What is the inverse of z(z/(z-1)) ?

The inverse of z(z/(z-1)) is (z+sqrt(z^2-4z))/2 ,(z-sqrt(z^2-4z))/2

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inverse of z(z/(z-1))

inverse

z (\frac{z}{z - 1})

\frac{z + z ^{2} - 4 z}{2}, \frac{z - z ^{2} - 4 z}{2}

Solution steps

y = z \frac{z}{z - 1}

Replace

z

with

y

z = y \frac{y}{y - 1}

Solve for

y, z = y \frac{y}{y - 1}

\frac{z + z ^{2} - 4 z}{2}, \frac{z - z ^{2} - 4 z}{2}

f (x) = \frac{- x}{2} - 5

tan (\frac{31}{71})

3 7 x - 9

g (x) = 3 x

f (x) = ln (\frac{1}{3 - x})

What is the inverse of z(z/(z-1)) ?
The inverse of z(z/(z-1)) is (z+sqrt(z^2-4z))/2 ,(z-sqrt(z^2-4z))/2