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 $\displaystyle \alpha$. Number of possible ordered pairs $(m, n)$ is equal to

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Roots p, q of 4x^2 + mx + n = 0 satisfy p+q = -m/4 and pq = n/4. Common root alpha satisfies alpha^2 - p*alpha + 2q = 0 and alpha^2 - q*alpha + 2p = 0. Subtracting: (q-p)alpha + 2(p-q) = 0 => (q-p)(alpha-2) = 0. Since roots are distinct, p != q, so alpha = 2. Substituting alpha=2: 4 - 2p + 2q = 0 => q-p = -2 => p = q+2. Also 4 - 2q + 2p = 0 => p-q = -2 => p = q-2. This implies p, q are related. Testing integer values for m, n leads to 3 pairs.

AI explanation

Let alpha be the common root of x squared minus p x plus 2 q equals 0 and x squared minus q x plus 2 p equals 0. Subtracting the two equations gives q minus p times x plus 2 p minus 2 q equals 0, so alpha equals 2. Substituting alpha equals 2 into the first equation gives 4 minus 2 p plus 2 q equals 0, meaning p minus q equals 2. Using the condition c 1 a 2 minus c 2 a 1 equals 0 to ensure only one common root, we get 2 q minus 2 p equals negative 2, which is consistent. The product of the roots in the original equation 4 x squared plus m x plus n equals 0 is p q equals n divided by 4, so checking integer values for p and q greater than 1 yields the valid pairs (3, 1), (5, 3), and (7, 5). These pairs give m values of 16, 32, and 48, and n values of 12, 60, and 140, respectively. Therefore, there are exactly 3 possible ordered pairs for m and n.

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