Multiple choice

If the equation $\displaystyle \left( { p }^{ 2 }+{ q }^{ 2 } \right) { x }^{ 2 }-2\left( pr+qs \right) x+{ r }^{ 2 }+{ s }^{ 2 }=0$ has equal roots then

  1. $\displaystyle pq=rs$
  2. $\displaystyle ps=rq$
  3. $\displaystyle ps=\sqrt { rq } $
  4. $\displaystyle ps=\sqrt { rs } $
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B Correct answer
Explanation

For the quadratic equation to have equal roots, its discriminant must be zero. This simplifies to (pr + qs)^2 - (p^2 + q^2)(r^2 + s^2) = 0, which expands and simplifies to (ps - rq)^2 = 0, yielding ps = rq.

AI explanation

For a quadratic equation to have equal roots, its discriminant must equal zero. The discriminant is [-2(pr + qs)]^2 - 4(p^2 + q^2)(r^2 + s^2) = 0. Expanding this gives 4(p^2r^2 + 2prqs + q^2s^2) - 4(p^2r^2 + p^2s^2 + q^2r^2 + q^2s^2) = 0, which simplifies to 8prqs - 4p^2s^2 - 4q^2r^2 = 0. Dividing by -4 and rearranging results in p^2s^2 - 2prqs + q^2r^2 = 0, which factors perfectly as a perfect square trinomial into (ps - qr)^2 = 0. Therefore, ps = qr, which is the same as ps = rq.

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