The roots of the polynomial s2+4s+8𝑠2+4𝑠+8 are−4−4 and −2−2−2j−2𝑗 and 2j2𝑗−2+2j−2+2𝑗 and −2−2j−2−2𝑗−2−2 and −2
Question
Solution
It seems like there's a mistake in the question. The roots of the polynomial s^2 + 4s + 8 can be found using the quadratic formula, which is:
s = [-b ± sqrt(b^2 - 4ac)] / (2a)
Here, a = 1, b = 4, and c = 8.
Substituting these values into the formula gives:
s = [-4 ± sqrt((4)^2 - 418)] / (2*1) s = [-4 ± sqrt(16 - 32)] / 2 s = [-4 ± sqrt(-16)] / 2 s = [-4 ± 4i] / 2 s = -2 ± 2i
So, the roots of the polynomial s^2 + 4s + 8 are -2 + 2i and -2 - 2i.
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