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(B) Given that $\alpha$ and $\beta$ are the roots of $ax^2+bx+c=0$. Then,$\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$. The roots of t

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$a+b+c$

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(B) Given that $\alpha$ and $\beta$ are the roots of $ax^2+bx+c=0$. Then,$\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$. The roots of the equation $px^2+qx+r=0$ are $x_1 = \frac{1-\alpha}{\alpha} = \frac{1}{\alpha}-1$ and $x_2 = \frac{1-\beta}{\beta} = \frac{1}{\beta}-1$. The equation with roots $x_1$ and $x_2$ is given by $x^2 - (x_1+x_2)x + x_1x_2 = 0$. Sum of roots: $x_1+x_2 = (\frac{1}{\alpha}-1) + (\frac{1}{\beta}-1) = \frac{\alpha+\beta}{\alpha\beta} - 2 = \frac{-b/a}{c/a} - 2 = -\frac{b}{c} - 2 = -\frac{b+2c}{c}$. Product of roots: $x_1x_2 = (\frac{1}{\alpha}-1)(\frac{1}{\beta}-1) = \frac{1}{\alpha\beta} - (\frac{1}{\alpha}+\frac{1}{\beta}) + 1 = \frac{1}{\alpha\beta} - \frac{\alpha+\beta}{\alpha\beta} + 1 = \frac{a}{c} - (\frac{-b/a}{c/a}) + 1 = \frac{a}{c} + \frac{b}{c} + 1 = \frac{a+b+c}{c}$. Substituting these into the quadratic equation: $x^2 - (-\frac{b+2c}{c})x + \frac{a+b+c}{c} = 0$. Multiplying by $c$: $cx^2 + (b+2c)x + (a+b+c) = 0$. Comparing this with $px^2+qx+r=0$,we get $r = a+b+c$.

If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$ and if $px^2+qx+r=0$ has roots $\frac{1-\alpha}{\alpha}$ and $\frac{1-\beta}{\beta}$,then $r$ is equal to

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