Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

what I have been realizing ( i feel its wrong now ) is that

x ( x - 1 ) < 0 Means

either x< 0 or x - 1 < 0 or x < 1

so x < 0 or x < 1 how did u get 0 < x < 1 ?

\(x(x-1)<0\) --> roots are \(x=0\) and \(x=1\) --> "<" sign indicates that the solution lies between the roots: \(0<x<1\):

Attachment:

1.gif [ 891 Bytes | Viewed 2491 times ]

If it were: \(x(x-1)>0\) --> the same roots: \(x=0\) and \(x=1\) --> ">" sign indicates that the solution lies to the left of the smaller root and to the right of the larger root: x<0 or x>1:

Attachment:

2.gif [ 997 Bytes | Viewed 2494 times ]

Follow the links in my previous post for more on this subject.

Now then factorizing we have: x^2 (x-1)(x+1) , we only need to take care of (x-1)(x+1) since x^2 is always >=0

But then I get the range -1<x<1 which is not correct given that for instance when x=0 clearly inequality does not hold

Would you mind showing me where I went wrong here?

Thanks! Cheers J

Note that the inequality is \(x^4 - x < 0\)

So on factoring you get \(x (x^3 - 1) < 0\) \(x(x - 1)(x^2 + x + 1) < 0\)

Now, (x^2 + x + 1) is always positive. It has no real roots. Hence we might as well assume it to be a positive number and ignore it for the sake of getting our transition points. Transition points are 0 and 1. Since we want the negative region, the inequality will be negative between 0 and 1.

0 < x < 1

Though, you should try to understand the relation between x, x^2 and x^3 in the ranges < -1, -1 < x < 0, 0 < x <1 and x > 1 so that you don't need to use much algebra for such questions.
_________________

Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t [#permalink]

Show Tags

21 Jun 2015, 02:31

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t [#permalink]

Show Tags

22 Jun 2015, 13:53

1

This post received KUDOS

Is x > x^2 ?

This would hold true if x<1

(1) x is greater than x^3 x can be a negative integer. For example x = -2. -2^3 = -8. But -2 is not greater than -2^2 = 4. Not sufficient. (2) x is greater than x^4 For this to hold true, x would have to be less than 1. sufficient

Re: Is x > x^2 ? (1) x is greater than x^3 (2) x is greater t [#permalink]

Show Tags

08 Jul 2017, 03:45

VeritasPrepKarishma wrote:

jlgdr wrote:

Good that you bumped onto this thread Bunuel

Here's my question

On second statement how do we do critical points?

So we have x>x^4

Therefore x^4-x<0

Now then factorizing we have: x^2 (x-1)(x+1) , we only need to take care of (x-1)(x+1) since x^2 is always >=0

But then I get the range -1<x<1 which is not correct given that for instance when x=0 clearly inequality does not hold

Would you mind showing me where I went wrong here?

Thanks! Cheers J

Note that the inequality is \(x^4 - x < 0\)

So on factoring you get \(x (x^3 - 1) < 0\) \(x(x - 1)(x^2 + x + 1) < 0\)

Now, (x^2 + x + 1) is always positive. It has no real roots. Hence we might as well assume it to be a positive number and ignore it for the sake of getting our transition points. Transition points are 0 and 1. Since we want the negative region, the inequality will be negative between 0 and 1.

0 < x < 1

Though, you should try to understand the relation between x, x^2 and x^3 in the ranges < -1, -1 < x < 0, 0 < x <1 and x > 1 so that you don't need to use much algebra for such questions.

Hello VeritasPrepKarishma, Its been a while since i studies all such formulaes like (a+1)^3 or a^3-1 or (a+b+c)^2 etc, so should I revisit each and every of these formulaes?

Coming back to the question, Can I say that since x>x^4 and since x^4>0 always therefore x MUST be >0 Thus, x>0 and x^4-x<0 => x(x^3-1)<0 => x^3-1<0 => x^3<1 Now, since x >0 then x^3 will be less than 1 only when 0<x<1; since for any real number greater than 1, the cube is greater than the number itself.

Now then factorizing we have: x^2 (x-1)(x+1) , we only need to take care of (x-1)(x+1) since x^2 is always >=0

But then I get the range -1<x<1 which is not correct given that for instance when x=0 clearly inequality does not hold

Would you mind showing me where I went wrong here?

Thanks! Cheers J

Note that the inequality is \(x^4 - x < 0\)

So on factoring you get \(x (x^3 - 1) < 0\) \(x(x - 1)(x^2 + x + 1) < 0\)

Now, (x^2 + x + 1) is always positive. It has no real roots. Hence we might as well assume it to be a positive number and ignore it for the sake of getting our transition points. Transition points are 0 and 1. Since we want the negative region, the inequality will be negative between 0 and 1.

0 < x < 1

Though, you should try to understand the relation between x, x^2 and x^3 in the ranges < -1, -1 < x < 0, 0 < x <1 and x > 1 so that you don't need to use much algebra for such questions.

Hello VeritasPrepKarishma, Its been a while since i studies all such formulaes like (a+1)^3 or a^3-1 or (a+b+c)^2 etc, so should I revisit each and every of these formulaes?

Coming back to the question, Can I say that since x>x^4 and since x^4>0 always therefore x MUST be >0 Thus, x>0 and x^4-x<0 => x(x^3-1)<0 => x^3-1<0 => x^3<1 Now, since x >0 then x^3 will be less than 1 only when 0<x<1; since for any real number greater than 1, the cube is greater than the number itself.

Thus i can say x<0<1. Hence sufficient.

Is my reasoning correct?

Regards

Some important algebraic identities:

\((x + y)^2 = x^2 + y^2 + 2xy\)

\((x - y)^2 = x^2 + y^2 - 2xy\)

\(x^2 - y^2 = (x + y)*(x - y)\)

Others, if needed, you can calculate there and then.

As for this question, yes you can arrive at 0 < x< 1 through that method too. No problem in the reasoning though I would suggest you to look at the other methods discussed above too. They are more generic.
_________________

As for this question, yes you can arrive at 0 < x< 1 through that method too. No problem in the reasoning though I would suggest you to look at the other methods discussed above too. They are more generic.

Thank you Karishma You guyz are doing a very good job. Will look at the other methods described above as well. Regards
_________________