# Solution?: Quintic Equation from Physical System

• A
First time in this forum, so greetings to everyone!

I am currently working with some physical models in the field of natural ventilation and I came across the following 5th order polynomial equation (quintic function):

$X^{5}+ C X - C =0$

This is the steady state solution of a physical system of coupled ODEs. And I am interested in finding a explicit relationship for X = f(), (the roots in terms of C), possibly analytically. The values of X (that makes physical sense) needs to be real and between 0 and 1. Additionally, C is a constant (a time scale of my system), which need to be larger or equal than 0 (and to infinity).

I can plot this equation numerically, but I was wondering if I can obtain a explicit/analytical solution of this equation for C>=0 and X[0 1]. I have been reading a little bit about quintic functions, and how some of them can be solved (this one seems to be in a similar way as Quintics in Bring–Jerrard form).

If anyone could give me some insight about this equations or if it is possible an explicit solution for the root with the constrains C>=0 and X[0 1], I would be extremely grateful!

Daniel

EDIT: improved explanation of one line.

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## Answers and Replies

jasonRF
Gold Member
I believe it might be possible to solve quintic equations using things like elliptic functions, but I expect it would be a ton of work, would not provide any extra insight, and at the end of the day would probably require at least as much computation as just solving the problem numerically to begin with.

Before jumping to numerics, we can learn a little about the real roots with very little work. Since the ##C=0## case is trivial, I will assume ##C>0##. From Descarte's rule of signs there is only 1 real positive root. If we let ##p(x) = x^5 + C x - C##, then it is clear that ##p(0) = -C## and ##p(1) = 1##, and since ##p## is continuous this tells us the real root is always between 0 and 1. So your constraint always holds.

Newton's method should converge pretty quickly as long as your initial guess is reasonable. For ##C<<1## we expect the root to be small, so ##p(x) \approx x^5 - C## and the root is approximately ##C^{1/5}##. For ##C>>1## we have ##p(x) \approx C x - C## and the root is approximately ##1##. So for an initial guess for Newton's method I would try ##x_0 = C^{1/5}/(1 + C^{1/5})##. I suggest you use Newton's method (or any other root-finding method) to find the root for values of ##C## that span the domain you care about and plot the curve. I suspect it is pretty well behaved.

jason

danielFiuza and berkeman
epenguin
Homework Helper
Gold Member
$$X^{5}+ C X - C =0$$

.
Latex needs a double \$ sign.

phyzguy
I don't think there is an analytic formula, especially with a unknown value of ##c##.
Here are some results from WolframAlpha for it: https://www.wolframalpha.com/input/?i=x^5+cx-c=0

Why would you not consider the Mathematica solution:

HypergeometricPFQ[{1/5, 2/5, 3/5, 4/5}, {1/2, 3/4, 5/4}, -3125/(256 c)]

to be an "analytic" solution.

jasonRF