. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them;3x^2 - 4√3x + 4 = 0

Dear Student

Comparing the equation with ax^2 + bx + c = 0,
we get a = 3, b = -4√3 and c = 4
We know, Discriminant = b^2 - 4ac = (-4√3)^2 - 4(3)(4)
= 48 - 48
= 0
As b^2 - 4ac = 0,
Real roots exist for the given equation and they are equal to each other.
Hence the roots will be -b/2a and -b/2a.
-b/2a = -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3
Therefore, the roots are 2/√3 and 2/√3.

Thanks