Answer
$\{\pm\sqrt{4+\sqrt{15}},\pm\sqrt{4-\sqrt{15}}\}$
Work Step by Step
Using the properties of equality, the given equation, $
x^4-8x^2=-1
,$ is equivalent to
\begin{align*}
x^4-8x^2+1&=0
.\end{align*}
Let $z=
x^2
$. Then the equation above is equivalent to
\begin{align*}
\left(x^2\right)^2-8x^2+1&=0
\\
z^2-8z^2+1&=0
.\end{align*}
Using $ax^2+bx+c=0,$ the equation above has $a=
1
$, $b=
-8
$ and $c=
1
$. Using $
x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}
$ or the Quadratic Formula, then
\begin{align*}\require{cancel}
z&=\dfrac{-(-8)\pm\sqrt{(-8)^2-4(1)(1)}}{2(1)}
\\\\&=
\dfrac{8\pm\sqrt{64-4}}{2}
\\\\&=
\dfrac{8\pm\sqrt{60}}{2}
\\\\&=
\dfrac{8\pm\sqrt{4\cdot15}}{2}
\\\\&=
\dfrac{8\pm2\sqrt{15}}{2}
\\\\&=
\dfrac{\cancelto48\pm\cancelto12\sqrt{15}}{\cancelto12}
\\\\&=
4\pm\sqrt{15}
.\end{align*}
Since $z=x^2$, then by back-substitution,
\begin{align*}
x^2&=4\pm\sqrt{15}
\\
x&=\pm\sqrt{4\pm\sqrt{15}}
&(\text{take square root of both sides})
.\end{align*}
Hence, the solution set of the equation $
x^4-8x^2=-1
$ is $\{\pm\sqrt{4+\sqrt{15}},\pm\sqrt{4-\sqrt{15}}\}$.
