SOLUTION: Systems are unstable if their characteristic equations have a positive root (solution). Determine whether each of the following characteristic equations represent a stable or unsta

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Question 168888: Systems are unstable if their characteristic equations have a positive root (solution). Determine whether each of the following characteristic equations represent a stable or unstable system.
(i) s^3+6s^2+11s+6=0
(ii) s^3+s^2-8s-12=0

Answer by kev82(151) About Me  (Show Source):
You can put this solution on YOUR website!
We have been told that a system is unstable if their equation has a positive root, so the question is really "do these equations have a positive root."
The second is easier, and we can immediately spot it has a positive root. The leading term is s^3, so for large enough s (denote s_0) it will be positive. You can find such a point if you wish, but we are guaranteed one exists. When s=0 the expression evaluates to -12. The function is continuous so by the intermediate value theorem must have a root between 0 and s_0.
If you are prepared to accept it, then a simple argument for the first problem is to say that the entire expression is made up of multiplication and addition. It is not possible to get 0 by only adding and multiplying positive numbers therefore no positive root.
To do it more rigourously, again, we know there is an s_0 for which the expression is positive, and we know at s=0 the value is 6, so the only way for there to be a positive root, is if there is a positive local minima that is below 0. We can differentiate to find the locations of the extrema. The extrema satisfy
3s%5E2%2B12s%2B11=0
Solving that quadratic (will leave it to you) gives two roots that are both negative, so the extrema are to the left of zero, hence there are no positive roots.