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Question 23 : Let m and n be positive integers, If x2 + mx + 2n = 0 and x2 + 2nx + m = 0 have real roots, then the smallest possible value of m + n is

  1. 8
  2. 6
  3. 5
  4. 7

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Explanatory Answer

The roots of a Quadratic Equation of the form, ax2 + bx + c = 0
are given by x = \\frac{-b ± √[b^{2} - 4ac]}{2a})

In order for these roots to be real, the portion under the square root should not be negative.
In other words 'b2 - 4ac', determines wether the roots are real or imaginary.
Hence, rightly, it is called the Determinat(D).

If the Determinant, D is greater than or equal to 0, then the equation has real roots.
For D to be greater than or equal to 0, b2 ≥ 4ac.

We are told that x2 + mx + 2n = 0 and x2 + 2nx + m = 0 have real roots,
That means, m2 ≥ 4(2n) and (2n)2 ≥ 4m.

Instead of trying to solve through equations, the two inequalities m2 ≥ 8n and n2 ≥ m.
Let's try to solve inputting specific numbers.

If n = 1; m2 ≥ 8; m ≥ 3
If m ≥ 3 at n = 1, n2 ≥ m stands invalid.

If n = 2; m2 ≥ 16; m ≥ 4
If m ≥ 4 at n = 2, n2 ≥ m stands valid, if m = 4 and n = 2

Hence 4,2 is the smallest(and the only) pair that m,n can take.
So, the minimum sum of m and n is 4+2 = 6.


The question is "Let m and n be positive integers, If x2 + mx + 2n = 0 and x2 + 2nx + m = 0 have real roots, then the smallest possible value of m + n is"

Hence, the answer is, "6"

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