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Find the values of k for roots are real and equal in equation:(k + 1)x2 – 2(3k + 1)x + 8k + 1 = 0 |
Answer» The given equation (k +1)x2 – 2(3k +1)x + 8k + 1 = 0 is in the form of ax2 + bx + c = 0 Where a = (k +1), b = -2(3k + 1), c = 8k + 1 For the equation to have real and equal roots, the condition is D = b2 – 4ac = 0 ⇒ (-2(3k + 1))2 – 4(k +1)(8k + 1) = 0 ⇒ 4(3k +1)2 – 4(k + 1)(8k + 1) = 0 ⇒ (3k + 1)2 – (k + 1)(8k + 1) = 0 ⇒ 9k2 + 6k + 1 – (8k2 + 9k + 1) = 0 ⇒ 9k2 + 6k + 1 – 8k2 – 9k – 1 = 0 ⇒ k2 – 3k = 0 ⇒ k(k – 3) = 0 Either k = 0 Or, k – 3 = 0 ⇒ k = 3, So, the value of k can either be 0 or 3 |
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