Multiple choice

If the roots of the equation $\dfrac{a}{x+a+k}+\dfrac{b}{x+b+k}=2$ are equal in magnitude but opposite in sign, then the value of $k$ is:

  1. $\dfrac{(a+b)}{4}$
  2. $\dfrac{(a+b)}{3}$
  3. $\dfrac{(a+b)}{2}$
  4. $\dfrac{-(a+b)}{4}$
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D Correct answer
Explanation

Let the roots be r and -r. The sum of roots is 0. Simplifying the equation a/(x+a+k) + b/(x+b+k) = 2 leads to a quadratic equation in x. For the sum of roots to be zero, the coefficient of the x term must be zero. Solving this condition yields k = -(a+b)/4.

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