The sum of the roots of the equation frac{1}{ | vedclass

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(D) Given equation: $\frac{1}{x+a} + \frac{1}{x+b} = \frac{1}{c}$Taking the common denominator on the left side:$\frac{(x+b) + (x+a)}{(x+a)(x+b)} = \f

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$-\frac{1}{2}\left(a^{2}+b^{2}\right)$

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(D) Given equation: $\frac{1}{x+a} + \frac{1}{x+b} = \frac{1}{c}$Taking the common denominator on the left side:$\frac{(x+b) + (x+a)}{(x+a)(x+b)} = \frac{1}{c}$$\frac{2x + a + b}{x^2 + x(a+b) + ab} = \frac{1}{c}$Cross-multiplying:$c(2x + a + b) = x^2 + x(a+b) + ab$$2cx + c(a+b) = x^2 + x(a+b) + ab$Rearranging into standard quadratic form $Ax^2 + Bx + C = 0$:$x^2 + x(a+b-2c) + (ab - c(a+b)) = 0$For a quadratic equation $Ax^2 + Bx + C = 0$,the sum of roots is $-B/A$. Given the sum of roots is $0$:$-(a+b-2c) = 0 \Rightarrow a+b = 2c$The product of the roots is $C/A = ab - c(a+b)$.Substituting $c = \frac{a+b}{2}$:Product $= ab - \left(\frac{a+b}{2}\right)(a+b) = ab - \frac{(a+b)^2}{2}$$= \frac{2ab - (a^2 + 2ab + b^2)}{2} = \frac{2ab - a^2 - 2ab - b^2}{2} = \frac{-(a^2 + b^2)}{2} = -\frac{1}{2}(a^2 + b^2)$

The sum of the roots of the equation $\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}$ is zero.

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