Question 35.PE.NAQ.4: The characteristic equation of a given system is s³ + (K + 4......

The Routh table for this system is

\begin{array}{c|ccc}s^3 & 1 & 6 & 0 \\s^2 & K+4 & 12 & 0 \\s^1 & \frac{6(K+4)-12}{K+4}& & 0 \\s^0 & 12 & &\end{array}

For the system to be stable,

K + 4 > 0

or                                                                       K > −4

Also

\begin{aligned}& \frac{6(K+4)-12}{K+4}>0 \\or\qquad\qquad & 6(K+4)-12>0 \\or \qquad\qquad & (K+4)>2 \\or \qquad\qquad& K>-2\end{aligned}

The condition K > −2 also satisfies the condition K > −4. Therefore, the system is stable for K > −2. The only value that satisfies this condition is K = −1.