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Is x^2 greater than x ? [#permalink]
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20 Jun 2012, 10:44
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38% (02:18) correct
62% (01:20) wrong based on 61 sessions
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Is x^2 greater than x ? (1) x^2 > x^3. (2) x^2 > x^4. I already solved it, but I tried to solve it by also the following method, and I don't know in which part I am wrong. (1) \(x^2 > x^3\). \(x^2  x^3 > 0\) \(x^2 * (1x) > 0\) So, we get: x= 0 and x = 1 (+)0()1(+) Therefore, the intervals are x<0 and x>1 However, if we solve it in this way: \(x^2 > x^3\) We divide both sides by \(x^2\): \(1 > x\) The interval is not the same. Please, tell me in which part of my first way I am wrong. Thanks!
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Re: Is x^2 greater than x ? [#permalink]
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20 Jun 2012, 12:32
metallicafan wrote: \(x^2 > x^3\). \(x^2  x^3 > 0\) \(x^2 * (1x) > 0\) So, we get: x= 0 and x = 1
(+)0()1(+)
Hi, When you say, \(x^2 (1x) > 0\) if \(x\neq 0\) then,\( x^2 >0\) for any real value of x. thus, inequality reduces to \(1x>0\) or \(x < 1\) Also, to use the number line method, you should change the coefficient of x as positive (for simplicity) \(x^2 (x1) < 0\) (multiply by 1 and reverse the sign) now, use the number line;  ()1 (+) Please go through the following post:Solving inequalitiesRegards,



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Re: Is x^2 greater than x ? [#permalink]
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20 Jun 2012, 12:41
Hi,
Is \(x^2>x\)? or \(x^2x>0\)? or \(x(x1)>0\)? or x<0 or x>1?
Detailed solution: Using (1), \(x^2>x^3\) or \(x^2(1x)>0\) or x < 1, for x = 0.5 \(x^2=0.25<x\) but for x=1, \(x^2=1>x\). Thus, Insufficient.
Using (2), \(x^2>x^4\) or \(x^2(1x^2)>0\) or \(x^2(1x)(1+x)>0\) or \(x^2(x1)(1+x)<0\) (multiplied by 1) (+)(1)()(+1)(+) or 1<x<1 Again, for x = 0.5 \(x^2=0.25<x\) but for x=0.5, \(x^2=0.25>x\). Thus, Insufficient.
Even after using (1) & (2), we have 1 < x <1, Insufficient.
Answer is (E),
Regards,



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Re: Is x^2 greater than x ? [#permalink]
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20 Jun 2012, 13:32
Thank you both! It seems that I just memorized the method and didn't analyze it well. I think that I get it now. And also I think that I have found another way that also can be helpful to you. Please, confirm if I am Ok. Please, Bunuel, take a look too. (1) \(x^2 > x^3\). \(x^2  x^3 > 0\) \(x^2 * (1x) > 0\) So, we get: x= 0 and x = 1 If \(x>1\) > \(x^2 * (1x) < 0\) If \(0<x<1\) > \(x^2 * (1x) > 0\) If \(x<0\) > \(x^2 * (1x) > 0\) So, we get:  (+)0 (+)1 () Therefore, the solution is x < 1.The same solution using this method: \(x^2 > x^3\) We divide by \(x^2\) \(1 > x\) \(x < 1\). What do you think?
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Re: Is x^2 greater than x ? [#permalink]
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20 Jun 2012, 13:43
metallicafan wrote: If \(x>1\) > \(x^2 * (1x) < 0\) If \(0<x<1\) > \(x^2 * (1x) > 0\) If \(x<0\) > \(x^2 * (1x) > 0\)
Hi, As you can see, the sign only depends on 1x, so, even powers can be ignored here. Regards,



