Question

Find the values of ‘k’ for which the equation x2 + 2(k - 1)x + (k + 5) = 0 has real and
equal roots.

Answer 1

Step-by-step explanation:

Where a = (2k +1), b = 2(k + 3), c = (k + 5) For the equation to have real and equal roots, the condition is D = b2 – 4ac = 0 ⇒ (2(k + 3))2 – 4(2k +1)(k + 5) = 0 ⇒ 4(k +3)2 – 4(2k2 + 11k + 5) = 0 ⇒ (k + 3)2 – (2k2 + 11k + 5) = 0 [dividing by 4 both sides] ⇒ k2 + 5k – 4 = 0 Now, solving for k by completing the square we have ⇒ k2 + 2 x (5252) x k + (5252)2 = 4 + (5252)2 ⇒ (k + 5252)2 = 4 + 254254 = 414−−√414 ⇒ k + (5252) = ± 41√2412 ⇒ k = (41√–5)2(41–5)2 or –(41√+5)2(41+5)2 So, the value of k can either be (41√–5)2(41–5)2 or –(41√+5)2(41+5)2.Read more on Sarthaks.com - https://www.sarthaks.com/647437/find-the-values-of-k-for-roots-are-real-and-equal-in-equation-2k-1-x-2-2-k-3-x-k-5-0