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Question

If the sum of the roots of the equation \(\dfrac{1}{2x+p} + \dfrac{1}{2x+q} = \dfrac{1}{r}\) is zero, then what is \(r\) equal to?

This question was previously asked in
NDA 2 2026 GAT Question Paper (13-Sep-2026)
The correct answer is

\(\dfrac{p+q}{2}\)

Combining the fractions gives \(r(4x+p+q) = (2x+p)(2x+q) = 4x^2+2x(p+q)+pq\), i.e. \(4x^2+[2(p+q)-4r]x+[pq-r(p+q)]=0\). The sum of roots is zero, so \(2(p+q)-4r=0\), giving \(r=\dfrac{p+q}{2}\).

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