College Algebra, Chapter 1, 1.3, Section 1.3, Problem 42

Find all real solutions of $\displaystyle 2y^2 - y - \frac{1}{2} = 0$.


$\begin{equation} \begin{aligned} 2y^2 - y - \frac{1}{2} =& 0 && \text{Given} \\ \\ 2y^2 - y =& \frac{1}{2} && \text{Add } \frac{1}{2} \\ \\ y^2 - \frac{y}{2} =& \frac{1}{4} && \text{Divide both sides by 2 to make the coefficient of $x^2$ equal to 1} \\ \\ y^2 - \frac{y}{2} + \frac{1}{16} =& \frac{1}{4} + \frac{1}{16} && \text{Complete the square: add } \left( \frac{\displaystyle \frac{-1}{2}}{2} \right)^2 = \frac{1}{16} \\ \\ \left( y - \frac{1}{4} \right)^2 =& \frac{5}{16} && \text{Perfect square, get the LCD of the right side of the equation} \\ \\ y - \frac{1}{4} =& \pm \sqrt{\frac{5}{16}} && \text{Take the square root} \\ \\ y =& \frac{1}{4} \pm \sqrt{\frac{5}{16}} && \text{Add } \frac{1}{4} \\ \\ y =& \frac{1}{4} + \frac{\sqrt{5}}{4} \text{ and } y = \frac{1}{4} - \frac{\sqrt{5}}{4} && \text{Solve for } y \\ \\ y =& \frac{1 + \sqrt{5}}{4} \text{ and } y = \frac{1 - \sqrt{5}}{4} && \text{Simplify} \end{aligned} \end{equation}$