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)\(2x^2 - 6x + 3 = 0\)

(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,

\(x = {{6 ± \sqrt{12}} \over {4}} = {{6 ± 2 \sqrt{3}} \over {4}} = {{3 ± \sqrt{3}} \over {2}}\)
\(x = {{3 + \sqrt{3}} \over {2}} , {{3 - \sqrt{3}} \over {2}}\)