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I have the following polynomial equation of 3th degree :

$$-\chi h\beta x^{3}+\chi h\beta x^{2}-2x+\left(1-\rho\right)=0$$

where $\chi$, $h$, $\beta $ and $\rho<1$ are constant parameters. I am trying to figure out the number of positive, negative and complex roots for this equation. For this, I use Descartes's rule. When I look for positive roots, there are 3 times of sign change. When I put $-x$ in the polynomial to see negative roots, there is no change of sign. If I am correct, this means that there is no negative real roots. By these information, I see that there are 3 or 1 positive roots. The case where there is 1 positive root, I suppose that there are two complex roots.

I wonder if it is possible to have a condition on constant parameters to know exactly in which case I have 3 positive real roots or 1 positive (2 complex roots) ?

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  • $\begingroup$ are $\chi, h, \beta$ positive? Are they also below $1 \; ?$ $\endgroup$ – Will Jagy Mar 15 '18 at 22:04
  • $\begingroup$ @Jagy yes, they are all positive constant parameters. They are not necessarily below 1. $\endgroup$ – optimal control Mar 15 '18 at 22:30
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Hint

The number of real or complex roots of a cubic equation are given by the sign of the discriminant, as you can see here . And using Vieta's formulas you can find if the real roots are all positive.

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with positive $A = \chi h \beta $ and $B = 1 - \rho,$ take $t = -x$ and you have $$ A t^3 + A t^2 + 2 t + B = 0. $$ There are no solutions with $t> 0.$ The first derivative is $$ 3A t^2 + 2 A t + 2 = \frac{1}{3A} \left((3At + A)^2 \; \; + \; \; A(6-A) \right) $$ so that when $A \leq 6$ the $t$ derivative is always at least $0$ and there is just a single real root with two complex roots.

If $A> 6$ you need to work more.

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  • $\begingroup$ Thanks for the answer! Maybe it is a dumb question but I don't see why you take the derivative of the equation. $\endgroup$ – optimal control Mar 16 '18 at 14:39

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