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Question

If the roots of the equation $$a\mathcal{x}^{2}+bx+c =0, a \in R^{+}$$ 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 $$|b|\geq 2a$$
Let $$\alpha ,\beta$$ be the roots$$\Rightarrow \alpha +\beta=\dfrac{-b}{c}  ; \alpha \beta=\dfrac{c}{a}$$
$$\alpha -\beta=2  \Rightarrow  \sqrt{\left(\dfrac{b}{a}\right)^{^{2}}-4\left(\dfrac{c}{a}\right)}=|\alpha -\beta|$$
$$\sqrt{\dfrac{b^{2}-4ac}{a^{2}}}=2.$$
$$b^{2}-4ac=4a^{2}$$
$$b^{2}=4a^{2}+4ac$$
As both $$\alpha$$  and $$\beta $$ are positive, $$ac$$ is positive
$$\Rightarrow b^{2}\geq 4a^{2}$$
$$\Rightarrow |b|\geq 2a$$

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