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

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$$\Rightarrow b^{2}\geq 4a^{2}$$$$\Rightarrow |b|\geq 2a$$Maths

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