1+3x+1/3x=96

Solution for 1+3x+1/3x=96 equation:

1+3x+1/3x=96

We move all terms to the left:

1+3x+1/3x-(96)=0


Domain of the equation: 3x!=0

x!=0/3

x!=0

x∈R


We add all the numbers together, and all the variables

3x+1/3x-95=0

We multiply all the terms by the denominator


3x*3x-95*3x+1=0

Wy multiply elements

9x^2-285x+1=0

a = 9; b = -285; c = +1;
Δ = b2-4ac
Δ = -2852-4·9·1
Δ = 81189
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:

$\sqrt{\Delta}=\sqrt{81189}=\sqrt{9*9021}=\sqrt{9}*\sqrt{9021}=3\sqrt{9021}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-285)-3\sqrt{9021}}{2*9}=\frac{285-3\sqrt{9021}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-285)+3\sqrt{9021}}{2*9}=\frac{285+3\sqrt{9021}}{18}
$