For the equation frac{1}{x + a} - frac{1}{x + | vedclass

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(D) Given the equation: $\frac{1}{x + a} - \frac{1}{x + b} = \frac{1}{x + c}$Simplify the left side: $\frac{(x + b) - (x + a)}{(x + a)(x + b)} = \frac

Vedclass · Accepted Answer

$\frac{-2bc}{b + c}$

Vedclass · Answer

(D) Given the equation: $\frac{1}{x + a} - \frac{1}{x + b} = \frac{1}{x + c}$Simplify the left side: $\frac{(x + b) - (x + a)}{(x + a)(x + b)} = \frac{1}{x + c}$$\frac{b - a}{x^2 + (a + b)x + ab} = \frac{1}{x + c}$Cross-multiply: $x^2 + (a + b)x + ab = (b - a)(x + c)$$x^2 + (a + b)x + ab = (b - a)x + (b - a)c$$x^2 + (a + b - b + a)x + ab - (b - a)c = 0$$x^2 + 2ax + ab - bc + ac = 0$Since the product of the roots is $0$,the constant term must be $0$:$ab + ac - bc = 0 \implies a(b + c) = bc \implies a = \frac{bc}{b + c}$The sum of the roots of a quadratic equation $x^2 + Bx + C = 0$ is $-B$.Sum of roots $= -2a = -2 \left( \frac{bc}{b + c} \right) = \frac{-2bc}{b + c}$.

For the equation $\frac{1}{x + a} - \frac{1}{x + b} = \frac{1}{x + c}$,if the product of the roots is zero,what is the sum of the roots?

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