CAT 2020 Slot 3 QA Question & Solution
AlgebraMedium
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
Options
7
6
8
5
Solution
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.