

InterviewSolution
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Find the values of k for which the quadratic equation (3k + 1)x2 + 2(k + 1)x + 1 = 0 has equal roots. Also, find the roots. |
Answer» The given equation (3k +1)x2 + 2(k +1)x + 1 = 0 is in the form of ax2 + bx + c = 0 Where a = (3k +1), b = 2(k + 1), c = 1 For the equation to have real and equal roots, the condition is D = b2 – 4ac = 0 ⇒ (2(k + 1))2 – 4(3k +1)(1) = 0 ⇒ (k +1)2 – (3k + 1) = 0 [After dividing by 4 both sides] ⇒ k2 + 2k + 1 – 3k – 1 = 0 ⇒ k2 – k = 0 ⇒ k(k – 1) = 0 Either k = 0 Or, k – 3 = 0 ⇒ k = 1, So, the value of k can either be 0 or 1 Now, using k = 0 in the given quadratic equation we get (3(0) + 1)x2 + 2(0 + 1)x + 1 = 0 x2 + 2x + 1 = 0 ⇒ (x + 1)2 = 0 Thus, x = -1 is the root of the given quadratic equation. Next, on using k = 1 in the given quadratic equation we get (3(1) + 1)x2 + 2(1 + 1)x + 1 = 0 4x2 + 4x + 1 = 0 ⇒ (2x + 1)2 = 0 Thus, 2x = -1 ⇒ x = -1/2 is the root of the given quadratic equation. |
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