What is the inverse of 2/(x^2-2x-3) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse of 2/(x^2-2x-3)

inverse

\frac{2}{x ^{2} - 2 x - 3}

\frac{2 x + 16 x ^{2} + 8 x}{2 x}, \frac{2 x - 16 x ^{2} + 8 x}{2 x}

Solution steps

y = \frac{2}{x ^{2} - 2 x - 3}

Replace

x

with

y

x = \frac{2}{y ^{2} - 2 y - 3}

Solve for

y, x = \frac{2}{y ^{2} - 2 y - 3}

\frac{2 x + 16 x ^{2} + 8 x}{2 x}, \frac{2 x - 16 x ^{2} + 8 x}{2 x}

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