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

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

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inverse of 1/(s^2+as)

inverse

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

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

Solution steps

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

Replace

s

with

y

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

Solve for

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

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

f (x) = 0.2

y = (x - 3)^{2} + 7, x > 3

f (x) = e^{x}, x \leq 0

\frac{2 x - 3}{5 x + 2}

y = 4 x - 3 + 5

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