# Real Positive Zeros of Equation

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

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.
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