What is the inverse of (16a^4+16a)/(a^2+1-a) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse of (16a^4+16a)/(a^2+1-a)

inverse

\frac{16 a ^{4} + 16 a}{a ^{2} + 1 - a}

\frac{- 2 + a + 4}{4}, - \frac{a + 4 + 2}{4}

Solution steps

y = \frac{16 a ^{4} + 16 a}{a ^{2} + 1 - a}

Replace

a

with

y

a = \frac{16 y ^{4} + 16 y}{y ^{2} + 1 - y}

Solve for

y, a = \frac{16 y ^{4} + 16 y}{y ^{2} + 1 - y}

\frac{- 2 + a + 4}{4}, - \frac{a + 4 + 2}{4}

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