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# If x is an integer, is |x|>1.

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If x is an integer, is |x|>1. [#permalink]

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27 Apr 2012, 16:14
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45% (02:47) correct 55% (01:30) wrong based on 58 sessions

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If x is an integer, is |x|>1.

(1) (1-2x)(1+x) < 0
(2) (1-x)(1+2x) < 0

Can somebody please explain this question?

Thanks
Vikram
[Reveal] Spoiler: OA
Math Expert
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Kudos [?]: 80121 [2] , given: 10027

Re: If x is an integer, is |x|>1. [#permalink]

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27 Apr 2012, 22:22
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Expert's post
If x is an integer, is |x|>1.

(1) (1-2x)(1+x) < 0
(2) (1-x)(1+2x) < 0

Can somebody please explain this question?

Thanks
Vikram

This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach": x2-4x-94661.html#p731476

If x is an integer, is |x| > 1?

First of all: is $$|x| > 1$$ means is $$x<-1$$ (-2, -3, -4, ...) or $$x>1$$ (2, 3, 4, ...), so for YES answer $$x$$ can be any integer but -1, 0, and 1.

(1) (1 - 2x)(1 + x) < 0 --> rewrite as $$(2x-1)(x+1)>0$$ (so that the coefficient of x^2 to be positive after expanding): roots are $$x=-1$$ and $$x=\frac{1}{2}$$ --> "$$>$$" sign means that the given inequality holds true for: $$x<-1$$ and $$x>\frac{1}{2}$$. $$x$$ could still equal to 1, so not sufficient.

(2) (1 - x)(1 + 2x) < 0 --> rewrite as $$(x-1)(2x+1)>0$$: roots are $$x=-\frac{1}{2}$$ and $$x=1$$ --> "$$>$$" sign means that the given inequality holds true for: $$x<-\frac{1}{2}$$ and $$x>1$$. $$x$$ could still equal to -1, so not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is $$x<-1$$ and $$x>1$$. Sufficient.

This question is also discussed here: m14-72785.html

Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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Re: If x is an integer, is |x|>1. [#permalink]

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10 May 2012, 06:48
roots are x=-1 and x=1/2 and --> ">" sign means that the given inequality holds true for: x<-1 and x>1/2 ... can you please help me with this concept and what will happen if sign was "<"..further, will it be right in stating that when there is a positive sign, x is greater than the positive root and x is less than the negative root?
Math Expert
Joined: 02 Sep 2009
Posts: 34527
Followers: 6313

Kudos [?]: 80121 [0], given: 10027

Re: If x is an integer, is |x|>1. [#permalink]

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10 May 2012, 06:49
Expert's post
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This post was
BOOKMARKED
pavanpuneet wrote:
roots are x=-1 and x=1/2 and --> ">" sign means that the given inequality holds true for: x<-1 and x>1/2 ... can you please help me with this concept and what will happen if sign was "<"..further, will it be right in stating that when there is a positive sign, x is greater than the positive root and x is less than the negative root?

Explained here:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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Re: If x is an integer, is |x|>1. [#permalink]

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19 Jul 2012, 19:23
Bunuel wrote:
If x is an integer, is |x|>1.

(1) (1-2x)(1+x) < 0
(2) (1-x)(1+2x) < 0

Can somebody please explain this question?

Thanks
Vikram

This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach": x2-4x-94661.html#p731476

If x is an integer, is |x| > 1?

First of all: is $$|x| > 1$$ means is $$x<-1$$ (-2, -3, -4, ...) or $$x>1$$ (2, 3, 4, ...), so for YES answer $$x$$ can be any integer but -1, 0, and 1.

(1) (1 - 2x)(1 + x) < 0 --> rewrite as $$(2x-1)(x+1)>0$$ (so that the coefficient of x^2 to be positive after expanding): roots are $$x=-1$$ and $$x=\frac{1}{2}$$ --> "$$>$$" sign means that the given inequality holds true for: $$x<-1$$ and $$x>\frac{1}{2}$$. $$x$$ could still equal to 1, so not sufficient.

(2) (1 - x)(1 + 2x) < 0 --> rewrite as $$(x-1)(2x+1)>0$$: roots are $$x=-\frac{1}{2}$$ and $$x=1$$ --> "$$>$$" sign means that the given inequality holds true for: $$x<-\frac{1}{2}$$ and $$x>1$$. $$x$$ could still equal to -1, so not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is $$x<-1$$ and $$x>1$$. Sufficient.

This question is also discussed here: m14-72785.html

Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.

I have a question regarding the above solution let's say in statement 1, when you solve the inequality why do you say that x<-1 AND x >1/2

why is this an AND condition ....why not OR? If this were a quadratic equation x (can be) = 1/2 OR -1 OR both
For inequality why is the same thing an AND as opposed to OR?
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Solving compounded inequalities - any efficient approach? [#permalink]

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23 Oct 2013, 19:58
If x is an integer, is |x|>1?

(1) (1−2x)(1+x)<0

(2) (1−x)(1+2x)<0

Hi all - I tried this problem on a GMAT club test and I didn't really understand the method. Any quick approaches to finding the solution to each inequality? Is there any quick method to figure out the range of values of x for which statement 1 and 2 will be accurate?

Help appreciated! I'd like to know the quickest, most efficient way to approach such problems!

Cheers
Math Expert
Joined: 02 Sep 2009
Posts: 34527
Followers: 6313

Kudos [?]: 80121 [0], given: 10027

Re: Solving compounded inequalities - any efficient approach? [#permalink]

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24 Oct 2013, 01:05
sidvish wrote:
If x is an integer, is |x|>1?

(1) (1−2x)(1+x)<0

(2) (1−x)(1+2x)<0

Hi all - I tried this problem on a GMAT club test and I didn't really understand the method. Any quick approaches to finding the solution to each inequality? Is there any quick method to figure out the range of values of x for which statement 1 and 2 will be accurate?

Help appreciated! I'd like to know the quickest, most efficient way to approach such problems!

Cheers

Merging similar topics. Please refer to the solution above.

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Re: Solving compounded inequalities - any efficient approach?   [#permalink] 24 Oct 2013, 01:05
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