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Find the values of k for roots are real and equal in equation:(3k +1)x2+ 2(k +1)x + k = 0 |
Answer» The given equation (3k +1)x2 + 2(k +1)x + k = 0 is in the form of ax2 + bx + c = 0 Where a = (3k +1), b = 2(k + 1), c = k For the equation to have real and equal roots, the condition is D = b2 – 4ac = 0 ⇒ (2(k + 1))2 – 4(3k +1)(k) = 0 ⇒ 4(k +1)2 – 4(3k2 + k) = 0 ⇒ (k + 1)2 – k(3k + 1) = 0 ⇒ 2k2 – k – 1 = 0 Now, solving for k by factorization we have ⇒ 2k2 – 2k + k – 1 = 0 ⇒ 2k(k – 1) + 1(k – 1) = 0 ⇒ (k – 1)(2k + 1) = 0, k = 1 and k = \(\frac{-1}{2}\), So, the value of k can either be 1 or\(\frac{-1}{2}\). |
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