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Is x + x^2 + x^3 > 0? (1) x + x^2 > 0 (2) x^2 + x^3 > 0 03 Oct 2014, 16:22
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C
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56% (02:10) correct 44% (01:58) wrong based on 237 sessions

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Is \(x + x^2 + x^3> 0\)?

(1) \(x + x^2 > 0\)

(2) \(x^2 + x^3 > 0\)
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C
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Is x + x^2 + x^3 > 0? (1) x + x^2 > 0 (2) x^2 + x^3 > 0 03 Oct 2014, 16:41
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Is \(x + x^2 + x^3> 0\)?

Is \(x + x^2 + x^3> 0\)? --> Is \(x(1 + x + x^2)> 0\)? Notice that x^2+x+1 can be written as \((x+\frac{1}{2})^2+\frac{3}{4}\), so it's positive. Hence for \(x(1 + x + x^2)=x*positive\) to be positive x must be positive.

(1) \(x + x^2 > 0\) --> this holds true when x is positive as well as when x is less than -1 (\(x>0\) or \(x<-1\)). Not sufficient.

(2) \(x^2 + x^3 > 0\) --> reduce by x^2: \(1 + x>0\) --> \(x>-1\) (\(x\neq{0}\)). Not sufficient.

(1)+(2) Intersection of possible ranges is x>0. Sufficient.

Answer: C.
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Is x + x^2 + x^3 > 0? (1) x + x^2 > 0 (2) x^2 + x^3 > 0 03 Oct 2014, 16:47
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Is \(x + x^2 + x^3> 0\)?

Is \(x + x^2 + x^3> 0\)? --> Is \(x(1 + x + x^2)> 0\)? Notice that x^2+x+1 can be written as \((x+\frac{1}{2})^2+\frac{3}{4}\), so it's positive. Hence for \(x(1 + x + x^2)=x*positive\) to be positive x must be positive.

(1) \(x + x^2 > 0\) --> this holds true when x is positive as well as when x is less than -1 (\(x>0\) or \(x<-1\)). Not sufficient.

(2) \(x^2 + x^3 > 0\) -->reduce by x^2: \(1 + x>0\) --> \(x>-1\) (\(x\neq{0}\)). Not sufficient.

(1)+(2) Intersection of possible ranges is x>0. Sufficient.

Answer: C.
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The highlighted part. Is this what you are implying by "reduce by \(x^2\)"

take \(x^2\) common so that makes: \(x^2 (1+x) >0\)
Since\(x^2\) has to be positive, so 1+x > 0
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Is x + x^2 + x^3 > 0? (1) x + x^2 > 0 (2) x^2 + x^3 > 0 03 Oct 2014, 16:51
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earnit
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Is \(x + x^2 + x^3> 0\)?

Is \(x + x^2 + x^3> 0\)? --> Is \(x(1 + x + x^2)> 0\)? Notice that x^2+x+1 can be written as \((x+\frac{1}{2})^2+\frac{3}{4}\), so it's positive. Hence for \(x(1 + x + x^2)=x*positive\) to be positive x must be positive.

(1) \(x + x^2 > 0\) --> this holds true when x is positive as well as when x is less than -1 (\(x>0\) or \(x<-1\)). Not sufficient.

(2) \(x^2 + x^3 > 0\) -->reduce by x^2: \(1 + x>0\) --> \(x>-1\) (\(x\neq{0}\)). Not sufficient.

(1)+(2) Intersection of possible ranges is x>0. Sufficient.

Answer: C.

The highlighted part. Is this what you are implying by "reduce by \(x^2\)"

take \(x^2\) common so that makes: \(x^2 (1+x) >0\)
Since\(x^2\) has to be positive, so 1+x > 0
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Yes. Since x^2 is positive you could simply divide both sides by it.
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Is x + x^2 + x^3 > 0? (1) x + x^2 > 0 (2) x^2 + x^3 > 0 03 Oct 2014, 22:50
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Bunuel
Is \(x + x^2 + x^3> 0\)?

Is \(x + x^2 + x^3> 0\)? --> Is \(x(1 + x + x^2)> 0\)? Notice that x^2+x+1 can be written as \((x+\frac{1}{2})^2+\frac{3}{4}\), so it's positive. Hence for \(x(1 + x + x^2)=x*positive\) to be positive x must be positive.

(1) \(x + x^2 > 0\) --> this holds true when x is positive as well as when x is less than -1 (\(x>0\) or \(x<-1\)). Not sufficient.

(2) \(x^2 + x^3 > 0\) --> reduce by x^2: \(1 + x>0\) --> \(x>-1\) (\(x\neq{0}\)). Not sufficient.

(1)+(2) Intersection of possible ranges is x>0. Sufficient.

Answer: C.
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i have a doubt in option B... here x cannot by zero since then the condition wont be satisfied then & X >-1 then X has to be positive rite..
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Is x + x^2 + x^3 > 0? (1) x + x^2 > 0 (2) x^2 + x^3 > 0 04 Oct 2014, 03:40
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Bunuel
Is \(x + x^2 + x^3> 0\)?

Is \(x + x^2 + x^3> 0\)? --> Is \(x(1 + x + x^2)> 0\)? Notice that x^2+x+1 can be written as \((x+\frac{1}{2})^2+\frac{3}{4}\), so it's positive. Hence for \(x(1 + x + x^2)=x*positive\) to be positive x must be positive.

(1) \(x + x^2 > 0\) --> this holds true when x is positive as well as when x is less than -1 (\(x>0\) or \(x<-1\)). Not sufficient.

(2) \(x^2 + x^3 > 0\) --> reduce by x^2: \(1 + x>0\) --> \(x>-1\) (\(x\neq{0}\)). Not sufficient.

(1)+(2) Intersection of possible ranges is x>0. Sufficient.

Answer: C.


i have a doubt in option B... here x cannot by zero since then the condition wont be satisfied then & X >-1 then X has to be positive rite..
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Yes, x cannot be 0, because in this case \(x^2 + x^3\) is 0, not greater than 0.

But as shown there x is not necessarily positive, it could be a negative fraction, for example, -1/2. That's what \(x>-1\) (\(x\neq{0}\)) mean.
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Is x + x^2 + x^3 > 0? (1) x + x^2 > 0 (2) x^2 + x^3 > 0 18 Feb 2016, 19:21
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is x + x^2 + x^3 > 0

(1) x+x^2>0
(2) x^2+x^3>0


In general, x^2+x+1>0 and x^2+1>0 are always valid. Also, when it comes to inequality DS questions, it is important that if range of que includes range of con, the con is sufficient.

When you modify the original condition and the question, 1+x+x^2>0 is always valid from x(1+x+x^2)>0?, which makes x>0? possible. Thus, there is 1 variable(x), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), x(1+x)>0 -> x<-1, 0<x, the range of que doesn’t include the range of con, which is not sufficient.
For 2), when dividing x^2(1+x)>0 with x^2(since x^2 is positive, the sign of inequality doesn’t change when dividing with it.). Then, 1+x>0 becomes x>-1. The range of que doesn’t include the range of con, which is not sufficient.
When 1) & 2), from x>0, the range of que includes the range of con, which is sufficient.
Therefore, the answer is C.


 For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Is x + x^2 + x^3 > 0? (1) x + x^2 > 0 (2) x^2 + x^3 > 0 17 Oct 2023, 18:36
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