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Is x^2 greater than x ?

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Is x^2 greater than x ? [#permalink] New post 20 Jun 2012, 09:44
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A
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D
E

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38% (01:59) correct 62% (01:52) wrong based on 13 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 > 0
x^2 * (1-x) > 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!
[Reveal] Spoiler: OA

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Re: Is x^2 greater than x ? [#permalink] New post 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---(+)---------

Please go through the following post:
Solving inequalities

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Re: Is x^2 greater than x ? [#permalink] New post 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.

Answer is (E),

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Re: Is x^2 greater than x ? [#permalink] New post 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?
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Re: Is x^2 greater than x ? [#permalink] New post 20 Jun 2012, 12:43
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.

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Re: Is x^2 greater than x ? [#permalink] New post 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).

Answer: E.

Hope it's clear.
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Re: Is x^2 greater than x ? [#permalink] New post 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.

Hope it's clear.
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Re: Is x^2 greater than x ? [#permalink] New post 21 Jun 2012, 00:14
Hi,

Thanks for pointing it out! :-D

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, 00:14
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