Is x + x² + x³ > 0? (1) x + x² > 0 (2) x² + x³ > 0

A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not. B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not. C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient. D. Either statement BY ITSELF is sufficient to answer the question. E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.

(C) Statement (1) tells us that x + x² > 0. We can transform it into x(1 + x) > 0. The product is positive if the both factors have the same sign. So the solution is x > 0, x < -1. If x is positive then statement (1) holds and the desired inequality is true. But what if x < -1? For example, if x = -10 the desired inequality does NOT hold, -10 + 100 – 1000 = -910, which is negative. Therefore statement (1) by itself is NOT sufficient.

Statement (2) tells us that x² + x³ > 0. We can transform it into x²(1 + x) > 0. Since x² is a non-negative number, then the equality is true when (1 + x) > 0. Besides x is NOT 0. Therefore the solution of the inequality (2) is x > 0, 0 > x > -1. If x is positive then statement (1) holds and the desired inequality is true. But what if 0 > x > -1? Let’s try a value, x = -1/2. The left side of the desired inequality becomes -1/2 + 1/4 – 1/8, which is negative. Therefore statement (2) by itself is NOT sufficient.

The both statements combined give us the solution, which must be true for the both. It is x > 0. The desired inequality is definitely true when x is positive. Therefore statements (1) and (2) taken together are sufficient to answer the question. The correct answer is C. ----------

[color=#FF0000]I am not sure of the solution provided here. From the first statement we get x > 0 and 1 + x > 0, which gives us x > -1 (if they both are positive), unlike stated here - "x < -1" in the explanation. This will only be possible if both the terms are less than 0.

Seems like, with this, I am still not able to come to C as a solution.[/color]

(1) x + x² > 0 , which transforms into (1) x(1 + x) > 0[quote][color=#FF0000]From the first statement we get x > 0 and 1 + x > 0, …[/color]

x(1 + x) > 0 holds true if the factors x and (x + 1) are: - both positive; - both negative.

So we have two possible situations: positive and negative, not just positive[quote][color=#FF0000]…, which gives us x > -1 (if they both are positive)[/color]Plug in x = -1/2 or x = 0 to see that x > -1 is NOT the solution of statement (1).

For the both factors to be positive: x must be positive (x > 0) [img]http://www.gmat-mba-prep.com/forum/download/file.php?id=109[/img] x + 1 must be positive (x > -1) [img]http://www.gmat-mba-prep.com/forum/download/file.php?id=110[/img]

The both factors must be positive at the same time. So we choose the intersection as the solution for the positive factors: [img]http://www.gmat-mba-prep.com/forum/download/file.php?id=109[/img] x > 0[quote][color=#FF0000]…, unlike stated here - "x < -1" in the explanation. This will only be possible if both the terms are less than 0.[/color]Statement (1) yields such situation of the both terms being negative. So x < -1 fits statement (1).

The complete solution of Statement (1) consists of two intervals: x < -1 and x > 0 [img]http://www.gmat-mba-prep.com/forum/download/file.php?id=111[/img]

[color=#FF0000]Can you please explain it to me better. The solution is not clear as to how 1 and 2 combined answer the question. I understand that these separately cannot answer the question. Thank you.[/color]

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