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

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) (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

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.
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