Roots for incomplete quartic

\(x^4-10x^2+1=0\)

Set \(x^4\) as \(\left(x^2\right)^2\), therefore:

\(\left(x^2\right)^2-10x^2+1=0\)

Now set \(x^2\) as any variable you like, let's choose the variable \(y\):

\(y^2-10y+1=0\)

Now use the quadratic formula: \(y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\), where:

\(a=1\)
\(b=-10\)
\(c=1\)

You should get the following roots:

\(y_1=\frac{5+\sqrt{22}}{3}\)
\(y_1=\frac{5-\sqrt{22}}{3}\)

Now replace \(y\) with \(x^2\):

\(x^2=\frac{5+\sqrt{22}}{3}\)
\(x^2=\frac{5-\sqrt{22}}{3}\)

For each, square-root both sides of the equation:

\(x=\pm \sqrt{\frac{5+\sqrt{22}}{3}}\)
\(x=\pm \sqrt{\frac{5-\sqrt{22}}{3}}\)

This means there are 4 solutions to the equation...

Good luck