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 350,000 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:

Re: For what range of values of 'x' ? [#permalink]
03 Sep 2010, 07:17

2

This post received KUDOS

15x - 2/x > 1

Case1 : x > 0 Multiply both sides by "x" => 15x^2 - 2 > x => 15x^2 - x - 2 > 0 (solve the quadratic eqn for 15x^2 - x- 2 = 0 , it gives you : x = -1/3 or 2/5) Since x > 0 , so x > 2/5

Case 2 : x < 0 Multiply both sides by "x" => 15x^2 - 2 < x => 15x^2 - x - 2 < 0 Since, x < 0 so, the range of permissible value is : -1/3 < x < 0.

Re: For what range of values of 'x' ? [#permalink]
03 Oct 2010, 12:15

I do not understand why 2 cases are considered in the above explanation ? Only if we had |x| should the 2 cases be considered, am I right ? _________________

Please give me kudos, if you like the above post. Thanks.

This works well, however I would suggest to rely on quadratic equation formula: \(-b(+-)\sqrt{D}/2a\), as you don't need to think, how to factorize an equation?

Try yourself to factorize the equation above vs using the formula (Unfortunately, if the equation doesn't has a fractional value (which I expect rarely could be the case in GMAT) , for example when D in above formula is irrational, then you might be trying to factorize the fraction forever.)

Re: For what range of values of 'x' ? [#permalink]
12 Aug 2013, 10:48

What I have learnt, if x is in denom, take LCM after bringing all element at one side.

If we have more than 2 roots, then we can use following method to know the range.

Attachment:

Solution.jpg [ 16.75 KiB | Viewed 2279 times ]

_________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Re: For what range of values of 'x' ? [#permalink]
26 Nov 2013, 19:46

anshumishra wrote:

15x - 2/x > 1

Case1 : x > 0 Multiply both sides by "x" => 15x^2 - 2 > x => 15x^2 - x - 2 > 0 (solve the quadratic eqn for 15x^2 - x- 2 = 0 , it gives you : x = -1/3 or 2/5) Since x > 0 , so x > 2/5

Case 2 : x < 0 Multiply both sides by "x" => 15x^2 - 2 < x => 15x^2 - x - 2 < 0 Since, x < 0 so, the range of permissible value is : -1/3 < x < 0.

In Case 1 where x > 0, I agree. I calculated x > (-1/3) and x > 2/5, and because x > 0, it must be x > 2/5.

In Case 2 where x < 0, I don't understand why you say that the range must be x > -1/3 and x < 0, since the values that satisfy the inequality "15x^2 - x - 2 < 0" show that x < -1/3 and x < 2/5.

Re: For what range of values of 'x' ? [#permalink]
02 Mar 2014, 00:22

TooLong150 wrote:

anshumishra wrote:

15x - 2/x > 1

Case1 : x > 0 Multiply both sides by "x" => 15x^2 - 2 > x => 15x^2 - x - 2 > 0 (solve the quadratic eqn for 15x^2 - x- 2 = 0 , it gives you : x = -1/3 or 2/5) Since x > 0 , so x > 2/5

Case 2 : x < 0 Multiply both sides by "x" => 15x^2 - 2 < x => 15x^2 - x - 2 < 0 Since, x < 0 so, the range of permissible value is : -1/3 < x < 0.

In Case 1 where x > 0, I agree. I calculated x > (-1/3) and x > 2/5, and because x > 0, it must be x > 2/5.

In Case 2 where x < 0, I don't understand why you say that the range must be x > -1/3 and x < 0, since the values that satisfy the inequality "15x^2 - x - 2 < 0" show that x < -1/3 and x < 2/5.

Re: For what range of values of 'x' ? [#permalink]
02 Mar 2014, 04:03

Expert's post

siriusblack1106 wrote:

TooLong150 wrote:

anshumishra wrote:

15x - 2/x > 1

Case1 : x > 0 Multiply both sides by "x" => 15x^2 - 2 > x => 15x^2 - x - 2 > 0 (solve the quadratic eqn for 15x^2 - x- 2 = 0 , it gives you : x = -1/3 or 2/5) Since x > 0 , so x > 2/5

Case 2 : x < 0 Multiply both sides by "x" => 15x^2 - 2 < x => 15x^2 - x - 2 < 0 Since, x < 0 so, the range of permissible value is : -1/3 < x < 0.

In Case 1 where x > 0, I agree. I calculated x > (-1/3) and x > 2/5, and because x > 0, it must be x > 2/5.

