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\(x^2 > x^3\). \(x^2 - x^3 > 0\) \(x^2 * (1-x) > 0\) So, we get: x= 0 and x = 1

-----(+)------0----(-)-------1------(+)------

Hi,

When you say, \(x^2 (1-x) > 0\) if \(x\neq 0\) then,\(x^2 >0\) for any real value of x. thus, inequality reduces to \(1-x>0\) or \(x < 1\)

Also, 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---(+)---------

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.

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) \(x^2 > x^3\). \(x^2 - x^3 > 0\) \(x^2 * (1-x) > 0\) So, we get: x= 0 and x = 1

If \(x>1\) -----> \(x^2 * (1-x) < 0\) If \(0<x<1\) -----> \(x^2 * (1-x) > 0\) If \(x<0\) -----> \(x^2 * (1-x) > 0\)

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?
_________________

"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."

My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html

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 \(x=0.5\)).

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.

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.

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?

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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