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Let $0 \lt \alpha \lt 1$ and $\beta,\gamma \gt 0$. Let $p(x) =x^{3}-\gamma x^{2}-\alpha x-\frac{\beta }{\gamma }$.

Can we choose $\alpha ,\beta ,\gamma $ such that $p(x)$ has one positive real root and two conjugate complex roots with negative real part?

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    Sorry, but I don't understand: if you try to apply the routh-hurwitz stability criterion, you obtain that this polynomial has to be, for any choice of the parameters with the given bounds, one root with positive real part and 2 roots with negative real parts, which means that there is one positive real root and two complex conjugate roots with negative real parts, for ANY choice of $\alpha,\beta,\gamma$ with the given bounds... maybe i'm missing something.2012-12-20
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    @wisefool: Perhaps the point of the exercise is doing things more concretely. Consider $p(x) = (x-r)(x^2+bx+c)$ where $r \gt 0$ is to be our positive root; also $b \gt 0$ and $b^2 < 4c$ to get the conjugate pair of roots with negative real part. The student can then work out real values for $b,c,r$ that give the required coefficients of the cubic.2012-12-20

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