Solving the quadratic formula to determine stability of a system I am trying to solve the $2\times 2$ matrix $$\begin{bmatrix} 0 &1 \\ -k &-b \end{bmatrix}$$ for a relationship between the variables $k$ and $b$ to determine when a system is stable. Stability means that the real component of the eigenvalues are all $0$ or less.
I have sought the eigenvalues and have gotten to the quadratic formula here: eigens $= \frac{-b \pm \sqrt{b^2-4k}}{2}$
I know that the eigenvalues must be less than $0$, but I am having trouble with the algebra to get to a clear relationship.
Any tips?
 A: Going off your formula for the eigenvalues: 
If $b^2 - 4k \leq 0$, then the real part of the eigenvalues will be $-\frac{b}{2}$. So in this case, the eigenvalues will have real part less than $0$ if $-\frac{b}{2} \leq 0$, or, equivalently, if $b \geq 0$
If $b^2 - 4k \geq 0$, then the eigenvalues are all real numbers of the form:
$$\frac{-b \pm \sqrt{b^2 -4k}}{2}$$
We always have that:
$$\frac{-b + \sqrt{b^2 -4k}}{2} \geq \frac{-b - \sqrt{b^2 -4k}}{2}$$
So it suffices to only check that $\frac{-b + \sqrt{b^2 -4k}}{2} \leq 0$
$$\frac{-b + \sqrt{b^2 -4k}}{2} \leq 0$$
$$\sqrt{b^2 - 4k} \leq b$$
Since we are considering the positive square root, this equation will never hold if $b$ is negative, so we assume $b$ is positive, allowing us to square both sides:
$$b^2 - 4k \leq b^2$$
$$-4k \leq 0$$
$$4k \geq 0$$
So your solutions: the eigenvalues have real part less than or equal to 0 if:
$b^2 \leq 4k$ and $b \geq0$
or
$b^2 \geq 4k \geq 0$ and $b \geq 0$
Which simplifies to the condition: $b, k \geq 0$
