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

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(C) 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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(C) 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 new equation are $\gamma = \frac{1-\alpha}{\alpha} = \frac{1}{\alpha}-1$ and $\delta = \frac{1-\beta}{\beta} = \frac{1}{\beta}-1$. The new equation is $x^2 - (\gamma+\delta)x + \gamma\delta = 0$. Sum of roots: $\gamma+\delta = (\frac{1}{\alpha}-1) + (\frac{1}{\beta}-1) = \frac{\alpha+\beta}{\alpha\beta} - 2 = \frac{-b/a}{c/a} - 2 = -\frac{b}{c} - 2 = -\frac{b+2c}{c}$. Product of roots: $\gamma\delta = (\frac{1}{\alpha}-1)(\frac{1}{\beta}-1) = \frac{1}{\alpha\beta} - (\frac{1}{\alpha}+\frac{1}{\beta}) + 1 = \frac{1}{\alpha\beta} - \frac{\alpha+\beta}{\alpha\beta} + 1 = \frac{a}{c} - \frac{-b/a}{c/a} + 1 = \frac{a}{c} + \frac{b}{c} + 1 = \frac{a+b+c}{c}$. The equation is $x^2 - (-\frac{b+2c}{c})x + \frac{a+b+c}{c} = 0$. Multiplying by $c$,we get $cx^2 + (b+2c)x + (a+b+c) = 0$. Comparing with $px^2+qx+r=0$,we find $r = a+b+c$.

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

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