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

The inverse of (3s)/(s^2+3s+2) is (-3s+3+sqrt(s^2-18s+9))/(2s),(-3s+3-sqrt(s^2-18s+9))/(2s)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse of (3s)/(s^2+3s+2)

inverse

\frac{3 s}{s ^{2} + 3 s + 2}

\frac{- 3 s + 3 + s ^{2} - 18 s + 9}{2 s}, \frac{- 3 s + 3 - s ^{2} - 18 s + 9}{2 s}

Solution steps

y = \frac{3 s}{s ^{2} + 3 s + 2}

Replace

s

with

y

s = \frac{3 y}{y ^{2} + 3 y + 2}

Solve for

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

\frac{- 3 s + 3 + s ^{2} - 18 s + 9}{2 s}, \frac{- 3 s + 3 - s ^{2} - 18 s + 9}{2 s}

What is the inverse of (3s)/(s^2+3s+2) ?
The inverse of (3s)/(s^2+3s+2) is (-3s+3+sqrt(s^2-18s+9))/(2s),(-3s+3-sqrt(s^2-18s+9))/(2s)