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3^x+1-1/3^x-2=0
Domain of the equation: 3^x!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
3^x-1/3^x-1=0
We multiply all the terms by the denominator
3^x*3^x-1*3^x-1=0
Wy multiply elements
9x^2-3x-1=0
a = 9; b = -3; c = -1;
Δ = b2-4ac
Δ = -32-4·9·(-1)
Δ = 45
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{5}}{2*9}=\frac{3-3\sqrt{5}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{5}}{2*9}=\frac{3+3\sqrt{5}}{18} $
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