What is the inverse of sqrt(x^2+7x),x>0 ?

The inverse of sqrt(x^2+7x),x>0 is (-7+sqrt(49+4x^2))/2 ,(-7-sqrt(49+4x^2))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of sqrt(x^2+7x),x>0

inverse

x^{2} + 7 x, x > 0

\frac{- 7 + 49 + 4 x ^{2}}{2}, \frac{- 7 - 49 + 4 x ^{2}}{2}

Solution steps

y = x^{2} + 7 x

Replace

x

with

y

x = y^{2} + 7 y

Solve for

y, x = y^{2} + 7 y

\frac{- 7 + 49 + 4 x ^{2}}{2}, \frac{- 7 - 49 + 4 x ^{2}}{2}

f (x) = \frac{x}{x ^{2} - 1}

f (x) = x^{2} - 5

f (y) = \frac{x ^{2} + 4}{x ^{2} - 1}

f (x) = x^{3} - 6 x^{2} + 7

y = - 3 x + 1, (3, 5)

What is the inverse of sqrt(x^2+7x),x>0 ?
The inverse of sqrt(x^2+7x),x>0 is (-7+sqrt(49+4x^2))/2 ,(-7-sqrt(49+4x^2))/2