What is the inverse of 1/(s(s+3)) ?

The inverse of 1/(s(s+3)) is (-3s+sqrt(9s^2+4s))/(2s),(-3s-sqrt(9s^2+4s))/(2s)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse of 1/(s(s+3))

inverse

\frac{1}{s ( s + 3 )}

\frac{- 3 s + 9 s ^{2} + 4 s}{2 s}, \frac{- 3 s - 9 s ^{2} + 4 s}{2 s}

Solution steps

y = \frac{1}{s ( s + 3 )}

Replace

s

with

y

s = \frac{1}{y ( y + 3 )}

Solve for

y, s = \frac{1}{y ( y + 3 )}

\frac{- 3 s + 9 s ^{2} + 4 s}{2 s}, \frac{- 3 s - 9 s ^{2} + 4 s}{2 s}

What is the inverse of 1/(s(s+3)) ?
The inverse of 1/(s(s+3)) is (-3s+sqrt(9s^2+4s))/(2s),(-3s-sqrt(9s^2+4s))/(2s)