Special case quintic equation analytic solution

Question

I've met this quintic equation in my research: $$x^5 - \frac{k}{k-1} \cdot x^3 +\frac{r}{k-1}=0$$ with the additional conditions: $$k>1; \quad 0<r<1; \quad 0<x<1$$ My background in math is limited. I know that it cannot be solved in radicals, but can be using the elliptic functions. However, elliptic function method requires transforming the equation to the Bring-Jerrard form. I just wondered if it is possible to find a analytic solution easier way. Basically, I need just one function x(r,k) in real positive number which satisfies the equation. Is it possible? How to get one?

Answer 1

Writing $x = (r/(k-1))^{1/5} y$, the equation becomes $$ y^5 - \frac{k}{(k-1)^{3/5} r^{2/5}} y^3+ 1 = 0$$ Let $c = k/((k-1)^{3/5} r^{2/5}$, so this is $y^5 - c y^3 + 1 = 0$.

Now, if you want a solution with $y > 0$, $c$ must not be too small. If $c > 0$, the minimum value of $y^5 - c y^3 + 1$ for $y \ge 0$ is $1 - 6 \sqrt{15} c^{5/2}/125$, and you want that $\le 0$, so you need $$c \ge \frac{5 \cdot 2^{3/5} \cdot 3^{2/5}}{6}$$ If $c$ is greater than this, there will be two positive real solutions for $y$ (and thus for $x$), if $c$ is less there will be none.