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## T he number of roots of the polynomial, 𝑠7 +𝑠6 + 7𝑠5 + 14𝑠4 + 31𝑠3 + 73𝑠2 + 25𝑠 + 200,in the open left half of the complex plane is

## (A) 3 (B) 4 (C) 5 (D) 6

Asked On2019-04-10 11:19:37 by:Rohit498

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Download question setAnswersThe characteristic equation is

s7 + s6 + 7s5 + 14s4 + 31s3 + 73s2 + 25s + 200 = 0

Auxiliary equation is A(s) = 8s4 + 48s2 + 200

Differentiating this w.r.t. s we get d /ds A (s ) = + 32s

^{3}+96sTotal number of poles = 7 Two sign changes above auxiliary equation = 2 poles in RHS

Two sign changes below auxiliary equation implies that out of 4 symmetric roots about origin, two poles are in LHS and two poles are in RHS. Therefore 3 poles in LHS and 4 poles in RHS.

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