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If $x_1$ and $x_2$ are roots of the equation $x^2+x+c=0\quad c\neq 0$ and $\large\frac{x_1^3}{2+x_2}+\frac{x_2^3}{2+x_1}=\frac{-1}{2}$ the value of -4c is

$\begin{array}{1 1}(A)\;1&(B)\;3\\(C)\;5&(D)\;7\end{array} $

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$\large\frac{x_1^3}{2+x_2}+\frac{x_2^3}{2+x_1}=\frac{-1}{2}$
$\Rightarrow 4(x_1^3+x_2^3)+2(x_1^4+x_2^4)+4+2(x_1+x_2)+x_1x_2=0$
But $x_1+x_2=-1$ and $x_1x_2=c$
$\Rightarrow 4c^2+5c=0$ as $c\neq 0$
$\Rightarrow -4c=5$
Hence (C) is the correct answer.
answered Apr 16, 2014 by sreemathi.v
 

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