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

The correct answer is 6.

We are given two quadratic equations where m and n are positive integers, and both equations must have real roots. For a quadratic to have real roots, its discriminant must be non-negative (≥ 0).

Step 1: Apply the discriminant condition to the first equation.

For $x^2+mx+2n=0$, the discriminant is:

$m^2-4\left(1\right)\left(2n\right)\ge 0$

$m^2\ge 8n$     — Condition (i)

Step 2: Apply the discriminant condition to the second equation.

For $x^2+2nx+m=0$, the discriminant is:

$(2n{)}^2-4(1\left)\right(m)\ge 0$

$4n^2\ge 4m$

$n^2\ge m$     — Condition (ii)

Step 3: Combine the two conditions to find the minimum value of m + n.

From Condition (ii), we know $m\le n^2$. Substituting this upper bound for m into Condition (i):

$(n^2{)}^2\ge 8n$

$n^4\ge 8n$

Since n is a positive integer, we can divide both sides by n:

$n^3\ge 8$

This gives us $n\ge 2$.

Step 4: Test n = 2 and find the corresponding m.

When n = 2:
• From Condition (i): $m^2\ge 8\left(2\right)=16$, so $m\ge 4$.
• From Condition (ii): $2^2\ge m$, so $m\le 4$.

Both conditions together require m = 4 exactly.

Step 5: Verify both equations have real roots with m = 4, n = 2.

Equation 1: $x^2+4x+4=0$ → Discriminant = 16 − 16 = 0 ✓ (repeated real root)
Equation 2: $x^2+4x+4=0$ → Discriminant = 16 − 16 = 0 ✓ (repeated real root)

Step 6: Conclusion

The smallest possible value of $m+n$ is:

$m+n=4+2=6$