(1/x)+(1/3x)=17/4

Solution for (1/x)+(1/3x)=17/4 equation:

(1/x)+(1/3x)=17/4

We move all terms to the left:

(1/x)+(1/3x)-(17/4)=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)+(+1/3x)-(+17/4)=0

We get rid of parentheses

1/x+1/3x-17/4=0

We calculate fractions


(-153x^2)/48x^2+48x/48x^2+16x/48x^2=0

We multiply all the terms by the denominator


(-153x^2)+48x+16x=0

We add all the numbers together, and all the variables

(-153x^2)+64x=0

We get rid of parentheses

-153x^2+64x=0

a = -153; b = 64; c = 0;
Δ = b2-4ac
Δ = 642-4·(-153)·0
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-64}{2*-153}=\frac{-128}{-306}
=64/153
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+64}{2*-153}=\frac{0}{-306}
=0
$