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Let $\text{m}$ and $\text{n}$ 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

1. $7$
2. $8$
3. $5$
4. $6$

Given that, $m,n > 0$

And,

• $x^{2} + mx + 2n = 0 \quad \longrightarrow (1)$
• $x^{2} + 2nx + m = 0 \quad \longrightarrow (2)$

We know that, $ax^{2} + bx + c = 0$ have real root.

Then, $\boxed{D \geq 0}$

$\Rightarrow b^{2} – 4ac \geq 0$

For equation $(1),$

$m^{2} – 8n \geq 0$

$\Rightarrow \boxed{ m^{2} \geq 8n} \; \longrightarrow (3)$

For equation $(2),$

$(2n)^{2} – 4m \geq 0$

$\Rightarrow 4n^{2} – 4m \geq 0$

$\Rightarrow \boxed{n^{2} \geq m}$

$\Rightarrow \boxed{ n^{4} \geq m^{2}} \; \longrightarrow (4)$

From equation $(3),$ and $(4).$

$n^{4} \geq m^{2} \geq 8n \; \longrightarrow {5}$

Now, taking first and last term.

$n^{4} \geq 8n$

$\Rightarrow n^{3} \geq 2^{3}$

$\Rightarrow \boxed{ n \geq 2}$

Thus, minimum value of $n = 2.$

So, $m^{2} \geq 16$

$\Rightarrow \boxed { m \geq 4}$

Thus, the minimum value of $m=4.$

Then, the minimum value of $m+n = 4 + 2 = 6.$

$\therefore$ The smallest possible value of $m+n$ is $6.$

Correct Answer$: \text{D}$

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