Multiple choice

Let $m, n$ be positive integers and the quadratic equation $\displaystyle 4x^2 + mx + n = 0$ has two distinct real roots $p$ and $q$ $(p \leq q)$. Also, the quadratic equations $\displaystyle x^2 - px + 2q = 0$ and $\displaystyle x^2 - qx + 2p = 0$ have a common root say $\displaystyle \alpha$. If $p$ and $q$ are rational, then uncommon root of the equation $\displaystyle x^2 - px + 2q = 0$ and $\displaystyle x^2 - qx + 2p = 0$ is equal to

  1. $-q$
  2. $\displaystyle \frac {-q}{4}$
  3. $\displaystyle \frac {-q}{2}$
  4. $\displaystyle \frac {q}{2}$
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A Correct answer
Explanation

Subtracting the two quadratic equations gives (q - p)(x + 2) = 0. Since p and q are distinct roots, the common root is -2. The other root of the first equation is p + 2, and the common-root condition gives p + q = -2, so this root equals -q.