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x+(1/3x)=80
We move all terms to the left:
x+(1/3x)-(80)=0
Domain of the equation: 3x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
x+(+1/3x)-80=0
We get rid of parentheses
x+1/3x-80=0
We multiply all the terms by the denominator
x*3x-80*3x+1=0
Wy multiply elements
3x^2-240x+1=0
a = 3; b = -240; c = +1;
Δ = b2-4ac
Δ = -2402-4·3·1
Δ = 57588
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{57588}=\sqrt{4*14397}=\sqrt{4}*\sqrt{14397}=2\sqrt{14397}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-240)-2\sqrt{14397}}{2*3}=\frac{240-2\sqrt{14397}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-240)+2\sqrt{14397}}{2*3}=\frac{240+2\sqrt{14397}}{6} $
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