Discuss the nature of the roots of the following quadratic equations :

(i) $3x^2 - 4\sqrt{3}x + 4 = 0$

(ii) $x^2 - \dfrac{1}{2}x + 4 = 0$

(iii) $-2x^2 + x + 1 = 0$

(iv) $2\sqrt{3}x^2 - 5x + \sqrt{3} = 0$

(i) The given equation is $3x^2 - 4\sqrt{3}x + 4 = 0$.

Comparing it with ax² + bx + c = 0, we get
a = 3 , b = $-4\sqrt{3}$ , c = 4
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(-4\sqrt{3})^2 - 4 \times 3 \times 4 \\[0.5em] = 48 - 48 \\[0.5em] = 0$

Hence the given equation has two equal real roots.

(ii) The given equation is $x^2 - \dfrac{1}{2}x + 4 = 0$.
Comparing it with ax² + bx + c = 0, we get
a = 1 , b = $-\dfrac{1}{2}$ , c = 4
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$\Big(-\dfrac{1}{2}\Big)^2 - 4 \times 1 \times 4 \\[1em] = \dfrac{1}{4} - 16 \\[1em] = \dfrac{1 - 64}{4} \text{ Taking L.C.M. } \\[1em] = -\dfrac{63}{4} \lt 0$

Hence the given equation has no real roots.

(iii)The given equation is -2x² + x + 1 = 0.
Comparing it with ax² + bx + c = 0, we get
a = -2 , b = 1 , c = 1
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(1)^2 - 4 \times -2 \times 1 \\[0.5em] = 1 + 8 \\[0.5em] = 9 \gt 0 \\[0.5em]$

Hence, the given equation has two distinct real roots.

(iv) The given equation is $2\sqrt{3}x^2 - 5x + \sqrt{3} = 0$.
Comparing it with ax² + bx + c = 0, we get
a = $2\sqrt{3}$ , b = -5 , c = $\sqrt{3}$
∴ Discriminant = b² - 4ac
Putting values of a, b, c in formula

$(-5)^2 - 4 \times 2\sqrt{3} \times \sqrt{3} \\[0.5em] = 25 - 24 \\[0.5em] = 1 \gt 0$

Hence, the given equation has two distinct real roots.