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)

(III) 2x² − 6x + 3 = 0

Answer 1

We know that for a quadratic equation ax² + bx + c = 0, discriminant is b² − 4ac.

(A) If b² − 4ac > 0 → two distinct real roots

(B) If b² − 4ac = 0 → two equal real roots

(C) If b² − 4ac < 0 → no real roots

(I) 2x² −3x + 5 = 0

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

a = 2, b = −3, c = 5

Discriminant = 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

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) 2x² − 6x + 3 = 0

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.

Therefore, the roots are or .