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$\alpha,\beta$ are roots of equation $ax^2+bx+c=0$.Equation whose roots are $\large\frac{1-\alpha}{\alpha}$ and $\large\frac{1-\beta}{\beta}$ are

$\begin{array}{1 1}(A)\;cx^2+(b+2c)x+(a+b+c)=0\\(B)\;bx^2+(a+2b)x+(a+b+c)=0\\(C)\;ax^2+(b+2c)x+(a+b+c)=0\\(D)\;\text{None of these}\end{array} $

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$\alpha+\beta=-\large\frac{b}{a}$
$\alpha\beta=\large\frac{c}{a}$
$\large\frac{1-\alpha}{\alpha}+\frac{1-\beta}{\beta}=\frac{\beta-\alpha\beta+\alpha-\alpha\beta}{\alpha\beta}$
$\Rightarrow \large\frac{-\Large\frac{b}{a}-\frac{2c}{a}}{\Large\frac{c}{a}}=-\frac{(b+2c)}{c}$
$\large\frac{1-\alpha}{\alpha}\frac{1-\beta}{\beta}=\frac{1-\alpha-\beta+\alpha\beta}{\alpha\beta}$
$\Rightarrow \large\frac{1+\Large\frac{b}{a}+\frac{c}{a}}{\Large\frac{c}{a}}$
$\Rightarrow \large\frac{a+b+c}{c}$
Equation:$x^2+(\large\frac{b+2c}{c})$$x+\large\frac{a+b+c}{c}=0$
$\equiv cx^2+(b+2c)x+(a+b+c)=0$
Hence (A) is the correct answer.
answered Apr 14, 2014 by sreemathi.v
 

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