1. ## Solution for equation

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.

2. ## Re: Solution for equation

It isn't.

$ax^4+bx^2+c=0$
Let $x_1=x^2$
$x_1=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\pm \sqrt{x_1}$

You will wind up with four solutions, not one.

3. ## Re: Solution for equation

Originally Posted by SlipEternal
It isn't.

$ax^4+bx^2+c=0$
Let $x_1=x^2$
$x_1=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$
$x=\pm \sqrt{x_1}$

You will wind up with four solutions, not one.
Why did you choose $$x_1 = x^2$$ and not $$x_1 = x^4$$?

4. ## Re: Solution for equation

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.