If \frac{\sqrt{p+x} + \sqrt{p-x}}{\sqrt{p+x} - \sqrt{p-x}} = p, then what is x equal to?

  1. A. \frac{p}{p^2+1}
  2. B. \frac{2p}{p^2+1}
  3. C. \frac{p^2}{p^2+1}
  4. D. \frac{2p^2}{p^2+1}

Correct Answer: D. \frac{2p^2}{p^2+1}

Explanation

Applying componendo and dividendo gives \frac{\sqrt{p+x}}{\sqrt{p-x}} = \frac{p+1}{p-1}. Squaring both sides yields \frac{p+x}{p-x} = \frac{(p+1)^2}{(p-1)^2} = \frac{p^2+2p+1}{p^2-2p+1}. Applying componendo and dividendo again results in \frac{2p}{2x} = \frac{2(p^2+1)}{4p}, which simplifies to x = \frac{2p^2}{p^2+1}.

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