How to solve equation $u^4 + bu + c=0 $

Question

To solve for the roots of $x^2 + bx^\frac{1}{2} + c=0$, I thought I might substitute $u = x^\frac{1}{2}$ to get $$u^4 + bu + c=0$$

But I don't really know what to do with that either. I plugged both into Wolfram Alpha, which provided some really long complicated solutions. However, I'm looking to implement this function in Julia and the translation to Julia would be ugly. I can determine a reasonable interval of interest, and can do a binary search for $0$ over that interval. So, I have that solution.

Is there a more elegant mathematical approach?

Answer 1

Hint: Writing $$bx^{1/2}=-c-x^2$$ so we get by squaring $$b^2x=c^2+x^4+2cx^2$$ and now good luck!

Answer 2

The equation $$u^4 + bu + c=0 $$ can be factorized as $$\left( u^2+2s u +\frac{8s^3-b}{4s} \right) \left( u^2-2s u +\frac{8s^3+b}{4s} \right)=0 $$ where $s$ satisfies $s^6-\frac c4s^2-\left(\frac b8\right)^2=0$, a cubic equation in $s^2$. The roots are then obtained via the familiar quadratic formula $$u = -s \pm \sqrt{-s^2+\frac b{4s}},\>\>\> s\pm \sqrt{-s^2-\frac b{4s}} $$

Take the example $u^4+4u-1=0$ and solve for $s^2=\frac1{2}$. Apply the formulas above to obtain the real roots

$$u=-\frac1{\sqrt2}\pm \sqrt{\sqrt2 -\frac12}$$