What is the inverse of ((3s+1))/((s^2+s-12)) ?

The inverse of ((3s+1))/((s^2+s-12)) is (-s+3+sqrt(49s^2-2s+9))/(2s),(-s+3-sqrt(49s^2-2s+9))/(2s)

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inverse of ((3s+1))/((s^2+s-12))

inverse

\frac{( 3 s + 1 )}{( s ^{2} + s - 12 )}

\frac{- s + 3 + 49 s ^{2} - 2 s + 9}{2 s}, \frac{- s + 3 - 49 s ^{2} - 2 s + 9}{2 s}

Solution steps

y = \frac{3 s + 1}{s ^{2} + s - 12}

Replace

s

with

y

s = \frac{3 y + 1}{y ^{2} + y - 12}

Solve for

y, s = \frac{3 y + 1}{y ^{2} + y - 12}

\frac{- s + 3 + 49 s ^{2} - 2 s + 9}{2 s}, \frac{- s + 3 - 49 s ^{2} - 2 s + 9}{2 s}

y = \frac{2 x}{x - 1}

f (x) = (3)

f (x) = \frac{8 x}{3 x - 2}

cos (\frac{18}{27})

x^{- 3}

What is the inverse of ((3s+1))/((s^2+s-12)) ?
The inverse of ((3s+1))/((s^2+s-12)) is (-s+3+sqrt(49s^2-2s+9))/(2s),(-s+3-sqrt(49s^2-2s+9))/(2s)