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

The inverse of 2/(s^2-s) is (s+sqrt(s^2+8s))/(2s),(s-sqrt(s^2+8s))/(2s)

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inverse of 2/(s^2-s)

inverse

\frac{2}{s ^{2} - s}

\frac{s + s ^{2} + 8 s}{2 s}, \frac{s - s ^{2} + 8 s}{2 s}

Solution steps

y = \frac{2}{s ^{2} - s}

Replace

s

with

y

s = \frac{2}{y ^{2} - y}

Solve for

y, s = \frac{2}{y ^{2} - y}

\frac{s + s ^{2} + 8 s}{2 s}, \frac{s - s ^{2} + 8 s}{2 s}

f (x) = ln (6 x - 8) - 2

\frac{2 x - 1}{3 x + 5}

f (x) = tan (19 x)

f (x) = (x - 9)

f (x) = (\frac{1}{x + 3})^{2} - 2

What is the inverse of 2/(s^2-s) ?
The inverse of 2/(s^2-s) is (s+sqrt(s^2+8s))/(2s),(s-sqrt(s^2+8s))/(2s)