Q1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them.
(i)$$2x^2 - 3x + 5 = 0$$
(ii)$$3x^2 -4 \sqrt{3} x + 4 = 0$$
(iii)$$2x^2 - 6x + 3 = 0$$

Ans.(i)$$2x^2 - 3x + 5 = 0$$
Comparing this equation with general equation $$ax^2 + bx + c$$,
We get a = 2, b = ?3 and c = 5
Discriminant = $$b^2 - 4ac = (-3)^2 - 4(2)(5) = 9 – 40 = ?31$$
Discriminant is less than 0 which means equation has no real roots.

(ii)$$3x^2 -4 \sqrt{3} x + 4 = 0$$
Comparing this equation with general equation $$ax^2 + bx + c$$,
We get a = 3, b = $$-4 \sqrt{3}$$ and c = 4 Discriminant = $$b^2 - 4ac = (-4 \sqrt{3})^2 - 4(3)(4) = 48 – 48 = 0$$
Discriminant is equal to zero which means equations has equal real roots.
Applying quadratic $$x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}$$ to find roots,
=>$$x = {{4 \sqrt{3} ± \sqrt{0}} \over {6}} = {{2 \sqrt{3}} \over {3}}$$
Because, equation has two equal roots, it means x = $${{2 \sqrt{3}} \over {3}} ,{{2 \sqrt{3}} \over {3}}$$

(iii)$$2x^2 - 6x + 3 = 0$$
Comparing this equation with general equation $$ax^2 + bx + c$$,
We get a = 2, b = ?6, and c = 3
Discriminant =$$b^2 - 4ac = (-6)^2 - 4(2)(3) = 36 – 24 = 12$$
Value of discriminant is greater than zero.
Therefore, equation has distinct and real roots.
Applying quadratic $$x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}$$ to find roots,
$$x = {{6 ± \sqrt{12}} \over {4}} = {{6 ± 2 \sqrt{3}} \over {4}} = {{3 ± \sqrt{3}} \over {2}}$$ $$x = {{3 + \sqrt{3}} \over {2}} , {{3 - \sqrt{3}} \over {2}}$$