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

The inverse of z/(z^2-3z+2) is (3z+1+sqrt(z^2+6z+1))/(2z),(3z+1-sqrt(z^2+6z+1))/(2z)

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inverse of z/(z^2-3z+2)

inverse

\frac{z}{z ^{2} - 3 z + 2}

\frac{3 z + 1 + z ^{2} + 6 z + 1}{2 z}, \frac{3 z + 1 - z ^{2} + 6 z + 1}{2 z}

Solution steps

y = \frac{z}{z ^{2} - 3 z + 2}

Replace

z

with

y

z = \frac{y}{y ^{2} - 3 y + 2}

Solve for

y, z = \frac{y}{y ^{2} - 3 y + 2}

\frac{3 z + 1 + z ^{2} + 6 z + 1}{2 z}, \frac{3 z + 1 - z ^{2} + 6 z + 1}{2 z}

x^{2} - 6 x - 8

\frac{x}{2} - 4

y = cos^{2} (x)

y = - csc (\frac{8}{9}) x

f (x) = - \frac{1}{5} x - 3

What is the inverse of z/(z^2-3z+2) ?
The inverse of z/(z^2-3z+2) is (3z+1+sqrt(z^2+6z+1))/(2z),(3z+1-sqrt(z^2+6z+1))/(2z)