Quadratic equation

Author Mahaveer prasad asked on 2019-07-22 13:26:00

Q. No. 8

(2k+1) x2 +2(k+3)x+(k+5 )=0 real and equal roots find the value of K

D = b² + 4ac = 0

{2(k + 3)}² - 4(k + 5)(2k + 1) = 0

4(k² + 6k + 9) - 4(2k² + k + 10k + 5) = 0

4k² + 24k + 36 - 8k² - 44k - 20 = 0

 - 4k² - 20k + 16 = 0

k² + 5k - 4 = 0

 k = {-5 ± √(25 + 16)}/2 = {- 5 ± √41}/2