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

• A. 7
• B. 8
• C. 5
• D. 6

Answer 1

The Correct Option is D

The correct answer is (D): \(6\)

Since the roots are real \(m^2-8n≥0\) and \((2n)^2-4m≥0 ⇒ n^2-m≥0\)

\(⇒ n^4≥m^2≥8n\)

\(⇒ n≥2\) and \(m≥4\)

Hence the least value of \(m+n=2+4=6\)

Answer 2

The discriminant must be larger than or equal to 0 in order to have true roots.
So,  \(m^ 2 −8n≥0 \&\  4n^ 2 −4m≥0.\)
\(m ^2 ≥8n \&\  n ^2 ≥m\)
The value of \(m+n\) will be at its minimum when \(m=4\) and \(n=2\), because m,n are positive integers. 
Hence, \(m+n=6.\)