Q1(ii):

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

Given: A quadratic equation $3x^2 - 4\sqrt{3}x + 4 = 0$.

To Find: The nature of the roots and the roots themselves if they exist.

Step 1: Identify the coefficients of the quadratic equation.
The standard form of a quadratic equation is $ax^2 + bx + c = 0$. Comparing the given equation $3x^2 - 4\sqrt{3}x + 4 = 0$ with the standard form, we identify:
$a = 3$
$b = -4\sqrt{3}$
$c = 4$

Step 2: Determine the nature of the roots using the Discriminant ($D$).
The discriminant is given by the formula $D = b^2 - 4ac$.
Substituting the values identified in Step 1:
$D = (-4\sqrt{3})^2 - 4(3)(4)$
$D = (16 \times 3) - 48$
$D = 48 - 48$
$D = 0$
[Since $D = 0$, the quadratic equation has two equal real roots.]

Step 3: Calculate the roots using the Quadratic Formula.
The quadratic formula is given by $x = \frac{-b \pm \sqrt{D}}{2a}$.
Since $D = 0$, the formula simplifies to $x = \frac{-b}{2a}$.
Substituting the values:
$x = \frac{-(-4\sqrt{3})}{2(3)}$
$x = \frac{4\sqrt{3}}{6}$

Step 4: Simplify the expression.
$x = \frac{4\sqrt{3}}{6}$
Dividing both the numerator and the denominator by their greatest common divisor, which is $2$:
$x = \frac{2\sqrt{3}}{3}$
Since the roots are equal, both roots are $\frac{2\sqrt{3}}{3}$.
[Note: $\frac{2\sqrt{3}}{3}$ can also be written as $\frac{2}{\sqrt{3}}$ by rationalizing the denominator.]

Final Answer: The roots are real and equal, and the roots are $\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3}$.