I would like to know if anything can be said about the number of roots of a polynomial whose coefficients depend on the $x$, particularly, $$x^2(f(x))^2-2xf(x)+g(x)=0$$

We further know that $f(x)$ and $g(x)$ are positive for all $x$.

From Descartes' rule of signs we know that there is no negative root and that there are at most 2 positive real roots. Can we say something more? Can we say under which conditions that polynomial has exactly one (double) real root?

  • $\begingroup$ It seems that you'll need further resrictions on $f$ and $g$ for this to even have one root. $\endgroup$ – AlexR Dec 4 '13 at 11:04
  • $\begingroup$ The roots should be: $$x=\frac{1}{f(x)}\left(1\pm\sqrt{1-g(x)}\right)$$ If I look at the discriminant we see that $D=1-g(x)$, so only $g(x)$ should influence the number of roots. But I am not sure if this is a correct step. (To me it looks weird that the root $x$ depends on $x$...it's like some recursion). Am I wrong? $\endgroup$ – pisoir Dec 4 '13 at 11:12
  • $\begingroup$ Just chose some $f(x), g(x)$ constant and then compute the solutions $\hat{x}_{1,2}$. Finally redefine $f(\hat{x}_{1,2})$ to be some other constant and you've destroyed your solution. $\endgroup$ – AlexR Dec 4 '13 at 11:20
  • $\begingroup$ Yeah, I agree. What I know more about the $f(x)$ and $g(x)$ is that they are both even functions (and positive). But I don't see how that can help me to estimate the number of roots.. $\endgroup$ – pisoir Dec 4 '13 at 11:31
  • $\begingroup$ It will still not be good to go, since you can redefine four values of $f$: $f(\pm\hat{x}_{1,2})$ to differ and still destroy it without destroying the symmetry of $f$. $\endgroup$ – AlexR Dec 4 '13 at 11:32

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