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Is x^2 greater than x ? [#permalink]
 20 Jun 2012, 10:44
Question Stats:
36% (02:13) correct
63% (01:55) wrong based on 0 sessions
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 > 0x^2 * (1-x) > 0So, 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^3We divide both sides by x^2: 1 > xThe 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]
20 Jun 2012, 12:32
metallicafan wrote: x^2 > x^3.
x^2 - x^3 > 0
x^2 * (1-x) > 0
So, we get: x= 0 and x = 1
-----(+)------0----(-)-------1------(+)------
Hi, When you say, x^2 (1-x) > 0if x\neq 0 then, x^2 >0 for any real value of x. thus, inequality reduces to 1-x>0or x < 1Also, to use the number line method, you should change the coefficient of x as positive (for simplicity) x^2 (x-1) < 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]
20 Jun 2012, 12:41
Hi, Is x^2>x? or x^2-x>0? or x(x-1)>0? or x<0 or x>1? Detailed solution: Using (1), x^2>x^3or x^2(1-x)>0or x < 1, for x = 0.5 x^2=0.25<xbut for x=-1, x^2=1>x. Thus, Insufficient. Using (2), x^2>x^4or x^2(1-x^2)>0or x^2(1-x)(1+x)>0or x^2(x-1)(1+x)<0 (multiplied by -1) - (+)-------(-1)--- (-)-----(+1)----- (+)------ or -1<x<1 Again, for x = 0.5 x^2=0.25<xbut 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]
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 > 0x^2 * (1-x) > 0So, we get: x= 0 and x = 1 If x>1 -----> x^2 * (1-x) < 0If 0<x<1 -----> x^2 * (1-x) > 0If x<0 -----> x^2 * (1-x) > 0So, we get: ----- (+)------0---- (+)-------1------ (-)------ Therefore, the solution is x < 1.The same solution using this method: x^2 > x^3We divide by x^21 > xx < 1. What do you think?
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Re: Is x^2 greater than x ? [#permalink]
20 Jun 2012, 13:43
metallicafan wrote: If x>1 -----> x^2 * (1-x) < 0
If 0<x<1 -----> x^2 * (1-x) > 0
If x<0 -----> x^2 * (1-x) > 0
Hi, As you can see, the sign only depends on 1-x, so, even powers can be ignored here. Regards,
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Re: Is x^2 greater than x ? [#permalink]
21 Jun 2012, 01:03
Is x^2 greater than x ?Question: is x^2>x? --> is x(x-1)>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]
21 Jun 2012, 01:07
cyberjadugar wrote: Hi,
Is x^2>x?
or x^2-x>0?
or x(x-1)>0?
or x<0 or x>1?
Detailed solution:
Using (1),
x^2>x^3
or x^2(1-x)>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(1-x^2)>0
or x^2(1-x)(1+x)>0
or x^2(x-1)(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]
21 Jun 2012, 01:14
Hi, Thanks for pointing it out!  Regards, Bunuel wrote: cyberjadugar wrote: Hi,
Is x^2>x?
or x^2-x>0?
or x(x-1)>0?
or x<0 or x>1?
Detailed solution:
Using (1),
x^2>x^3
or x^2(1-x)>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(1-x^2)>0
or x^2(1-x)(1+x)>0
or x^2(x-1)(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]
21 Jun 2012, 01:14
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