College Algebra, Chapter 1, 1.5, Section 1.5, Problem 60

Find all real solutions of the equation $\displaystyle \sqrt{1 + \sqrt{x + \sqrt{2x +1}}} = \sqrt{5 + \sqrt{x}}$


$\begin{equation} \begin{aligned} \sqrt{1 + \sqrt{x + \sqrt{2x +1}}} =& \sqrt{5 + \sqrt{x}} && \text{Given} \\ \\ 1 + \sqrt{x + \sqrt{2x + 1}} =& 5 + \sqrt{x} && \text{Square both sides} \\ \\ \sqrt{x + \sqrt{2x + 1}} =& 4 + \sqrt{x} && \text{Subtract } 1 \\ \\ x + \sqrt{2x + 1} =& 16 + 8 \sqrt{x} + x && \text{Square both sides} \\ \\ \sqrt{2x + 1} =& 16 + 8 \sqrt{x} && \text{Cancel out } x \\ \\ 2x + 1 =& 256 + 256 \sqrt{x} + 64x && \text{Square both sides} \\ \\ 256 \sqrt{x} + 62x + 255 =& 0 && \text{Combine like terms} \\ \\ 256 \sqrt{x} + 62 (\sqrt{x})^2 + 255 =& 0 && \text{If we let } w = \sqrt{x} \\ \\ 256w + 62w^2 + 255 =& 0 && \text{Subtract } 255 \\ \\ 62w^2 + 256 w =& -255 && \text{Divide both sides by } 62 \\ \\ w^2 + \frac{256}{62} w + \frac{16384}{3844} =& \frac{-255}{62} + \frac{16384}{3844} && \text{Complete the square: add } \left( \frac{\displaystyle \frac{256}{62}}{2} \right)^2 = \frac{16384}{3844} \\ \\ \left( w + \frac{128}{62} \right)^2 =& \frac{287}{1922} && \text{Perfect Square} \\ \\ w + \frac{128}{62} =& \pm \sqrt{\frac{287}{1922}} && \text{Take the square root} \\ \\ w =& \frac{-128}{62} \pm \frac{\sqrt{287}}{31 \sqrt{2}} && \text{Subtract } \frac{128}{62} \text{ and simplify} \\ \\ w =& \frac{-128 + \sqrt{574}}{62} \text{ and } w = \frac{-128 - \sqrt{574}}{62} && \text{Solve for } w \\ \\ \sqrt{x} =& \frac{-128 + \sqrt{574}}{62} \text{ and } \sqrt{x} = \frac{-128 - \sqrt{574}}{62} && \text{Substitute } w = \sqrt{x} \\ \\ x =& \left( \frac{-128 + \sqrt{574}}{62} \right)^2 \text{ and } x = \left( \frac{-128 - \sqrt{574}}{62} \right)^2 && \text{Solve for } x \end{aligned} \end{equation}$