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Re: Is x^2 greater than x ? [#permalink]
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21 Jun 2012, 01:03
Is x^2 greater than x ?Question: is \(x^2>x\)? > is \(x(x1)>0\)? is \(x<0\) or \(x>1\)? (1) x^2 > x^3. Now, from this statement we know that \(x\neq{0}\), so \(x^2>0\) and we can safely reduce by it to get \(1>x\) (\(x<1\)). So, finally we have that given inequality holds true for \(x<0\) and \(0<x<1\) (remember we should exclude 0 from the range, since if \(x=0\) then the given inequality doesn't hold true). Not sufficient. (2) x^2 > x^4. Apply the same logic here: we know that \(x\neq{0}\), so \(x^2>0\) and we can safely reduce by it to get \(1>x^2\) (\(x^2<1\)) > \(1<x<1\). So, finally we have that given inequality holds true for \(1<x<0\) and \(0<x<1\) (remember we should exclude 0 from the range, since if \(x=0\) then the given inequality doesn't hold true). Not sufficient.Not sufficient. (1)+(2) Intersection of the ranges from (1) and (2) gives: \(1<x<0\) and \(0<x<1\), so we can still have an YES answer (consider \(x=0.5\)) as well as a NO answer (consider \(x=0.5\)). Answer: E. Hope it's clear.
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Re: Is x^2 greater than x ? [#permalink]
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21 Jun 2012, 01:07
cyberjadugar wrote: Hi,
Is \(x^2>x\)? or \(x^2x>0\)? or \(x(x1)>0\)? or x<0 or x>1?
Detailed solution: Using (1), \(x^2>x^3\) or \(x^2(1x)>0\) or x < 1, for x = 0.5 \(x^2=0.25<x\) but for x=1, \(x^2=1>x\). Thus, Insufficient.
Using (2), \(x^2>x^4\) or \(x^2(1x^2)>0\) or \(x^2(1x)(1+x)>0\) or \(x^2(x1)(1+x)<0\) (multiplied by 1) (+)(1)()(+1)(+) or 1<x<1 Again, for x = 0.5 \(x^2=0.25<x\) but for x=0.5, \(x^2=0.25>x\). Thus, Insufficient.
Even after using (1) & (2), we have 1 < x <1, Insufficient.
Answer is (E),
Regards, The red parts are not correct. You should exclude zero, from all the ranges, since if \(x=0\), then neither of statements hold true. Hope it's clear.
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Re: Is x^2 greater than x ? [#permalink]
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21 Jun 2012, 01:14
Hi, Thanks for pointing it out! Regards, Bunuel wrote: cyberjadugar wrote: Hi,
Is \(x^2>x\)? or \(x^2x>0\)? or \(x(x1)>0\)? or x<0 or x>1?
Detailed solution: Using (1), \(x^2>x^3\) or \(x^2(1x)>0\) or x < 1, for x = 0.5 \(x^2=0.25<x\) but for x=1, \(x^2=1>x\). Thus, Insufficient.
Using (2), \(x^2>x^4\) or \(x^2(1x^2)>0\) or \(x^2(1x)(1+x)>0\) or \(x^2(x1)(1+x)<0\) (multiplied by 1) (+)(1)()(+1)(+) or 1<x<1 Again, for x = 0.5 \(x^2=0.25<x\) but for x=0.5, \(x^2=0.25>x\). Thus, Insufficient.
Even after using (1) & (2), we have 1 < x <1, Insufficient.
Answer is (E),
Regards, The red parts are not correct. You should exclude zero, from all the ranges, since if \(x=0\), then neither of statements hold true. Hope it's clear.



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Is x² greater than x? [#permalink]
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30 Dec 2016, 06:27
Is \(x^2\) greater than \(x\)? (1) \(x^2\) is greater than \(x^3\). (2) \(x^2\) is greater than \(x^4\). OA:
Last edited by paddy41 on 30 Dec 2016, 06:38, edited 1 time in total.



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Re: Is x² greater than x? [#permalink]
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30 Dec 2016, 06:36
Quote: Explanation: \(x^2\) is greater than x for all numbers except for those values of x between 0 and 1. Thus, we need to know whether or not x falls in that range.
Statement (1) is insucient. To simplify,divide both sides by \(x^2\), resulting in 1 > x. If that's true, x could be between 0 and 1, but it could also be less than 0.
Statement (2) is also insucient. Again, simplify by dividing by \(x^2\), which gives you 1 > \(x^2\). Thus, x could be any number between 1 and 1. Again, it could be between 0 and 1, but it could also be between 1 and 0.
Taken together, it's still insucient. Both statements allow for the possibility that x is between 0 and 1, but both statements also make it possible that x is between 1 and 0. Choice (E) is correct. Just one question about the highlighted part of the explanation: It is clear that \(x^2\) is always positive. However, is the division by \(x^2\) a valid inequality operation without knowing whether x is positive or negative?



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Re: Is x^2 greater than x ? [#permalink]
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30 Dec 2016, 08:18



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Re: Is x^2 greater than x ? [#permalink]
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30 Dec 2016, 08:20
paddy41 wrote: Quote: Explanation: \(x^2\) is greater than x for all numbers except for those values of x between 0 and 1. Thus, we need to know whether or not x falls in that range.
Statement (1) is insucient. To simplify,divide both sides by \(x^2\), resulting in 1 > x. If that's true, x could be between 0 and 1, but it could also be less than 0.
Statement (2) is also insucient. Again, simplify by dividing by \(x^2\), which gives you 1 > \(x^2\). Thus, x could be any number between 1 and 1. Again, it could be between 0 and 1, but it could also be between 1 and 0.
Taken together, it's still insucient. Both statements allow for the possibility that x is between 0 and 1, but both statements also make it possible that x is between 1 and 0. Choice (E) is correct. Just one question about the highlighted part of the explanation: It is clear that \(x^2\) is always positive. However, is the division by \(x^2\) a valid inequality operation without knowing whether x is positive or negative? When dividing/multiplying an inequality by a variable we need to know its sign. If it's positive we should keep the sign and if its negative we should flip the sign. That's the rule. We know that x^2 is positive, so we can safely multiply/divide an inequality by it (so in this case it does not matter whether x itself is positive or negative).
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Re: Is x^2 greater than x ?
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