If x is a not-zero number, is x^2 + 2x > x^2 + x?

1.Simplify the original question by factoring:
x^2 + 2x > x^2 + x
2x > x
x > 0
Simplified question: is x > 0?

Evaluate Statement (1) alone.

When dealing with a number that is raised to an even exponent, it is important to remember that the sign of the base number can be either positive or negative (i.e., if x2 = 16, x = -4 and 4). Moreover, it is important to remember that raising a fraction to a larger exponent makes the resulting number smaller:
(1/2)2 > (1/2)3 > (1/2)4
There are three possible cases:
Case (1): x < 0
If x were negative, xodd would be negative while xeven would be positive. This would make xodd {=negative} < xeven {=positive}, which is an explicit contradiction of Statement (1). As a result, we know x cannot be negative. Statement (1) is SUFFICIENT. At this point, you should not keep evaluating since you know that Statement (1) provides enough information to answer the question "is x > 0?"

Case (2): 0 < x < 1
In this case, based upon what was shown above, for xodd integer > xeven integer to hold true, odd integer must be less than even integer.

Case (3): x > 1
This case is the opposite of Case (2). In other words, for xodd integer > xeven integer, the odd integer must be greater than the even integer.
Since Statement (1) eliminates the possibility of x being a negative number, we can definitively answer the question: is x > 0?
Statement (1) alone is SUFFICIENT.

Evaluate Statement (2) alone.

Factor x2 + x - 12 = 0
(x - 3)(x + 4) = 0
x = 3, -4
Since x can be either positive or negative, Statement (2) is not sufficient.
Statement (2) alone is NOT SUFFICIENT.
Since Statement (1) alone is SUFFICIENT but Statement (2) alone is NOT SUFFICIENT, answer A is correct