Question

Let $m$ and $n$ be positive integers. If $x^2 + mx + 2n = 0$ and $x^2 + 2nx + m = 0$ have real roots, then the smallest possible value of $m + n$ is

Accepted Answer

B. For real roots: 1) $m^2 - 8n \ge 0 \implies m^2 \ge 8n$ 2) $4n^2 - 4m \ge 0 \implies n^2 \ge m$ Test small integer values for $n$: - If $n = 1$: $m^2 \ge 8 \implies m \ge 3$, and $n^2 \ge m \implies 1 \ge m$ (impossible). - If $n = 2$: $m^2 \ge 16 \implies m \ge 4$, and $n^2 \ge m \implies 4 \ge m$. Thus $m = 4$. $m + n = 4 + 2 = 6$. Check if smaller sum is possible: $m+n=6$ works. Hence smallest possible $m + n = 6$.

Minimum m + n for real roots

Explanation

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