Multiple choice

If the equation $px^{2}+qx+r=0$ and $rx^{2}+qx+p=0$, where $p\neq r$, have a negative common root, then value of $p-q$ is

  1. $r$
  2. $2r$
  3. $-r$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let alpha be the common root. p(alpha)^2 + q(alpha) + r = 0 and r(alpha)^2 + q(alpha) + p = 0. Subtracting: (p-r)(alpha)^2 + (r-p) = 0. (p-r)(alpha^2 - 1) = 0. Since p != r, alpha^2 = 1. Since the root is negative, alpha = -1. Substitute alpha = -1 into the first equation: p - q + r = 0. Thus, p - q = -r.

AI explanation

Let the negative common root of both equations be x. We have px^2 + qx + r = 0 and rx^2 + qx + p = 0, so subtracting the second from the first yields (p - r)x^2 + (r - p) = 0. Factoring this gives (p - r)(x^2 - 1) = 0, and since p is not equal to r, the common root must be x = -1. Substituting x = -1 into the first equation gives p(1) - q + r = 0, which simplifies to p - q = -r.

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