3x+1/3x-4=0

Solution for 3x+1/3x-4=0 equation:

3x+1/3x-4=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We multiply all the terms by the denominator


3x*3x-4*3x+1=0

Wy multiply elements

9x^2-12x+1=0

a = 9; b = -12; c = +1;
Δ = b2-4ac
Δ = -122-4·9·1
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{3}}{2*9}=\frac{12-6\sqrt{3}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{3}}{2*9}=\frac{12+6\sqrt{3}}{18}
$