CAT 2020 Set-3 | Question: 61

Question

Let $\text{m}$ and $\text{n}$ 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

Given that, $m,n > 0$

And,

• $x^{2} + mx + 2n = 0 \quad \longrightarrow (1) $
• $x^{2} + 2nx + m = 0 \quad \longrightarrow (2) $

We know that, $ax^{2} + bx + c = 0$ have real root.

Then, $ \boxed{D \geq 0} $

$ \Rightarrow b^{2} – 4ac \geq 0 $

For equation $(1),$

$ m^{2} – 8n \geq 0 $

$ \Rightarrow \boxed{ m^{2} \geq 8n} \; \longrightarrow (3) $

For equation $(2),$

$ (2n)^{2} – 4m \geq 0 $

$ \Rightarrow 4n^{2} – 4m \geq 0 $

$ \Rightarrow \boxed{n^{2} \geq m} $

$ \Rightarrow \boxed{ n^{4} \geq m^{2}} \; \longrightarrow (4) $

From equation $(3),$ and $(4).$

$ n^{4} \geq m^{2} \geq 8n \; \longrightarrow {5} $

Now, taking first and last term.

$ n^{4} \geq 8n $

$ \Rightarrow n^{3} \geq 2^{3} $

$ \Rightarrow \boxed{ n \geq 2} $

Thus, minimum value of $ n = 2.$

So, $m^{2} \geq 16 $

$ \Rightarrow \boxed { m \geq 4} $

Thus, the minimum value of $m=4.$

Then, the minimum value of $m+n = 4 + 2 = 6.$

$\therefore$ The smallest possible value of $m+n$ is $6.$

Correct Answer$: \text{D}$