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If the equation $ax^2+bx+c=0$ $(a > 0)$ has roots $\alpha$ and $\beta$ such that $ \alpha < -2$ and $\beta > 2$ then,

$\begin{array}{1 1}(A)\;b^2-4ac < 0&(B)\;c > 0\\(C)\;a+|b|+c < 0&(D)\;4a+2|b|+c > 0\end{array} $

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Since equation has two real roots $\alpha,\beta$
$\Rightarrow b^2-4ac > 0$
$\alpha < -2$ and $\beta > 2$ and $a > 0$
Product of roots < 0
$\Rightarrow \alpha\beta < 0\Rightarrow \large\frac{c}{a}$$ < 0$
$a > 0$
$\therefore C < 0$
$f(-1)=a-b+c < 0$
$f(1) =a+b+c < 0$
$\Rightarrow a+|b|+c < 0$
$f(2)=4a+2b+c < 0$
$f(-2)=4a-2b+c < 0$
$\Rightarrow 4a+2|b|+c < 0$
Hence (C) is the correct answer.
answered Apr 16, 2014 by sreemathi.v
 

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