What is the inverse of [(z^2)/((z-1)(z-3))] ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of [(z^2)/((z-1)(z-3))]

inverse

[\frac{z ^{2}}{( z - 1 ) ( z - 3 )}]

\frac{2 z + z ( z + 3 )}{z - 1}, \frac{2 z - z ( z + 3 )}{z - 1}

Solution steps

y = \frac{z ^{2}}{( z - 1 ) ( z - 3 )}

Replace

z

with

y

z = \frac{y ^{2}}{( y - 1 ) ( y - 3 )}

Solve for

y, z = \frac{y ^{2}}{( y - 1 ) ( y - 3 )}

\frac{2 z + z ( z + 3 )}{z - 1}, \frac{2 z - z ( z + 3 )}{z - 1}

f (x) = (1, 5) (- 3, 2) (8, - 6)

f (x) = \frac{400}{x ^{2}}

f (x) = 2 x^{(2)} - 3 x - 1

f (x) = \frac{- 6 x + 2}{- 8 x + 9}

f (x) = \frac{9}{- 8 x - 9}

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