1.

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»

(3k + 1) x2 + 2(k + 1) x + 1 = 0

Compare given equation with the general form of quadratic equation, which is ax2 + bx + c = 0

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

Find discriminant:

D = b2 – 4 ac

= (2(k + 1))2 – 4(3k + 1) x 1

= 4k2 + 4 + 8k – 12k – 4

= 4k (k – 1)

Since roots are real and equal (given)

Put D = 0

4k (k – 1) = 0

Either, k = 0 or k – 1 = 0

k = 0, k = 1



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