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If $\alpha,\beta$ are the roots of $ax^2+bx+c=0$ then equation whose roots are $2+\alpha,2+\beta$ are

$\begin{array}{1 1}(A)\;ax^2+x(4a-b)+4a-2b+c=0\\(B)\;ax^2+x(4a-b)+4a+2b+c=0\\(C)\;ax^2+x(b-4a)+4a+2b+c=0\\(D)\;ax^2+x(b-4a)+4a-2b+c=0\end{array} $

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$(2+\alpha)+(2+\beta)=4+(-\large\frac{b}{a})=\frac{4a-b}{a}$
$(2+\alpha)(2+\beta)=4+2(\alpha+\beta)+\alpha\beta$
$\Rightarrow 4-\large\frac{2b}{a}+\frac{c}{a}$
$\Rightarrow \large\frac{4a-2b+c}{a}$
Equation:$x^2+(\large\frac{4a-b}{a})$$x+(\large\frac{4a-2b+c}{a})=0$
$\equiv ax^2+x(4a-b)+4a-2b+c=0$
Hence (A) is the correct answer.
answered Apr 14, 2014 by sreemathi.v
 

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