Answer
$x=±\sqrt {\frac{5+\sqrt{33}}{2}}$ or $x=±\sqrt {\frac{5-\sqrt{33}}{2}}$
Work Step by Step
$x^4 - 5x^2 - 2 = 0$
Let $u=x^2$
Substitute.
$u^2 - 5u - 2 = 0$
Solve using the quadratic formula: $u = \frac{-b±\sqrt{b^2-4ac}}{2a}$
$a=1$, $b=-5$, $c=-2$
$u = \frac{-(-5)±\sqrt{(-5)^2-(4⋅1⋅-2)}}{2⋅1}$
$u = \frac{5±\sqrt{25+8}}{2}$
$u = \frac{5±\sqrt{33}}{2}$
Refer to the original substitution to get the value of $x$.
$u=x^2$
$\frac{5±\sqrt{33}}{2}=x^2$
$x^2=\frac{5+\sqrt{33}}{2}$ or $x^2=\frac{5-\sqrt{33}}{2}$
$x=±\sqrt {\frac{5+\sqrt{33}}{2}}$ or $x=±\sqrt {\frac{5-\sqrt{33}}{2}}$