In Case 2 where x < 0, I don't understand why you say that the range must be x > -1/3 and x < 0, since the values that satisfy the inequality "15x^2 - x - 2 < 0" show that x < -1/3 and x < 2/5.

Re: For what range of values of 'x' ? [#permalink]
02 Mar 2014, 17:02

Expert's post

TooLong150 wrote:

anshumishra wrote:

15x - 2/x > 1

Case1 : x > 0 Multiply both sides by "x" => 15x^2 - 2 > x => 15x^2 - x - 2 > 0 (solve the quadratic eqn for 15x^2 - x- 2 = 0 , it gives you : x = -1/3 or 2/5) Since x > 0 , so x > 2/5

Case 2 : x < 0 Multiply both sides by "x" => 15x^2 - 2 < x => 15x^2 - x - 2 < 0 Since, x < 0 so, the range of permissible value is : -1/3 < x < 0.

In Case 1 where x > 0, I agree. I calculated x > (-1/3) and x > 2/5, and because x > 0, it must be x > 2/5.

In Case 2 where x < 0, I don't understand why you say that the range must be x > -1/3 and x < 0, since the values that satisfy the inequality "15x^2 - x - 2 < 0" show that x < -1/3 and x < 2/5.

After you determine boundary points, you still need to test the potential +ve/-ve ranges per the inequality _________________

Both denominator and nominator positive --> \(x>\frac{2}{5}\); Both denominator and nominator negative --> \(-\frac{1}{3}<x<0\);

So inequality holds true in the ranges: \(-\frac{1}{3}<x<0\) and \(x>\frac{2}{5}\).

Answer: D.

Hi Bunuel, I thought that we shouldn't multiply an inequality with a variable whose range of value could be +ive and/or -ive.

I didn't multiply.

\(15x-\frac{2}{x}>1\) --> subtract 1: \(15x-\frac{2}{x}-1>0\) --> put in common denominator: \(\frac{15x^2-x-2}{x}>0\) --> factorize: \(\frac{(x+\frac{1}{3})(x-\frac{2}{5})}{x}>0\).

Re: For what range of values of 'x' ? [#permalink]
10 Jun 2014, 07:51

anshumishra wrote:

15x - 2/x > 1

Case1 : x > 0 Multiply both sides by "x" => 15x^2 - 2 > x => 15x^2 - x - 2 > 0 (solve the quadratic eqn for 15x^2 - x- 2 = 0 , it gives you : x = -1/3 or 2/5) Since x > 0 , so x > 2/5

Case 2 : x < 0 Multiply both sides by "x" => 15x^2 - 2 < x => 15x^2 - x - 2 < 0 Since, x < 0 so, the range of permissible value is : -1/3 < x < 0.

A graphical approach supplements this solution quite well: The roots are the blue dots, the yellow line separates the two cases, and the red lines are where the inequality holds true.

We assume two cases: x>0 and x<0.

For the first case, x>0, we want to see where 15x^2 -x-2 >0. Solving for the roots using the quadratic formula, we find roots when x = 2/5 and x = -1/3. Thus, for x>0, 15x^2 -x-2 >0 when x>2/5.

For the second case, x<0, we want to see where 15x^2 -x-2 <0. Again, using the roots, we know that the 15x^2 -x-2 < 0 when x > -1/3. We also assumed x<0 so we must include that in our solution.

Both denominator and nominator positive --> \(x>\frac{2}{5}\); Both denominator and nominator negative --> \(-\frac{1}{3}<x<0\);

So inequality holds true in the ranges: \(-\frac{1}{3}<x<0\) and \(x>\frac{2}{5}\).

Answer: D.

Hi Sorry for asking basic question. But do you have detail link to solve complicated quadratic equations like this: 15x2-x-2 ? I don't know how to solve this????

Both denominator and nominator positive --> \(x>\frac{2}{5}\); Both denominator and nominator negative --> \(-\frac{1}{3}<x<0\);

So inequality holds true in the ranges: \(-\frac{1}{3}<x<0\) and \(x>\frac{2}{5}\).

Answer: D.

Hi Sorry for asking basic question. But do you have detail link to solve complicated quadratic equations like this: 15x2-x-2 ? I don't know how to solve this????

Michigan Ross: Center for Social Impact : The Center for Social Impact provides leaders with practical skills and insight to tackle complex social challenges and catalyze a career in...

The Importance of Financial Regulation : Before immersing in the technical details of valuing stocks, bonds, derivatives and companies, I always told my students that the financial system is...

The following pictures perfectly describe what I’ve been up to these days. MBA is an extremely valuable tool in your career, no doubt, just that it is also...