College Algebra, Chapter 1, 1.5, Section 1.5, Problem 28

Find all real solutions of the equation $\displaystyle \sqrt{4 - 6x} = 2x$


$
\begin{equation}
\begin{aligned}

\sqrt{4 - 6x} =& 2x
&& \text{Given}
\\
\\
(\sqrt{4 - 6x})^2 =& (2x)^2
&& \text{Square both sides}
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\\
4 - 6x =& 4x^2
&& \text{Divide both sides by } 4
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1 - \frac{3}{2} x =& x^2
&& \text{Add } \frac{3}{2} x \text{ and subtract } 1
\\
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x^2 + \frac{3}{2} x - 1 =& 0
&& \text{Factor out}
\\
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(x + 2)\left( x - \frac{1}{2} \right) =& 0
&& \text{Zero Product Property}
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x + 2 =& 0 \text{ and } x - \frac{1}{2} = 0
&& \text{Solve for } x
\\
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x =& -2 \text{ and } x = \frac{1}{2}
&&
\\
\\
x =& \frac{1}{2}
&& \text{The only solution that satisfy the equation } \sqrt{4 - 6x} = 2x

\end{aligned}
\end{equation}
$