(1/x)+(2/3x)=9/8

Solution for (1/x)+(2/3x)=9/8 equation:

(1/x)+(2/3x)=9/8

We move all terms to the left:

(1/x)+(2/3x)-(9/8)=0


Domain of the equation: x)!=0

x!=0/1

x!=0

x∈R



Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+1/x)+(+2/3x)-(+9/8)=0

We get rid of parentheses

1/x+2/3x-9/8=0

We calculate fractions


(-81x^2)/192x^2+192x/192x^2+128x/192x^2=0

We multiply all the terms by the denominator


(-81x^2)+192x+128x=0

We add all the numbers together, and all the variables

(-81x^2)+320x=0

We get rid of parentheses

-81x^2+320x=0

a = -81; b = 320; c = 0;
Δ = b2-4ac
Δ = 3202-4·(-81)·0
Δ = 102400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{102400}=320$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(320)-320}{2*-81}=\frac{-640}{-162}
=3+77/81
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(320)+320}{2*-81}=\frac{0}{-162}
=0
$