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

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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}

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