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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 $k\in\{-1,1\}/\mathbb{\mathbb{Q}}$ ?

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    @Pragabhava : and solving Cubic equation? it's quite complicated to say that this is rough upper bound, maybe other ideas?2012-10-26

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So you want $y(r)=r^p+Ar+B$ cross the axis on $r>0$, right? Note that this function has at most one positive critical point $R$ found from $pR^{p-1}+A=0$. You can also figure out easily what happens at $r\to 0$ and $r\to+\infty$. Now you get 3 interesting places to look at: $0$, $+\infty$ and $R$ (if the latter exists). If there is a sign change between some 2 of them, you are guaranteed a positive root. The good news is that if all three have the same sign, there is no positive root. So the criterion is complete.

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    Well, all three terms have different growth rates near $0$ and $\infty$, and when one term is twice larger than any of the other two, you certainly don't have a root. This gives you SOME bounds to start with.2012-10-31