Multiple choice

lf $pr=2(q+s)$, then among the equations $\mathrm{x}^{2}+ px +\mathrm{q}=0$ and $\mathrm{x}^{2}+ rx +\mathrm{s}=0$

  1. both have real roots

  2. both have imaginary roots

  3. at least one has real roots

  4. at least one has imaginary roots
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Discriminants are D1 = p^2 - 4q and D2 = r^2 - 4s. Sum of discriminants = p^2 + r^2 - 4(q+s). Since pr = 2(q+s), sum = p^2 + r^2 - 2pr = (p-r)^2. Since the sum of discriminants is >= 0, at least one discriminant must be >= 0.

AI explanation

Assume for contradiction that both equations have imaginary roots, which requires their discriminants to be strictly negative: p^2 - 4q < 0 and r^2 - 4s < 0. This implies p^2 < 4q and r^2 < 4s, so p^2 + r^2 < 4(q + s). Since we are given pr = 2(q + s), we substitute to get p^2 + r^2 < 2pr, or (p - r)^2 < 0. Since a perfect square cannot be negative, this is a contradiction, proving at least one equation must have real roots.

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