Multiple choice

If the roots of the equation $px^2+2qx+r=0$ and $qx^2-2\sqrt{pr}x+q=0$ be real, then

  1. $p = q$
  2. $q^2=pr$
  3. $p^2=qr$
  4. $r^2=pq$
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B Correct answer
Explanation

For the roots to be real, the discriminant must be non-negative. For px^2 + 2qx + r = 0, D1 = (2q)^2 - 4pr = 4(q^2 - pr) >= 0, so q^2 >= pr. For qx^2 - 2sqrt(pr)x + q = 0, D2 = (-2sqrt(pr))^2 - 4(q)(q) = 4pr - 4q^2 >= 0, so pr >= q^2. Since q^2 >= pr and pr >= q^2, it must be that q^2 = pr.

AI explanation

For the first quadratic px^2 + 2qx + r = 0 to have real roots, the discriminant condition gives (2q)^2 - 4pr >= 0, meaning q^2 >= pr. For the second quadratic qx^2 - 2*sqrt(pr)x + q = 0 to have real roots, the discriminant condition requires (-2*sqrt(pr))^2 - 4q^2 >= 0, which simplifies to pr >= q^2. The only way both conditions q^2 >= pr and pr >= q^2 can be true simultaneously is if q^2 = pr.

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