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

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


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