Question

(2k +1)x²+2(k+3)x +(k+5)
find the values for k for which the roots are real and equal.
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Answer 1

Step-by-step explanation:

which the roots are real and equal.

Answer 2

Answer:

(2k + 1)x² + 2(k + 3)x + (k + 5) = 0

has real and equal roots only when discriminant , 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

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.

Step-by-step explanation:

Hope this helps you :-)