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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
To have real roots the discriminant should be greater than or equal to 0.
So, $$m^2-8n\ge0\ \&\ 4n^2-4m\ge0$$
=> $$m^2\ge8n\ \&\ n^2\ge m$$
Since m,n are positive integers the value of m+n will be minimum when m=4 and n=2.
.'. m+n=6.
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