(3x+1/3x)=5

Solution for (3x+1/3x)=5 equation:

(3x+1/3x)=5

We move all terms to the left:

(3x+1/3x)-(5)=0


Domain of the equation: 3x)!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

(+3x+1/3x)-5=0

We get rid of parentheses

3x+1/3x-5=0

We multiply all the terms by the denominator


3x*3x-5*3x+1=0

Wy multiply elements

9x^2-15x+1=0

a = 9; b = -15; c = +1;
Δ = b2-4ac
Δ = -152-4·9·1
Δ = 189
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{189}=\sqrt{9*21}=\sqrt{9}*\sqrt{21}=3\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{21}}{2*9}=\frac{15-3\sqrt{21}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{21}}{2*9}=\frac{15+3\sqrt{21}}{18}
$