Find the discriminant of the equation $3x^{2}-2x+\frac{1}{3}=0$ and hence find the nature of its roots. Find them,if they are real.
(D) Given equation: $3x^{2}-2x+\frac{1}{3}=0$.
Comparing with $ax^{2}+bx+c=0$,we get $a=3, b=-2, c=\frac{1}{3}$.
Discriminant $D = b^{2}-4ac = (-2)^{2}-4(3)(\frac{1}{3}) = 4-4 = 0$.
Since $D=0$,the equation has two equal real roots.
The roots are given by $x = \frac{-b}{2a}$.
$x = \frac{-(-2)}{2(3)} = \frac{2}{6} = \frac{1}{3}$.
Thus,the roots are $\frac{1}{3}, \frac{1}{3}$.
Comparing with $ax^{2}+bx+c=0$,we get $a=3, b=-2, c=\frac{1}{3}$.
Discriminant $D = b^{2}-4ac = (-2)^{2}-4(3)(\frac{1}{3}) = 4-4 = 0$.
Since $D=0$,the equation has two equal real roots.
The roots are given by $x = \frac{-b}{2a}$.
$x = \frac{-(-2)}{2(3)} = \frac{2}{6} = \frac{1}{3}$.
Thus,the roots are $\frac{1}{3}, \frac{1}{3}$.
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