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Q2. Find the roots of the following Quadratic Equations by applying quadratic formula.
(i) \(2x^2 - 7x + 3 = 0\)
(ii) \(2x^2 + x - 4 = 0\)
(iii) \(4x^2 + 4\sqrt{3}x + 3 = 0\)
(iv) \(2x^2 + x + 4 = 0\)

Answer :

(i) \(2x^2 - 7x + 3 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{7 ± \sqrt{(-7)^2 - 4(2)(3)}} \over {(2)(2)}}\)
=>\(x = {{7 ± \sqrt{49 - 24}} \over {4}}\)
=>\(x = {{7 ± 5} \over {4}}\)
=>\(x = {{7 + 5} \over {4}} , {{7 - 5} \over {4}}\)
=>\(x = 3 , {{1} \over {2}}\)

(ii) \(2x^2 + x - 4 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{-1 ± \sqrt{(1)^2 - 4(2)(-4)}} \over {(2)(2)}}\)
=>\(x = {{-1 ± \sqrt{33}} \over {4}}\)
=>\(x = {{-1 + \sqrt{33}} \over {4}} , {{-1 - \sqrt{33}} \over {4}}\)

(iii) \(4x^2 + 4\sqrt{3}x + 3 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{-4\sqrt{3} ± \sqrt{(4\sqrt{3})^2 - 4(4)(3)}} \over {(2)(4)}}\)
=>\(x = {{-4\sqrt{3} ± \sqrt{0}} \over {8}}\)
=>\(x = {{-\sqrt{3}} \over {2}} , {{-\sqrt{3}} \over {2}}\)

(iv) \(2x^2 + x + 4 = 0\)
The general form of equation is \(ax^2 + bx + c = 0\)
Quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)
=>\(x = {{-1 ± \sqrt{(1)^2 - 4(2)(-4)}} \over {(2)(2)}}\)
=>\(x = {{-1 ± \sqrt{-31}} \over {4}}\)
since square root of negative number does not exist.
Therefore, there is no solution.