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1/3f=f
We move all terms to the left:
1/3f-(f)=0
Domain of the equation: 3f!=0We add all the numbers together, and all the variables
f!=0/3
f!=0
f∈R
-1f+1/3f=0
We multiply all the terms by the denominator
-1f*3f+1=0
Wy multiply elements
-3f^2+1=0
a = -3; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-3)·1
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{3}}{2*-3}=\frac{0-2\sqrt{3}}{-6} =-\frac{2\sqrt{3}}{-6} =-\frac{\sqrt{3}}{-3} $$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{3}}{2*-3}=\frac{0+2\sqrt{3}}{-6} =\frac{2\sqrt{3}}{-6} =\frac{\sqrt{3}}{-3} $
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