Let m and n be positive integers, If $$x^{2}+mx+2n=0$$ and $$x^{2}+2nx+m=0$$ have real roots, then the smalles

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

6. To have real roots the discriminant should be greater than or equal to 0. So, m 2 − 8 n ≥ 0 & a m p ; 4 n 2 − 4 m ≥ 0 m^2-8n\ge0\ \&amp;\ 4n^2-4m\ge0 m 2 − 8 n ≥ 0 & am p ; 4 n 2 − 4 m ≥ 0 => m 2 ≥ 8 n & a m p ; n 2 ≥ m m^2\ge8n\ \&amp;\ n^2\ge m m 2 ≥ 8 n & am p ; n 2 ≥ m

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

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

Let m and n be positive integers, If x2+mx+2n=0x^{2}+mx+2n=0x2+mx+2n=0 and x2+2nx+m=0x^{2}+2nx+m=0x2+2nx+m=0 have real roots, then the smallest possible value of m+nm+nm+n is

Let m and n be positive integers, If x2+mx+2n=0x^{2}+mx+2n=0x2+mx+2n=0 and x2+2nx+m=0x^{2}+2nx+m=0x2+2nx+m=0 have real roots, then the smallest possible value of m+nm+nm+n is

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

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