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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.

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?

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?

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.

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?

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!

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.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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