Math, asked by naiksubha84, 9 months ago

Please send fast answer

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Answered by manassunil28
0

Answer:

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

Answered by ajiteshwarpathak12
0

Answer:

Step-by-step explanation:

K= 1/3

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