What is the inverse of 1/((z-1/2)(z-1/4)) ?

The inverse of 1/((z-1/2)(z-1/4)) is (3z+sqrt(z(z+64)))/(8z),(3z-sqrt(z(z+64)))/(8z)

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inverse of 1/((z-1/2)(z-1/4))

inverse

\frac{1}{( z - \frac{1}{2} ) ( z - \frac{1}{4} )}

\frac{3 z + z ( z + 64 )}{8 z}, \frac{3 z - z ( z + 64 )}{8 z}

Solution steps

y = \frac{1}{( z - \frac{1}{2} ) ( z - \frac{1}{4} )}

Replace

z

with

y

z = \frac{1}{( y - \frac{1}{2} ) ( y - \frac{1}{4} )}

Solve for

y, z = \frac{1}{( y - \frac{1}{2} ) ( y - \frac{1}{4} )}

\frac{3 z + z ( z + 64 )}{8 z}, \frac{3 z - z ( z + 64 )}{8 z}

f (x) = \frac{x + 2}{x + 1}

ln (2 + e^{x - 3})

3 - x + 2

2 x^{2} - x

f (x) = \frac{( 2 x + 5 )}{3}

What is the inverse of 1/((z-1/2)(z-1/4)) ?
The inverse of 1/((z-1/2)(z-1/4)) is (3z+sqrt(z(z+64)))/(8z),(3z-sqrt(z(z+64)))/(8z)