Bunuel wrote:

In the equation x^2 + mx + n = 0, x is a variable and m and n are constants. What is the value of n ?

(1) x + 2 is a factor of x^2 + mx + n

(2) The equation x^2 + mx + n = 0 has only one root

We need to determine the value of n.

Statement One Alone:

x + 2 is a factor of x^2 + mx + n.

Let f(x) = x^2 + mx + n. If x + 2 is a factor of x^2 + mx + n, then by the factor theorem, f(-2) = 0. That is, (-2)^2 + m(-2) + n = 0 or 4 - 2m + n = 0. However, since we don’t know the value of m, we can’t determine the value of n. Statement one alone is not sufficient to answer the question.

Statement Two Alone:

The equation x^2 + mx + n = 0 has only one root.

Let r be the root. Then we can express x^2 + mx + n as (x - r)^2:

x^2 + mx + n = (x - r)^2

x^2 + mx + n = x^2 - 2rx + r^2

We see that m = -2r and n = r^2. Since we don’t know the value of m or r, we can’t determine the value of n. Statement two alone is not sufficient to answer the question.

Statements One and Two Together:

Since statement one states that x + 2 is a factor of x^2 + mx + n, x = -2 is a root of the equation x^2 + mx + nx = 0. Furthermore, since statement two states that there is only one root, x = -2 must be a double root. That is, x^2 + mx + n = (x + 2)^2:

x^2 + mx + n = (x + 2)^2

x^2 + mx + n = x^2 + 4x + 4

We see that m = 4 and n = 4.

Answer: C

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