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

The product of the roots for the equation px^2 + qx + r = 0 is given by r/p. We are given the roots as (1 - alpha)/alpha and (1 - beta)/beta, so their product is ((1 - alpha)(1 - beta))/(alpha * beta). Expanding the numerator gives 1 - (alpha + beta) + (alpha * beta). From the first equation ax^2 + bx + c = 0, the sum of the roots alpha + beta is -b/a and the product is c/a. Substituting these yields (1 - (-b/a) + c/a) / (c/a), which simplifies to ((a + b + c)/a) * (a/c) = (a + b + c)/c. Setting this equal to r/p, we get r/p = (a + b + c)/c. Assuming the standard proportionality where the transformed equation corresponds to p = c, we find r = a + b + c.

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