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If x is an integer, is x>1.
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27 Apr 2012, 15:14
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If x is an integer, is x>1. (1) (12x)(1+x) < 0 (2) (1x)(1+2x) < 0 Can somebody please explain this question? Thanks Vikram
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Re: If x is an integer, is x>1.
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27 Apr 2012, 21:22
vdadwal wrote: If x is an integer, is x>1.
(1) (12x)(1+x) < 0 (2) (1x)(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": x24x94661.html#p731476If 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 \((2x1)(x+1)>0\) (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 \((x1)(2x+1)>0\): 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. Answer: C. This question is also discussed here: m1472785.htmlSolving inequalities: x24x94661.html#p731476 ( check this one first) inequalitiestrick91482.htmldatasuffinequalities109078.htmlrangeforvariablexinagiveninequality109468.html?hilit=extreme#p873535everythingislessthanzero108884.html?hilit=extreme#p868863Hope it helps.
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Re: If x is an integer, is x>1.
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10 May 2012, 05:48
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?



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Re: If x is an integer, is x>1.
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10 May 2012, 05:49



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Re: If x is an integer, is x>1.
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19 Jul 2012, 18:23
Bunuel wrote: vdadwal wrote: If x is an integer, is x>1.
(1) (12x)(1+x) < 0 (2) (1x)(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": x24x94661.html#p731476If 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 \((2x1)(x+1)>0\) (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 \((x1)(2x+1)>0\): 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. Answer: C. This question is also discussed here: m1472785.htmlSolving inequalities: x24x94661.html#p731476 ( check this one first) inequalitiestrick91482.htmldatasuffinequalities109078.htmlrangeforvariablexinagiveninequality109468.html?hilit=extreme#p873535everythingislessthanzero108884.html?hilit=extreme#p868863Hope 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?
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23 Oct 2013, 18:58
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



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Re: Solving compounded inequalities  any efficient approach?
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24 Oct 2013, 00:05



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Re: If x is an integer, is x>1.
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23 Sep 2018, 08:57
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