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16+1/3x=3x
We move all terms to the left:
16+1/3x-(3x)=0
Domain of the equation: 3x!=0We add all the numbers together, and all the variables
x!=0/3
x!=0
x∈R
-3x+1/3x+16=0
We multiply all the terms by the denominator
-3x*3x+16*3x+1=0
Wy multiply elements
-9x^2+48x+1=0
a = -9; b = 48; c = +1;
Δ = b2-4ac
Δ = 482-4·(-9)·1
Δ = 2340
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2340}=\sqrt{36*65}=\sqrt{36}*\sqrt{65}=6\sqrt{65}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-6\sqrt{65}}{2*-9}=\frac{-48-6\sqrt{65}}{-18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+6\sqrt{65}}{2*-9}=\frac{-48+6\sqrt{65}}{-18} $
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