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(A) Given the equation $ax^2+bx+c=0$ has roots $\alpha$ and $\beta$,we have $\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.Consider the

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$\frac{\alpha}{\alpha-1}, \frac{\beta}{\beta-1}$

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(A) Given the equation $ax^2+bx+c=0$ has roots $\alpha$ and $\beta$,we have $\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.Consider the equation $ax^2-bx(x-1)+c(x-1)^2=0$.Divide the entire equation by $(x-1)^2$ (assuming $x 
eq 1$):$a(\frac{x}{x-1})^2 - b(\frac{x}{x-1}) + c = 0$.Let $y = \frac{x}{x-1}$.Then the equation becomes $ay^2 - by + c = 0$.This is not quite the original form. Let us rewrite the original equation as $a(\frac{x}{x-1})^2 + b(\frac{x}{x-1}) + c = 0$ if we had $ax^2+bx(x-1)+c(x-1)^2=0$.For $ax^2-bx(x-1)+c(x-1)^2=0$,dividing by $(x-1)^2$ gives $a(\frac{x}{x-1})^2 - b(\frac{x}{x-1}) + c = 0$.Let $Y = \frac{x}{x-1}$. Then $aY^2 - bY + c = 0$.Since $a(-Y)^2 + b(-Y) + c = 0$ corresponds to roots $\alpha, \beta$,then $Y = -\alpha$ and $Y = -\beta$ are roots of $aY^2+bY+c=0$.Actually,the roots of $aY^2-bY+c=0$ are $-\alpha$ and $-\beta$ is incorrect. The roots of $aY^2-bY+c=0$ are $\alpha$ and $\beta$ if we compare $aY^2-bY+c=0$ with $aY^2+bY+c=0$ by replacing $Y$ with $-Y$.Thus,$\frac{x}{x-1} = \alpha \implies x = \alpha x - \alpha \implies x(1-\alpha) = -\alpha \implies x = \frac{\alpha}{\alpha-1}$.Similarly,$x = \frac{\beta}{\beta-1}$.Therefore,the roots are $\frac{\alpha}{\alpha-1}$ and $\frac{\beta}{\beta-1}$.

If $\alpha$ and $\beta$ are the roots of $ax^2+bx+c=0$,then the roots of $ax^2-bx(x-1)+c(x-1)^2=0$ are

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