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Re: Is x^2 greater than x ? [#permalink]
20 Jun 2012, 11:32

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) > 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---(+)---------

Re: Is x^2 greater than x ? [#permalink]
20 Jun 2012, 11:41

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^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.

Re: Is x^2 greater than x ? [#permalink]
20 Jun 2012, 12:32

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 * (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."

Re: Is x^2 greater than x ? [#permalink]
21 Jun 2012, 00:03

Expert's post

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).

Re: Is x^2 greater than x ? [#permalink]
21 Jun 2012, 00:07

Expert's post

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.

Re: Is x^2 greater than x ? [#permalink]
21 Jun 2012, 00:14

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.