If the roots of the equation $a x^{2} + b x + c = 0$ are real and distinct, then

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both roots are greater than $\frac{- b}{2 a}$

none of these

both roots are less than $\frac{- b}{2 a}$

one of the roots exceeds $\frac{- b}{2 a}$

answer is C.

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Detailed Solution

The roots of the given equation are

$α = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} and β = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}$

Since $α, β$ are real and distinct, therefore $b^{2} - 4 a c > 0$

It is evident from Fig 1 that $β_{1} < - \frac{b}{2 a} < α$

So, one root is less than - $\frac{b}{2 a}$ and other exceeds - $\frac{b}{2 a}$

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