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Question

If the roots of the equation ax2+bx+c=0,aR+ are two consecutive odd positive integers, then

A
|b|4a
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B
|b|4a
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C
|b|2a
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D
|b|a
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Solution

The correct option is B |b|2a
Let α,β be the rootsα+β=bc;αβ=ca
αβ=2(ba)24(ca)=|αβ|
b24aca2=2.
b24ac=4a2
b2=4a2+4ac
As both α and β are positive, ac is positive
b24a2
|b|2a

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