If one of the zeroes of the quadratic polynomial (2k – 1)x² + 3kx² + 1 is - 5, then the value of k is
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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 .
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