If $α$ and $β (α < β)$ be two different real roots of the equation $a x^{2} + b x + c = 0,$ then

α > - \frac{b}{2 a}

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β < - \frac{b}{2 a}

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α < - \frac{b}{2 a} < β

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β < - \frac{b}{2 a} < α

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Solution

The correct option is C $α < - \frac{b}{2 a} < β$
Let $f (x) = a x^{2} + b x + c .$
Then, $f (α) = 0 = f (β) .$
Also, $f (x)$ is continuous and differentiable in $[α, β]$ as it is a polynomial function of $x$
Hence, by Rolle's theorem, there exits a $k$ in $(α, β)$ , such that
$f^{'} (k) = 0 \Rightarrow 2 a k + b = 0 \Rightarrow k = - \frac{b}{2 a}$
$∴ α < - \frac{b}{2 a} < β$

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