During my research on physical problem, I faced the following simple equation:
$r^{2k+1}+ab\,r-a=0$
With:
$-1\leq k\leq1\:,\:0
I need to put bounders on $a,b,k$ such that this equation will have always at least one positive root, also I need to find some rough upper & lower boundary estimation for this positive zero.
We see that if $b=0$ it will be very easy to solve it, anyway this zero will not give us an upper estimation because $ab$ can be positive as much as negative. Also I thought of putting :
$k=\frac{m}{n}\:,\: m,n\in\mathbb{Z}\:,\: m\leqslant n\:$
and we get the polynomial:
$z^{2m+n}+ab\,z^{n}-a=0\:,\: z=\sqrt[n]{r}$
And using Descartes' Sign Rule tells us only that there will be no positive roots at all only when $a,b<0$ , that's not very useful for me.
So my question is if you can think of any additional tricks/transforms to extract this conditions and the upper bound? also I'm wondering if there is any argumentation on checking the behavior of the polynomial roots (As I did above) and suppose that it will be also true for $k\in\{-1,1\}/\mathbb{\mathbb{Q}}$ ?