Why is the solution for:
$$ ax^{4} + bx^{2} + c = 0$$
$$x_1 = \frac{-b + \sqrt{b^2-4ac}}{2a}$$
$$x_{final} = \sqrt[4]{x_1}$$?
Thank you.
Because if $x_1=x^4$, then you have $ax_1+bx^2+c=0$, which is an equation in two variables, so it is not at all helpful. We can also say $ax_1+b\sqrt{x_1}+c=0$, but this is not a quadratic equation. By saying $x_1=x^2$, I wind up with a quadratic equation of $x_1$ that is solvable.