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}
$