Find the nature of the roots of the following quadratic equation. If real roots exist,find them:
$3x^2 - 4\sqrt{3}x + 4 = 0$

(N/A) We know that for a quadratic equation $ax^2 + bx + c = 0$,the discriminant is $D = b^2 - 4ac$.
$(A)$ If $D > 0$,there are two distinct real roots.
$(B)$ If $D = 0$,there are two equal real roots.
$(C)$ If $D < 0$,there are no real roots.
Given equation: $3x^2 - 4\sqrt{3}x + 4 = 0$.
Comparing this with $ax^2 + bx + c = 0$,we get $a = 3$,$b = -4\sqrt{3}$,and $c = 4$.
Discriminant $D = b^2 - 4ac = (-4\sqrt{3})^2 - 4(3)(4)$.
$D = 48 - 48 = 0$.
Since $D = 0$,the equation has two equal real roots.
The roots are given by $x = \frac{-b}{2a}$.
$x = \frac{-(-4\sqrt{3})}{2(3)} = \frac{4\sqrt{3}}{6} = \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}$.
Thus,the roots are $\frac{2}{\sqrt{3}}$ and $\frac{2}{\sqrt{3}}$.