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

We need to find when both quadratic equations have real roots, then minimize m + n.

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For any quadratic equation ax² + bx + c = 0 to have real roots, its discriminant (b² - 4ac) must be ≥ 0.

The discriminant tells us about the nature of roots. When it's negative, we get complex roots; when it's non-negative, we get real roots.

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For the first equation $x^2 + xmx + 2n = 0$:

Discriminant $= m^2 - 4(1)(2n) = m^2 - 8n$

For real roots: $m^2 - 8n \geq 0$

This gives us: $m^2 \geq 8n$ ... (1)

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For the second equation $x^2 + 2nx + m = 0$:

Discriminant $= (2n)^2 - 4(1)(m) = 4n^2 - 4m$

For real roots: $4n^2 - 4m \geq 0$

Dividing by 4: $n^2 - m \geq 0$

This gives us: $m \leq n^2$ ... (2)

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From conditions (1) and (2): $8n \leq m^2$ and $m \leq n^2$

This means: $8n \leq m^2 \leq (n^2)^2 = n^4$

So we need: $8n \leq n^4$

Dividing by n (since n > 0): $8 \leq n^3$

Since n is a positive integer, we need $n^3 \geq 8$, which means $n \geq 2$.

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When $n = 1$:

Condition (1): $m^2 \geq 8(1) = 8$, so $m \geq 3$

Condition (2): $m \leq 1^2 = 1$, so $m \leq 1$

These contradict each other! No solution for $n = 1$.

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When $n = 2$:

Condition (1): $m^2 \geq 8(2) = 16$, so $m \geq 4$

Condition (2): $m \leq 2^2 = 4$, so $m \leq 4$

Both conditions satisfied when $m = 4$

Sum: $m + n = 4 + 2 = 6$

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When $n = 3$:

Condition (1): $m^2 \geq 8(3) = 24$, so $m \geq 5$ (since $5^2 = 25 > 24$)

Condition (2): $m \leq 3^2 = 9$, so $m \leq 9$

Smallest valid m is 5

Sum: $m + n = 5 + 3 = 8$

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When $n = 4$:

Condition (1): $m^2 \geq 8(4) = 32$, so $m \geq 6$ (since $6^2 = 36 > 32$)

Condition (2): $m \leq 4^2 = 16$, so $m \leq 16$

Smallest valid m is 6

Sum: $m + n = 6 + 4 = 10$

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As n increases, the minimum required value of m increases, making $m + n$ larger.

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Therefore, the smallest possible value of $m + n$ is 6.