Find k if zeroes a, b of the polynomial 5x^2 + (2k+1)x + (k-2) are such that 2a+5b=1
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alpha and beta are the zeroes of the polynomial
now,
alpha + beta = -(2k +1)/5
=> 5alpha + 5beta = -(2k +1)--------(2)
alpha × beta = (k -2)/5------(2)
given that:-
2alpha + 5beta = 1
5beta = (1 - 2alpha)----(3) put in--(1)
we get,
5alpha + 1 - 2alpha = -(2k +1)
=> 3alpha = -2k - 2
=> alpha = (-2k - 2)/3-----(4) put in --(3)
we get,
5beta = 1 - 2(-2k - 2)/3
=> 5beta = (3 + 4k + 4)/3
=> beta = (4k + 7)/15---(5)
from---(2) ,(4) and (5)
we get,
(-2k - 2)/3 × (4k + 7)/15 = (k - 2)/5
=> (-8k² -14k - 8k - 14)/3×3 = (k -2)
=> -8k² - 22k - 14 = 9k - 18
=> -8k² - 22k - 9k - 14 + 18 = 0
=> -8k² - 31k + 4 = 0
=> 8k² + 31k - 4 = 0
=> 8k² + 32k - k - 4 = 0
=> 8k(k + 4) - 1(k + 4) = 0
=> (8k -1)(k + 4) = 0
=> k = 1/8 or k = -4
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