1/2z+4=9/z

Solution for 1/2z+4=9/z equation:

1/2z+4=9/z

We move all terms to the left:

1/2z+4-(9/z)=0


Domain of the equation: 2z!=0

z!=0/2

z!=0

z∈R



Domain of the equation: z)!=0

z!=0/1

z!=0

z∈R


We add all the numbers together, and all the variables

1/2z-(+9/z)+4=0

We get rid of parentheses

1/2z-9/z+4=0

We calculate fractions


z/2z^2+(-18z)/2z^2+4=0

We multiply all the terms by the denominator


z+(-18z)+4*2z^2=0

Wy multiply elements

8z^2+z+(-18z)=0

We get rid of parentheses

8z^2+z-18z=0

We add all the numbers together, and all the variables

8z^2-17z=0

a = 8; b = -17; c = 0;
Δ = b2-4ac
Δ = -172-4·8·0
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-17}{2*8}=\frac{0}{16}
=0
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+17}{2*8}=\frac{34}{16}
=2+1/8
$