If the roots of the equation frac{alpha}{x - | vedclass

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(A) Given equation: $\frac{\alpha}{x - \alpha} + \frac{\beta}{x - \beta} = 1$Multiplying by $(x - \alpha)(x - \beta)$,we get:$\alpha(x - \beta) + \bet

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(A) Given equation: $\frac{\alpha}{x - \alpha} + \frac{\beta}{x - \beta} = 1$Multiplying by $(x - \alpha)(x - \beta)$,we get:$\alpha(x - \beta) + \beta(x - \alpha) = (x - \alpha)(x - \beta)$$\alpha x - \alpha \beta + \beta x - \alpha \beta = x^2 - \alpha x - \beta x + \alpha \beta$$(\alpha + \beta)x - 2\alpha \beta = x^2 - (\alpha + \beta)x + \alpha \beta$Rearranging terms to form a quadratic equation:$x^2 - 2(\alpha + \beta)x + 3\alpha \beta = 0$Let the roots be $k$ and $-k$. For a quadratic equation $ax^2 + bx + c = 0$,the sum of roots is $-b/a$.Sum of roots: $k + (-k) = 0$From the equation,the sum of roots is $2(\alpha + \beta)$.Therefore,$2(\alpha + \beta) = 0$,which implies $\alpha + \beta = 0$.

If the roots of the equation $\frac{\alpha}{x - \alpha} + \frac{\beta}{x - \beta} = 1$ are equal in magnitude but opposite in sign,then $\alpha + \beta = $

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