What is the inverse of 1/(s^2+s+1) ?

The inverse of 1/(s^2+s+1) is (-s+sqrt(-3s^2+4s))/(2s),(-s-sqrt(-3s^2+4s))/(2s)

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inverse of 1/(s^2+s+1)

inverse

\frac{1}{s ^{2} + s + 1}

\frac{- s + - 3 s ^{2} + 4 s}{2 s}, \frac{- s - - 3 s ^{2} + 4 s}{2 s}

Solution steps

y = \frac{1}{s ^{2} + s + 1}

Replace

s

with

y

s = \frac{1}{y ^{2} + y + 1}

Solve for

y, s = \frac{1}{y ^{2} + y + 1}

\frac{- s + - 3 s ^{2} + 4 s}{2 s}, \frac{- s - - 3 s ^{2} + 4 s}{2 s}

y = - lo g_{10} (x)

f (x) = \frac{2 x - 4}{x + \frac{3}{2}}

f (x) = - 1 - x

f (x) = 6 x^{2} + 66 x + 60

f (x) = 10 x - 2^{3}

What is the inverse of 1/(s^2+s+1) ?
The inverse of 1/(s^2+s+1) is (-s+sqrt(-3s^2+4s))/(2s),(-s-sqrt(-3s^2+4s))/(2s)