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

(A) 0

(B) 1

(C) 2

(D) None of these

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Best answer

SOLUTION —

Given equation $\frac{\alpha}{x-\alpha}+\frac{\beta}{x-\beta}=1$ can be rewritten as

$x^2-2(\alpha+\beta) x+3 \alpha \beta=0$

Let roots be $\alpha^{\prime}$ and $-\alpha^{\prime}$.

$\therefore \quad \alpha^{\prime}+\left(-\alpha^{\prime}\right)=2(\alpha+\beta) \Rightarrow \alpha+\beta=0$

So, The correct option will be (A). 

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