What is the inverse of (4s+12)/(s^2+8s+16) ?

The inverse of (4s+12)/(s^2+8s+16) is (2(-2s+sqrt(-s+1)+1))/s ,(2(-2s-sqrt(-s+1)+1))/s

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse of (4s+12)/(s^2+8s+16)

inverse

\frac{4 s + 12}{s ^{2} + 8 s + 16}

\frac{2 ( - 2 s + - s + 1 + 1 )}{s}, \frac{2 ( - 2 s - - s + 1 + 1 )}{s}

Solution steps

y = \frac{4 s + 12}{s ^{2} + 8 s + 16}

Replace

s

with

y

s = \frac{4 y + 12}{y ^{2} + 8 y + 16}

Solve for

y, s = \frac{4 y + 12}{y ^{2} + 8 y + 16}

\frac{2 ( - 2 s + - s + 1 + 1 )}{s}, \frac{2 ( - 2 s - - s + 1 + 1 )}{s}

What is the inverse of (4s+12)/(s^2+8s+16) ?
The inverse of (4s+12)/(s^2+8s+16) is (2(-2s+sqrt(-s+1)+1))/s ,(2(-2s-sqrt(-s+1)+1))/s