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During my research on physical problem, I faced the following simple equation:

$r^{2k+1}+ab\,r-a=0$

With:

$-1\leq k\leq1\:,\:0<r\:,\: a,b\in\mathbb{R}$

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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Why don't you try to work with $r^{2k+1} = a -ab r$ in the same way one finds eigenvalues of the mixed bounadry conditions (i.e. $\tan \lambda = \lambda$)? –  Pragabhava Oct 25 '12 at 23:10
    
Can you please advice where I can read on how to convert it to eigenvalues problems? –  TMS Oct 25 '12 at 23:16
    
Oh, you misunderstood me. What I meant is to attack the problem in the same way as, for example $\tan \lambda = \lambda$. If you plot $r^{2k+1}$ and $a-ab r$ you can see where are the intersections, which are the roots. For example if $k < -1/2$, then you have an asymptote at $r=0$, and it goes to zero from above as $r\rightarrow \infty$, hence if the straight line $a-ab r$ has a positive slope, there must be a positive solution. –  Pragabhava Oct 25 '12 at 23:43
    
I see now what you mean, still that can't give me the boundaries... they are more important for me.. –  TMS Oct 25 '12 at 23:49
    
For $k > 0$, you know that $r < r^{2k+1} < r^3$, and you can dominate with the line and the cubic, depending on the sign of $ab$ and if $0 < r < 1$ or $1 < r < \infty$. Negative $k$ is not so straight forward, but the same arguments can be made. –  Pragabhava Oct 25 '12 at 23:58
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1 Answer

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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Oh yes I forget cheeking critical points, that nice, anyway this still can't give me upper & lower bounders.... –  TMS Oct 25 '12 at 23:47
    
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. –  fedja Oct 31 '12 at 21:19
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