College Algebra, Chapter 1, 1.5, Section 1.5, Problem 36

Find all real solutions of the equation $\displaystyle \left( \frac{x + 1}{x} \right)^2 + 4 \left( \frac{x + 1}{x} \right) + 3 = 0$


$
\begin{equation}
\begin{aligned}

\left( \frac{x + 1}{x} \right)^2 + 4 \left( \frac{x + 1}{x} \right) + 3 =& 0
&& \text{Given}
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w^2 + 4w + 3 =& 0
&& \text{Let } w = \frac{x + 1}{x}
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(w + 3)(w + 1) =& 0
&& \text{Factor out}
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w + 3 =& 0 \text{ and } w + 1 = 0
&& \text{Zero Product Property}
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w =& -3 \text{ and } w = -1
&& \text{Solve for } w
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\frac{x + 1}{x} =& -3 \text{ and } \frac{x + 1}{x} = -1
&& \text{Substitute } w = \frac{x + 1}{x}
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x + 1 =& -3x \text{ and } x + 1 = -x
&& \text{Solve for } x
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4x =& -1 \text{ and } 2x = -1
&&
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x =& \frac{-1}{4} \text{ and } x = \frac{-1}{2}
&&

\end{aligned}
\end{equation}
$