If alpha and beta are the roots of the equati | vedclass

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(B) Given that $\alpha$ and $\beta$ are the roots of $a x^2+b x+c=0$. $\alpha+\beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$. Let the roots of

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$a+b+c$

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(B) Given that $\alpha$ and $\beta$ are the roots of $a x^2+b x+c=0$. $\alpha+\beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$. Let the roots of $p x^2+q x+r=0$ be $\gamma = \frac{1-\alpha}{\alpha} = \frac{1}{\alpha}-1$ and $\delta = \frac{1-\beta}{\beta} = \frac{1}{\beta}-1$. Then $\alpha = \frac{1}{1+\gamma}$ and $\beta = \frac{1}{1+\delta}$. Since $\alpha$ is a root of $a x^2+b x+c=0$,we have $a(\frac{1}{1+x})^2 + b(\frac{1}{1+x}) + c = 0$. $a + b(1+x) + c(1+x)^2 = 0$. $a + b + bx + c(1 + 2x + x^2) = 0$. $c x^2 + (b+2c)x + (a+b+c) = 0$. Comparing this with $p x^2+q x+r=0$,we get $r = a+b+c$.

If $\alpha$ and $\beta$ are the roots of the equation $a x^2+b x+c=0$ and if $p x^2+q x+r=0$ has roots $\frac{1-\alpha}{\alpha}$ and $\frac{1-\beta}{\beta}$,then $r$ is equal to

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