

InterviewSolution
Saved Bookmarks
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 |
|