Question

(2k +1) x² + 2 (k + 3) x + (k+5)=0
solve for k​

Answer 1

Answer:

Step-by-step explanation:

(2k + 1)x² + 2(k + 3)x + (k + 5) = 0 has real and equal roots only when discriminant ,D = b² + 4ac = 0

or, {2(k + 3)}² - 4(k + 5)(2k + 1) = 0

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

or, 4k² + 24k + 36 - 8k² - 44k - 20 = 0

or, - 4k² - 20k + 16 = 0

or, k² + 5k - 4 = 0

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

hence, values of k = (-5 ± √41}/2 in which the equation (2k + 1)x² + 2(k + 3)x + (k + 5) = 0 has real and equal roots .