3x+(1/3x)+25=100

Solution for 3x+(1/3x)+25=100 equation:

3x+(1/3x)+25=100

We move all terms to the left:

3x+(1/3x)+25-(100)=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)+25-100=0

We add all the numbers together, and all the variables

3x+(+1/3x)-75=0

We get rid of parentheses

3x+1/3x-75=0

We multiply all the terms by the denominator


3x*3x-75*3x+1=0

Wy multiply elements

9x^2-225x+1=0

a = 9; b = -225; c = +1;
Δ = b2-4ac
Δ = -2252-4·9·1
Δ = 50589
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:

$\sqrt{\Delta}=\sqrt{50589}=\sqrt{9*5621}=\sqrt{9}*\sqrt{5621}=3\sqrt{5621}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-225)-3\sqrt{5621}}{2*9}=\frac{225-3\sqrt{5621}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-225)+3\sqrt{5621}}{2*9}=\frac{225+3\sqrt{5621}}{18}
$