What is the inverse of (8x)/(x^2+25) ?

The inverse of (8x)/(x^2+25) is (4+sqrt(-25x^2+16))/x ,(4-sqrt(-25x^2+16))/x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse of (8x)/(x^2+25)

inverse

\frac{8 x}{x ^{2} + 25}

\frac{4 + - 25 x ^{2} + 16}{x}, \frac{4 - - 25 x ^{2} + 16}{x}

Solution steps

y = \frac{8 x}{x ^{2} + 25}

Replace

x

with

y

x = \frac{8 y}{y ^{2} + 25}

Solve for

y, x = \frac{8 y}{y ^{2} + 25}

\frac{4 + - 25 x ^{2} + 16}{x}, \frac{4 - - 25 x ^{2} + 16}{x}

What is the inverse of (8x)/(x^2+25) ?
The inverse of (8x)/(x^2+25) is (4+sqrt(-25x^2+16))/x ,(4-sqrt(-25x^2+16))/x