College Algebra, Chapter 2, 2.3, Section 2.3, Problem 56

Find all real solutions of the equation $x^4 - 8x^2 + 2 = 0$

$
\begin{equation}
\begin{aligned}
x^4 - 8x^2 + 2 &= 0\\
\\
w^2 - 8w + 2 &= 0 && \text{Let } w = x^2\\
\\
w^2 - 8w &= -2 && \text{Subtract } 2\\
\\
w^2 - 8w + 16 &= -2 + 16 && \text{Complete the square: Add } \left( -\frac{8}{2} \right)^2 = 16\\
\\
(w - 4)^2 &= 14 && \text{Perfect Square}\\
\\
(w - 4) &= \pm \sqrt{14} && \text{Take the square root}\\
\\
w &= 4 \pm \sqrt{14} && \text{Add }4\\
\\
w = 4 + \sqrt{14} \text{ and } w &= 4 - \sqrt{14} && \text{Solve for } w\\
\\
x^2 = 4 + \sqrt{14} \text{ and } x^2 &= 4 - \sqrt{14} && \text{Substitute } w = x^2\\
\\
x = \pm \sqrt{4+\sqrt{14}} \text{ and } x &= \pm \sqrt{4 - \sqrt{4}} && \text{Take the square root}
\end{aligned}
\end{equation}
$


Thus, the solutions are
$x = \sqrt{4+\sqrt{14}},\quad x = - \sqrt{4+\sqrt{14}},\quad x = \sqrt{4-\sqrt{14}},$ and $ x = -\sqrt{4-\sqrt{14}}$