Q. 84.3( 15 Votes )

# Find the values of k for which the quadratic equation (3k + 1) x2 + 2(k + 1)x + 1 = 0 has real and equal roots.

Answer :

Given equation is (3k + 1) x2 + 2(k + 1)x + 1 = 0

Comparing with standard quadratic equation ax2 + bx + c = 0

a = (3k + 1) b = 2(k + 1) c = 1

Given that the roots of equation are real and equal

Thus D = 0

Discriminant D = b2 – 4ac = 0

(2k + 2)2 – 4.(3k + 1).1 = 0 using (a + b)2 = a2 + 2ab + b2

4k2 + 8k + 4 – 12k – 4 = 0

4k2 – 4k = 0

4k(k – 1) = 0

k = 0 (k – 1) = 0

k = 0 k = 1

The values of k are 0, 1 for which roots of the quadratic equation are real and equal.

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