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I have a quartic polynomial in $x$ (too long to write here)

$f(x,c_1, c_2, c_2)$

where $c_1, c_2, c_3$ are constants which I have complete freedom over how to fix their values, as long as they are all greater than zero.

As this is a quartic, I have 4 roots to $f=0$ (two are usually imaginary and two real). I want to fix my constants so that I have two real, positive values of $x$.

At the moment I'm just plotting $f(x)$ with different values of $c_1, c_2, c_3$ and trying to pick these values so that my graph crosses the real positive axis twice. However so far I am unsuccessful.

Is there some analytical method of doing this?

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  • $\begingroup$ Maybe this helps to make a step forward: You may check if your polynomial has two real roots using the discriminant, see en.wikipedia.org/wiki/Quartic_function#Nature_of_the_roots, but I do not know how to check if these real roots are positive without computing them. $\endgroup$ – gammatester Jul 17 '14 at 11:20
  • $\begingroup$ @gammatester thanks for the advice, but I know it has two real roots (for some constants). In fact, I've tried hundreds of combinations and it always has two real and two imaginary roots. $\endgroup$ – Phibert Jul 17 '14 at 11:22
  • $\begingroup$ Descartes' rule of signs gives a necessary condition for two positive roots: the number of sign changes of the coefficients must be even. $\endgroup$ – Antonio Vargas Jul 17 '14 at 14:58

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