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1/3x^2-36=0
Domain of the equation: 3x^2!=0We multiply all the terms by the denominator
x^2!=0/3
x^2!=√0
x!=0
x∈R
-36*3x^2+1=0
Wy multiply elements
-108x^2+1=0
a = -108; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-108)·1
Δ = 432
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{432}=\sqrt{144*3}=\sqrt{144}*\sqrt{3}=12\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{3}}{2*-108}=\frac{0-12\sqrt{3}}{-216} =-\frac{12\sqrt{3}}{-216} =-\frac{\sqrt{3}}{-18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{3}}{2*-108}=\frac{0+12\sqrt{3}}{-216} =\frac{12\sqrt{3}}{-216} =\frac{\sqrt{3}}{-18} $
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