What is the inverse of 2/((s^2-2s+9)) ?

The inverse of 2/((s^2-2s+9)) is (2s+sqrt(-32s^2+8s))/(2s),(2s-sqrt(-32s^2+8s))/(2s)

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inverse of 2/((s^2-2s+9))

inverse

\frac{2}{( s ^{2} - 2 s + 9 )}

\frac{2 s + - 32 s ^{2} + 8 s}{2 s}, \frac{2 s - - 32 s ^{2} + 8 s}{2 s}

Solution steps

y = \frac{2}{s ^{2} - 2 s + 9}

Replace

s

with

y

s = \frac{2}{y ^{2} - 2 y + 9}

Solve for

y, s = \frac{2}{y ^{2} - 2 y + 9}

\frac{2 s + - 32 s ^{2} + 8 s}{2 s}, \frac{2 s - - 32 s ^{2} + 8 s}{2 s}

f (x) = x^{2} - 4 x - 1

f (x) = 3 x + 4 x + 6

f (x) = (x - 7) (x - 9)

f (x) = \frac{2 x + 5}{x + 1}

What is the inverse of 2/((s^2-2s+9)) ?
The inverse of 2/((s^2-2s+9)) is (2s+sqrt(-32s^2+8s))/(2s),(2s-sqrt(-32s^2+8s))/(2s)