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?
Best Answer
Given cubic polynomial $p(x)$ with coefficients as shown,Descartes Rule of Signs tells us there will be one positive root and either two or no negative roots. If only one real root exists, then of course a conjugate pair of complex roots must exist.
So what is required is to choose:
$$ p(x)=(x-r)(x^2 + bx + c) $$
in a compatible way, i.e. that $r \gt 0$, that the quadratic factor above has a conjugate pair of roots with negative real part, and that the expanded product has coefficients as prescribed.
The conjugate pair of roots is guaranteed by a negative discriminant, $b^2 \lt 4c$, while their real part being negative implies $b \gt 0$. For the rest let's multiply out:
$$ p(x) = x^3- (r-b)x^2- (rb- c)x- rc $$
We've already deduced that $r$ and $c$ are positive, so a negative constant term of $p(x)$ is certain.
Likewise for the coefficient of $x^2$ to have the proper sign means $b \lt r$. Finally we want $0 \lt rb- c \lt 1$ to meet the condition on $\alpha$ the coefficient of $x$.
It's not hard to find a suitable choice for $r,b,c$. Taking $b = 1$, we need $r$ greater than that, say $r = 1.1$. Then $c$ needs to be big enough to get the negative discriminant but small enough that $\alpha = rb- c$ is between $0$ and $1$. So $c = 1$ will work:
$$ p(x) = (x-1.1)(x^2 + x + 1) = x^3 - 0.1x^2 - 0.1x - 1.1 $$