College Algebra, Chapter 1, 1.3, Section 1.3, Problem 48

Find all real solutions of $\displaystyle \sqrt{6} x^2 + 2x - \sqrt{\frac{3}{2}} = 0$.


$\begin{equation} \begin{aligned} \sqrt{6} x^2 + 2x - \sqrt{\frac{3}{2}} =& 0 && \text{Given} \\ \\ \sqrt{6} x^2 + 2x =& \sqrt{\frac{3}{2}} && \text{Add } \sqrt{\frac{3}{2}} \\ \\ x^2 + \frac{2}{\sqrt{6}} x =& \frac{\displaystyle \sqrt{\frac{3}{2}}}{\sqrt{6}} && \text{Divide both sides of the equation by $\sqrt{6}$ to make the coefficient of $x^2$ equal to 1} \\ \\ x^2 + \frac{2}{\sqrt{6}} + \frac{1}{6} =& \frac{\displaystyle \sqrt{\frac{3}{2}}}{\sqrt{6}} + \frac{1}{6} && \text{Complete the square: add } \left( \frac{\displaystyle \frac{2}{\sqrt{6}}}{2} \right)^2 = \frac{1}{6} \\ \\ \left( x + \frac{1}{\sqrt{6}} \right)^2 =& \frac{2}{3} && \text{Perfect square, simplify the right side of the equation} \\ \\ x + \frac{1}{\sqrt{6}} =& \pm \sqrt{\frac{2}{3}} && \text{Take the square root} \\ \\ x =& \frac{-1}{\sqrt{6}} \pm \sqrt{\frac{2}{3}} && \text{Subtract } \frac{1}{\sqrt{6}} \\ \\ x =& \frac{-1}{\sqrt{6}} + \sqrt{\frac{2}{3}} \text{ and } x = \frac{-1}{\sqrt{6}} - \sqrt{\frac{2}{3}} && \text{Solve for } x \\ \\ x =& \frac{\sqrt{6}}{6} \text{ and } x = \frac{- \sqrt{6}}{2} && \text{Simplify} \end{aligned} \end{equation}$