If α,β are the zeroes of the polynomial f(x)=ax2+bx+c, then find 1α+1β−2αβ
A
b+2ac2a
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B
−(ba+cab)
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C
−(bc+2ca)
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D
(bc+ab)
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Solution
The correct option is A−(bc+2ca) Given, α,β are the zeroes of the polynomial f(x)=ax2+bx+c ⇒α+β=−ba ⇒αβ=ca Now, 1α+1β−2αβ=β+αβα−2αβ =−baca−2ca =−bc−2ca =−(bc+2ca) Option C is correct.