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Question:

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: (i) 2x² - 3x + 5 = 0 (ii) 3x² - 4√3x + 4 = 0 (iii) 2x² - 6x + 3 = 0

Solution:

i) 2x² - 3x + 5 = 0
a = 2, b = -3, c = 5
Discriminant, D = b² - 4ac = 9 - 40 = -31
Since D < 0, the roots are imaginary for the given equation.
ii) 3x² - 4√3x + 4 = 0
a = 3, b = -4√3, c = 4
Discriminant, D = b² - 4ac = 48 - 48 = 0
Since D = 0, the roots are real and equal for the given equation.
3x² - 4√3x + 4 = 0
a = 3, b = -4√3, c = 4
x = [-b ± √(b² - 4ac)] / 2a
x = 4√3 / 6 = 2√3 / 3
iii) 2x² - 6x + 3 = 0
a = 2, b = -6, c = 3
Discriminant, D = b² - 4ac = 36 - 24 = 12
Since D > 0, roots are real but not equal.
x = [-b ± √(b² - 4ac)] / 2a
x = 6 ± √12 / 4
x = 6 ± 2√3 / 4
x = 3 ± √3 / 2