac{34}{9}}\checkmark.
\end{aligned}
\end{equation} The solution is $z= -\frac{1}{9}$.
