Multiple choice

If the roots of the equation $\displaystyle{\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}}$ are negatives of each other, then $r = $

  1. $p + q$
  2. $p - q$
  3. $\displaystyle \frac { p+q }{ 2 } $
  4. $ \displaystyle \frac { p-q }{ 2 } $
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C Correct answer
Explanation

If the roots are x and -x, the equation 1/(x+p) + 1/(x+q) = 1/r becomes (2x + p + q) / (x^2 + x(p+q) + pq) = 1/r. Cross-multiplying gives r(2x + p + q) = x^2 + x(p+q) + pq. For the roots to be symmetric around zero, the coefficient of the x term must be zero, implying 2r = p + q, or r = (p+q)/2.

AI explanation

Multiplying the given equation by the common denominator yields r times (x plus q plus x plus p) equals (x plus p)(x plus q). Expanding and rearranging terms gives the quadratic equation x squared plus x times (p plus q minus 2r) plus (pq minus r times p minus r times q) equals 0. For the roots to be negatives of each other, the sum of the roots must be zero. The sum of the roots is the negative coefficient of x, so p plus q minus 2r equals 0. Solving this gives r equals the fraction (p plus q) divided by 2.

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