If the roots of the equation (x - a)(x - b) = | vedclass

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(C) Given the equation $(x - a)(x - b) = c$,we can rewrite it as $(x - a)(x - b) - c = 0$. Since $\alpha$ and $\beta$ are the roots of $(x - a)(x - b)

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$a$ and $b$

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(C) Given the equation $(x - a)(x - b) = c$,we can rewrite it as $(x - a)(x - b) - c = 0$. Since $\alpha$ and $\beta$ are the roots of $(x - a)(x - b) - c = 0$,we can express the quadratic as $(x - \alpha)(x - \beta) = (x - a)(x - b) - c$. Rearranging this,we get $(x - \alpha)(x - \beta) + c = (x - a)(x - b)$. Therefore,the equation $(x - \alpha)(x - \beta) + c = 0$ is equivalent to $(x - a)(x - b) = 0$. The roots of this equation are clearly $x = a$ and $x = b$.

If the roots of the equation $(x - a)(x - b) = c$ (where $c \neq 0$) are $\alpha$ and $\beta$,then what are the roots of the equation $(x - \alpha)(x - \beta) + c = 0$?

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