CAT 2020 Exam Shift 3 Solved Paper

Question : 61 of 76

Marks: +1, -0

Let

m

and

n

be positive integers, If

x^{2}+m x+2 n=0

and

x^{2}+2 n x+m=0

have real roots, then the smallest possible value of

m+n

Solution:

Since the roots are real

m^{2}-8 n ≥ 0

and

(2 n)^{2}-4 m ≥ 0 ⇒ n^{2}-m ≥ 0

⇒ n^{4} ≥ m^{2} ≥ 8 n

⇒ n ≥ 2

and

m ≥ 4

Hence the least value of

m+n=2+4=6

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