Question

If (4 - k)x² +2(k+2)x +(8k + 1) = 0 has
equal roots, then find the value of k.​

Answer 1

Given, (4 – k)x2 + (2k + 4)x + (8k + 1) = 0 It is in the form of ax2 + bx + c = 0 Where, a = 4 – k, b = 2k + 4, c = 8k + 1

Calculating the discriminant, D = b2 – 4ac = (2k + 4)2 – 4(4 – k)(8k + 1) = 4k2 + 16 + 4k – 4(32 + 4 – 8k2 – k) = 4(k2 + 4 + k – 32 – 4 + 8k2 + k) = 4(9k2 – 27k)

As the given equation is a perfect square, then D = 0 ⇒ 4(9k2 – 27k) = 0 ⇒ (9k2 – 27k) = 0 ⇒ 3k(k – 3) = 0 Thus, 3k = 0 ⇒ k = 0 Or, k-3 = 0 ⇒k = 3

Hence, the value of k should be 0 or 3 for the given to be perfect square.Read more on Sarthaks.com - https://www.sarthaks.com/647624/for-what-value-of-k-4-k-x-2-2k-4-x-8k-1-0-is-a-perfect-square