Negative roots of a cubic equation Under what conditions will the cubic equation
$ax^3 + bx^2 + cx + d$ where $a,b,c,d \in \mathbb R$ yield roots which have negative real parts? (All roots must have negative real parts)
Motivation:
I am studying a dynamical system i.e, Chua circuit, in $ \mathbb R^3 $and wish to analyze it's stability. For stability analysis, one needs to find out eigen values of the Jacobian matrix. If the eigen values have negative real parts, then the system will have a stable fixed point. 
I wish to synchronize the system and vary certain parameters so as to ensure that the system always has negative eigen values. 
For equations, visit http://www.chuacircuits.com/diagram.php
 A: Assume we have the monic polynomial
$$ p(x) = x^3 - s_1 x^2 + s_2 x -s_3 = (x-\xi_1)(x-\xi_2)(x-\xi_3).$$
Since $s_1=\xi_1+\xi_2+\xi_3$, $s_1< 0$ is a necessary condition for the roots of $p(x)$ to lie in $\text{Re}(z)<0$.

If the discriminant of $p(x)$ is positive, then $p(x)$ has three real roots and they are all negative iff $p(x)$ is increasing on $\mathbb{R}^+$, or:
$$ \forall x\geq 0, \qquad  p'(x) = 3x^2-2s_1 x + s_2 \geq 0,$$
but $s_1< 0$  implies that the last line is equivalent to $s_2\geq 0$. 

If the discriminant of $p(x)$ is negative, $p(x)$ has only one real root (say $\xi_1$) and two conjugated complex roots $\xi_2,\xi_3 = \sigma\pm it$. It happens that:
$$ s_1 = \xi_1 + 2\sigma,\qquad s_2=2\xi_1\sigma+(\sigma^2+t^2),\qquad s_3 = \xi_1(\sigma^2+t^2) $$
so $s_3 < 0$ and $s_2>0$ are necessary conditions to have $\xi_1,\sigma<0$.
It happens that $\sigma < 0$ is equivalent to $\xi_1 > s_1$, so $p(x)$ must have a root in $(s_1,+\infty)$.
That can be checked through Sturm's theorem or by locating the stationary points of $p(x)$.
A: There is always a real root and you have three possibilities: 
Let $F(x)= ax^3+bx^2+cx+d$ ; the discriminant  $\Delta$ is defined by
$$\Delta=(bc)^2+18abcd-4ac^3-4db^3-27(ad)^2$$
(1) $\Delta>0$: $F(x)=a(x-x_1)(x-x_2)(x-x_3)$;  the three roots, $x_1, x_2, x_3,$ are real.
(2) $\Delta=0$: $F(x)=a(x-x_1)(x-x_2)^2$; at least two roots coincide (eventually the three coincide); the three roots are real.
(3) $\Delta<0$: $F(x)=a(x-x_1)[(x-\alpha)^2+(\beta)^2]$ when $x_2=\alpha+i\beta$ with $\beta\neq0$; one (necessarily) root is real and the other complex conjugates.
From this,I think you can deduce a condition in order that “All roots must have negative real parts” as you desire.
A: You can separate $x$ into the real and imaginary parts, $x=u+Iv$, to get two equations that must be satisfied. For one of them we can solve for $v$ and substitute to get the equation $$u^3+(b/a)u^2+((2ac+2b^2)/8a^2)u+(1/8)(bc-da)/a^2$$ for $u$. So you want this to have three negative roots $u$. To answer that you can try to use Descartes' rule of signs to give some information.  
