If the roots of ax^2 + bx + c = 0 are alpha, | vedclass

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(C) For the equation $ax^2 + bx + c = 0$,the discriminant is $D_1 = b^2 - 4ac$. The difference of roots is given by $(\alpha - \beta)^2 = (\alpha + \b

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$(\frac{A}{a})^2$

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(C) For the equation $ax^2 + bx + c = 0$,the discriminant is $D_1 = b^2 - 4ac$. The difference of roots is given by $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = (\frac{-b}{a})^2 - 4(\frac{c}{a}) = \frac{b^2 - 4ac}{a^2}$.For the equation $Ax^2 + Bx + C = 0$,the roots are $\alpha - k$ and $\beta - k$. The difference of these roots is $(\alpha - k) - (\beta - k) = \alpha - \beta$.Thus,the square of the difference of the roots is $(\alpha - \beta)^2 = \frac{B^2 - 4AC}{A^2}$.Equating the two expressions for $(\alpha - \beta)^2$,we get $\frac{b^2 - 4ac}{a^2} = \frac{B^2 - 4AC}{A^2}$.Rearranging the terms,we find $\frac{B^2 - 4AC}{b^2 - 4ac} = \frac{A^2}{a^2} = (\frac{A}{a})^2$.

If the roots of $ax^2 + bx + c = 0$ are $\alpha, \beta$ and the roots of $Ax^2 + Bx + C = 0$ are $\alpha - k, \beta - k$,then $\frac{B^2 - 4AC}{b^2 - 4ac}$ is equal to

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