What is the inverse of 2pi(x)^2+8pi(x) ?

The inverse of 2pi(x)^2+8pi(x) is (-8pi+sqrt(64pi^2+8pix))/(4pi),(-8pi-sqrt(64pi^2+8pix))/(4pi)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse of 2pi(x)^2+8pi(x)

inverse

2 π (x)^{2} + 8 π (x)

\frac{- 8 π + 64 π ^{2} + 8 π x}{4 π}, \frac{- 8 π - 64 π ^{2} + 8 π x}{4 π}

Solution steps

y = 2 π x^{2} + 8 π x

Replace

x

with

y

x = 2 π y^{2} + 8 π y

Solve for

y, x = 2 π y^{2} + 8 π y

\frac{- 8 π + 64 π ^{2} + 8 π x}{4 π}, \frac{- 8 π - 64 π ^{2} + 8 π x}{4 π}

What is the inverse of 2pi(x)^2+8pi(x) ?
The inverse of 2pi(x)^2+8pi(x) is (-8pi+sqrt(64pi^2+8pix))/(4pi),(-8pi-sqrt(64pi^2+8pix))/(4pi)