What is the inverse of (s-3)/(s^2+2s-8) ?

The inverse of (s-3)/(s^2+2s-8) is (-2s+1+sqrt(36s^2-16s+1))/(2s),(-2s+1-sqrt(36s^2-16s+1))/(2s)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse of (s-3)/(s^2+2s-8)

inverse

\frac{s - 3}{s ^{2} + 2 s - 8}

\frac{- 2 s + 1 + 36 s ^{2} - 16 s + 1}{2 s}, \frac{- 2 s + 1 - 36 s ^{2} - 16 s + 1}{2 s}

Solution steps

y = \frac{s - 3}{s ^{2} + 2 s - 8}

Replace

s

with

y

s = \frac{y - 3}{y ^{2} + 2 y - 8}

Solve for

y, s = \frac{y - 3}{y ^{2} + 2 y - 8}

\frac{- 2 s + 1 + 36 s ^{2} - 16 s + 1}{2 s}, \frac{- 2 s + 1 - 36 s ^{2} - 16 s + 1}{2 s}

What is the inverse of (s-3)/(s^2+2s-8) ?
The inverse of (s-3)/(s^2+2s-8) is (-2s+1+sqrt(36s^2-16s+1))/(2s),(-2s+1-sqrt(36s^2-16s+1))/(2s)