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Is x < 0? (1) x^3 + x^2 + x + 2 = 0 (2) x^2 - x - 2 < 0

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Is x < 0? (1) x^3 + x^2 + x + 2 = 0 (2) x^2 - x - 2 < 0 [#permalink]

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Is x < 0?


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

(2) x^2 - x - 2 < 0
[Reveal] Spoiler: OA

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Last edited by idontknowwhy94 on 07 Oct 2016, 09:34, edited 1 time in total.

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Re: Is x < 0? (1) x^3 + x^2 + x + 2 = 0 (2) x^2 - x - 2 < 0 [#permalink]

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New post 07 Oct 2016, 09:32
Guys IHO ,

the answer is
[Reveal] Spoiler:
A
as the only way for the equation to be 0 is when x is -ve

for st 2 ; we have 2 answers
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Re: Is x < 0? (1) x^3 + x^2 + x + 2 = 0 (2) x^2 - x - 2 < 0 [#permalink]

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New post 07 Oct 2016, 09:47
idontknowwhy94 wrote:
Is x<0?

1) \(x^{3} + x^{2}\) + x + 2 = 0
2) \(x^{2}\)-x-2 < 0


Keep the Kudos coming in and let the questions come out


1) if x>=0 then the expression will never be 0.

thus x<0.........suff

2) X^2-x-2<0
(x+1)(x-2)<0
X satisfies for range -1<x<2
as x could be >0 or <0 ..........insuff....

Ans A

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Re: Is x < 0? (1) x^3 + x^2 + x + 2 = 0 (2) x^2 - x - 2 < 0 [#permalink]

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New post 08 Oct 2016, 02:15
You can solve the equation algebraically as well.

1) x^3+x^2+x+2=0

on solving you get the following values of x :- -2/-3,-2.In all cases x<0. Therefore x must be <0. Statement is Sufficient.

2) on Solving statement number 2, we get 2 values of x, x<2 OR x<-1. Two different solns. i.e X can be +ve and X can be -ve. Therefore,statement is not sufficient.

Ans: A

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Re: Is x < 0? (1) x^3 + x^2 + x + 2 = 0 (2) x^2 - x - 2 < 0 [#permalink]

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New post 13 Oct 2016, 02:00
idontknowwhy94 wrote:
Is x<0?

1) \(x^{3} + x^{2}\) + x + 2 = 0
2) \(x^{2}\)-x-2 < 0


Keep the Kudos coming in and let the questions come out


Responding to a pm:
Quote:
For statement 1 it is clear that x >= 0 but when I tested -ve values I didn't find any which can satisfy the equation. Though I knew that x < 0, I continuously tried to find some -ve value that satisfied the equation. But it was a waste of time to find the value. How can I crack these type of traps on GMAT or it is my lack of understanding of concepts.


In stmnt 1, note that the point is that x cannot be positive or 0. So it has to be negative (since it has to be real).
If x is positive, all terms are positive and their sum cannot be 0.
If x is 0, you get 2 = 0 which is not valid.
So x has to be negative. No need to look for the actual value. It will not be an integer.

\(x^{3} + x^{2}+ x + 2 = 0\)
If you plug in x = -1, you get 1 = 0
If you plug in x = -2, you get -4 = 0
So somewhere in between -1 to -2, x will take a value such that you will get 0 = 0
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Re: Is x < 0? (1) x^3 + x^2 + x + 2 = 0 (2) x^2 - x - 2 < 0 [#permalink]

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New post 19 Oct 2017, 19:02
idontknowwhy94 wrote:
Is x<0?

1) \(x^{3} + x^{2}\) + x + 2 = 0
2) \(x^{2}\)-x-2 < 0


Keep the Kudos coming in and let the questions come out


1) \(x^{3} + x^{2}\) + x + 2 = 0
=> x^2(x+1) + (x + 1) + 1 =0
=> (x+1)(x^2+1) + 1 = 0
=> (x+1)(x^2+1) = -1

RHS= -1 < 0 => LHS <0. As x^2+1 > 0, then x +1 <0 => x < -1 => x <0 (suff)

2)2) \(x^{2}\)-x-2 < 0
=> -1 < x < 2
=> Insuff

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Re: Is x < 0? (1) x^3 + x^2 + x + 2 = 0 (2) x^2 - x - 2 < 0   [#permalink] 19 Oct 2017, 19:02
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