4/3z+8/z=1

Solution for 4/3z+8/z=1 equation:

4/3z+8/z=1

We move all terms to the left:

4/3z+8/z-(1)=0


Domain of the equation: 3z!=0

z!=0/3

z!=0

z∈R



Domain of the equation: z!=0

z∈R


We calculate fractions


4z/3z^2+24z/3z^2-1=0

We multiply all the terms by the denominator


4z+24z-1*3z^2=0

We add all the numbers together, and all the variables

28z-1*3z^2=0

Wy multiply elements

-3z^2+28z=0

a = -3; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·(-3)·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28}{2*-3}=\frac{-56}{-6}
=9+1/3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+28}{2*-3}=\frac{0}{-6}
=0
$