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

There are three types of roots that are possible for a quadratic equation:

For a quadratic equation: a x2 + b x + c = 0, we know that its roots is given by the formula

(i) Comparing this equation with ax2 + bx + c = 0, we obtain,

a = 2, b = −3, c = 5

Discriminant D = b2 − 4ac = (− 3)2 − 4 (2) (5) = 9 − 40 = −31

As, b2 − 4ac < 0,

Therefore, no real root is possible for the given equation.

(ii) Comparing this equation with ax2 + bx + c = 0,

we obtain,

a = 3

b = -4√3

c = 4

Discriminant = 48 − 48 = 0

As, b2 − 4ac = 0,

Therefore, real roots exist for the given equation and they are equal to each other.

And the roots will be and .

Therefore, the roots are and .

(iii) Comparing this equation with ax2+bx + c = 0,

we obtain,

a = 2, b = −6, c = 3

Discriminant = b2 − 4ac = (− 6)2 − 4 (2) (3) = 36 − 24 = 12

As, b2 − 4ac > 0,

Therefore, distinct real roots exist for this equation as follows.

So, the roots are or .

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