GATE 2017 EE (EE2) – Question 44

The root locus with characteristic equation $s^2+6Ks+2s+5=0$, K>0, enters the real axis at

Show answer and explanation

Correct answer: (B) $s=-\sqrt5$

Explanation

The roots become real when their quadratic discriminant vanishes: $(6K+2)^2-20=0$. Positive K selects 6K+2=2√5.

At the repeated root, $s=-(6K+2)/2=-\sqrt5$. Thus the complex branches meet the real axis at **−√5, B**.

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