College Algebra, Chapter 1, 1.5, Section 1.5, Problem 38

Find all real solutions of the equation $\displaystyle \left( \frac{x}{x + 2} \right)^2 = \frac{4x}{x + 2} - 4$


$
\begin{equation}
\begin{aligned}

\left( \frac{x}{x + 2} \right)^2 =& \frac{4x}{x + 2} - 4
&& \text{Given}
\\
\\
\left( \frac{x}{x + 2} \right)^2 - \left( \frac{4x}{x + 2} \right) + 4 =& 0
&& \text{Subtract } \frac{4x}{x + 2} \text{ and add } 4
\\
\\
w^2 - 4w + 4 =& 0
&& \text{Let } w = \frac{x}{x + 2}
\\
\\
(w - 2)^2 =& 0
&& \text{Factor out}
\\
\\
w - 2 =& 0
&& \text{Take the square root}
\\
\\
w =& 2
&& \text{Add } 2
\\
\\
\frac{x}{x + 2} =& 2
&& \text{Substitute } w = \frac{x}{x + 2}
\\
\\
x =& 2x + 4
&& \text{Apply cross multiplication}
\\
\\
x =& -4
&& \text{Solve for } x


\end{aligned}
\end{equation}
$