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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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Manager
Joined: 03 Oct 2016
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Concentration: Technology, General Management
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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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Updated on: 07 Oct 2016, 10:34
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Question Stats:

64% (02:13) correct 36% (02:29) wrong based on 113 sessions

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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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Originally posted by idontknowwhy94 on 07 Oct 2016, 09:56.
Last edited by idontknowwhy94 on 07 Oct 2016, 10:34, 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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07 Oct 2016, 10:32
Guys IHO ,

the answer is 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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07 Oct 2016, 10:47
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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08 Oct 2016, 03:15
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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13 Oct 2016, 03:00
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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19 Oct 2017, 20:02
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]

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26 Mar 2019, 06:34
mohit_w wrote:
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

Can you provide the solution for x^3+x^2+x+2=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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26 Mar 2019, 20:42
lavverma wrote:
mohit_w wrote:
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

Can you provide the solution for x^3+x^2+x+2=0

This equation is not conducive to being solved easily. On the Veritas blog, see how you solve a third degree equation, if needed: https://www.veritasprep.com/blog/2013/0 ... rd-degree/

You cannot get the first root by easy hit n trial here.

You only need to observe here that all terms are positive. So the root must be negative as explained in my comment above: https://gmatclub.com/forum/is-x-0-1-x-3 ... l#p1747591
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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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31 Mar 2019, 00:37
idontknowwhy94 wrote:
Is x < 0?

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

(2) x^2 - x - 2 < 0

#1
x^3 + x^2 + x + 2 = 0
x has to be in between -1 to -2 so sufficient
#2
solving expression we get x=-1 and x=2
insufficient
IMO A
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Re: Is x < 0? (1) x^3 + x^2 + x + 2 = 0 (2) x^2 - x - 2 < 0   [#permalink] 31 Mar 2019, 00:37
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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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