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:

For everyone interested in modulus practices, I have listed, [#permalink]

Show Tags

03 Dec 2006, 05:22

2

This post received KUDOS

13

This post was BOOKMARKED

For everyone interested in modulus practices, I have listed, at the bottom of this topic, several threads of modulus in a graduated level of difficulty.

Keep in mind that it's for trainning purposes and that the most advanced posts are not reprensative of the real GMAT.

FOR THE THIRD PROBLEM , I BELIEVE THE ANSWER MISS THE RANGE

-2<X<1

I'm ok thanks... it follows the way

Great catch .... Yes, I did a "silly" mistake 2-4 = 6 and ended up with a wrong x^2 + x + 6 .... I have corrected my post following your catch... very good :D

o If x < 3, then the inequation becomes: -(x-3)<2x-4 <=> 7 < 3x <x> 7/3 => 7/3 < x <3>= 3, then the inequation becomes:[/b] x-3<2x-4 <x> 1 => x >= 3

Finally, we have : x > 7/3

Hi Fig,

Why have you mentioned that x >=3 and not that x > 3.. I see absolute Questions which sometimes have the >= sign and sometimes the > sign only.. Might be a very mindless question but hope you would answer..

o If x < 3, then the inequation becomes: -(x-3)<2x-4 <=> 7 < 3x <x> 7/3 => 7/3 < x <3>= 3, then the inequation becomes:[/b] x-3<2x-4 <x> 1 => x >= 3

Finally, we have : x > 7/3

Hi Fig,

Why have you mentioned that x >=3 and not that x > 3.. I see absolute Questions which sometimes have the >= sign and sometimes the > sign only.. Might be a very mindless question but hope you would answer..

Not meaningless but rather precised question ...

Actually, it's just to avoid adding 1 solution alone from another interval... and so to meet a higher degree of "perfection" by making it a smoother way to the final solution and a straightforward simplicity to understand

Note that nothing is wrong if we say > or >= as soon as we do not miss to consider all values of x, especially here the ones that limit 2 intervals

As we are working positive value and as the function f(x) = 1/x decreases when x increases on the domain of x > 0, we can imply that:

|x-2| =< 1/4

can someone please explain why this happens. Why can't I just use |1/x-2| >=4. However, when x<2, I get x<=7/4. Also I agree with everyone else as the range include 2, doesn't lie outside of our limitations.

Since the sign of a*x^2+b*x+c is that of a for any values of x not between the 2 roots, solutions of the equation a*x^2+b*x+c = 0 if they exist.

Thus, o Sign((x+2)(x-2)) > 0 when x > 2 or x < -2 o Sign((x+2)(x-2)) < 0 when -2 < x < 2

Trying hard to grasp this concept, but don't think I'm understanding what you're trying to convey here. Ideally, I would try to solve this as such:

Given that |x^2 - 4| is same as |(x+2)(x-2)|

Solving for x in |x^2 - 4| > x + 2 in the following two scenarios: 1. When (x+2)(x-2) >= 0 ==> x^2 - 4 > x + 2 ==> x>-1 and x>2

2. When (x+2)(x-2) < 0 ==> -(x^2 - 4) > x + 2 ==> 4 - x^2 > x + 2 ==> x<1 and x<-2

Does the combination of these solutions for x represent the final answer? Am I missing a step?

First of all you won't see such equation on the GMAT, but anyway:

\(|x^2-4|>x+2\).

We should consider the cases when \(x^2-4<0\) and \(x^2-4\geq{0}\):

\(x^2-4<0\) for \(-2<x<2\) --> \(-(x^2-4)>x+2\) --> \(x^2+x-2<0\) --> \((x+2)(x-1)<0\)--> \(-2<x<1\). Since we are considering the range \(-2<x<2\) then the solution for this case will be intersections of these two (common part): \(-2<x<1\);

\(x^2-4\geq{0}\) for \(x\leq{-2}\) or \(x\geq{2}\) --> \((x^2-4)>x+2\) --> \(x^2-x-6>0\) --> \((x+2)(x-3)>0\)--> \(x<-2\) or \(x>3\). Since we are considering the range \(x\leq{-2}\) or \(x\geq{2}\) then the solution for this case will be intersections of these four (common parts): \(x<-2\) or \(x>3\);

Finally the ranges of \(x\) for which \(|x^2-4|>x+2\) true are: \(x<-2\), \(-2<x<1\) and \(x>3\).

Re: For everyone interested in modulus practices, I have listed, [#permalink]

Show Tags

27 Aug 2014, 04:40

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: For everyone interested in modulus practices, I have listed, [#permalink]

Show Tags

02 Feb 2016, 23:21

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: For everyone interested in modulus practices, I have listed, [#permalink]

Show Tags

31 Aug 2017, 08:18

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.
_________________

Today we’re excited to announce two new premium themes: Small Business and Photo Blog . Small Business Small Business is a new premium theme for your entrepreneurial endeavors. At...

As part of our commitment to privacy and transparency, we’re updating our Privacy Policy . We want to give you more and clearer information about how we collect and use...

Product Leadership is a book that distills the best practices of Product Management into one source. Mark Ericsson is the founder of Product Tank and Mind the Product and...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...