Find all School-related info fast with the new School-Specific MBA Forum

It is currently 22 Aug 2014, 15:37

Close

GMAT Club Daily Prep

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

modulus inequalities

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
User avatar
Status: Still Struggling
Joined: 02 Nov 2010
Posts: 139
Location: India
GMAT Date: 10-15-2011
GPA: 3.71
WE: Information Technology (Computer Software)
Followers: 5

Kudos [?]: 11 [0], given: 8

modulus inequalities [#permalink] New post 19 Nov 2010, 08:48
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions
Hi Guys,

came across this problem without a OA. Please help me find a solution:

x^2 - |x| - 12
--------------- >= 2x
x-3

Please post your solutions along with explanation.

---------------------------------------------
Consider KUDOS if you like my post!!!
_________________

Appreciation in KUDOS please!
Knewton Free Test 10/03 - 710 (49/37)
Princeton Free Test 10/08 - 610 (44/31)
Kaplan Test 1- 10/10 - 630
Veritas Prep- 10/11 - 630 (42/37)
MGMAT 1 - 10/12 - 680 (45/34)

Expert Post
4 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4668
Location: Pune, India
Followers: 1075

Kudos [?]: 4788 [4] , given: 163

Re: modulus inequalities [#permalink] New post 20 Nov 2010, 12:07
4
This post received
KUDOS
Expert's post
krishnasty wrote:
Hi Guys,

came across this problem without a OA. Please help me find a solution:

x^2 - |x| - 12
--------------- >= 2x
x-3

Please post your solutions along with explanation.

---------------------------------------------
Consider KUDOS if you like my post!!!


I am assuming your question is: \frac{{x^2 - |x| - 12}}{{x - 3}} >= 2x
First of all, remember that we cannot cross multiply in this case. The reason is that we do not know the sign of (x - 3). To handle the 2x on the right hand side, bring it to the left hand side to get:

\frac{{x^2 - |x| - 12}}{{x - 3}} - 2x >= 0

\frac{{-x^2 - |x| + 6x - 12}}{{x - 3}} >= 0

Now take 2 cases.

Case 1: If x >= 0, |x| = x
\frac{{-x^2 - x + 6x - 12}}{{x - 3}} >= 0

\frac{{x^2 - 5x + 12}}{{x - 3}} <= 0 (We flipped the sign of left hand side and hence the inequality flipped too.)

Now, x^2 - 5x + 12 has no real roots. It will always be positive. (because b^2 - 4ac is negative so we know that it has no real roots. Put x = 0, you get 12 which is positive. So this quadratic will never touch the x axis and will always remain above it.)

So for the left hand side to be negative, x < 3. (x cannot be 3 because in that case the denominator will be 0.)
Since x has to be greater than 0, the only solution here is 0 <= x < 3.

Case 2: If x <= 0, |x| = -x
\frac{{-x^2 + x + 6x - 12}}{{x - 3}} >= 0

\frac{{x^2 - 7x + 12}}{{x - 3}} <= 0 (We flipped the sign of left hand side and hence the inequality flipped too.)

\frac{{(x - 4)(x - 3)}}{{x - 3}} <= 0
or x <= 4
Since x has to be less than(or equal to) 0, the only solution here is x <= 0.

Combining the solutions above, we get x < 3 as the final solution.

(I have assumed that you are comfortable with quadratic expressions and solving for inequalities once you have factors. If you are not, let me know and I will point you to the related theory.)
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Intern
Intern
User avatar
Joined: 27 Aug 2010
Posts: 30
Followers: 0

Kudos [?]: 3 [0], given: 4

Re: modulus inequalities [#permalink] New post 20 Nov 2010, 12:13
VeritasPrepKarishma wrote:
krishnasty wrote:
Hi Guys,

came across this problem without a OA. Please help me find a solution:

x^2 - |x| - 12
--------------- >= 2x
x-3

Please post your solutions along with explanation.

---------------------------------------------
Consider KUDOS if you like my post!!!


I am assuming your question is: \frac{{x^2 - |x| - 12}}{{x - 3}} >= 2x
First of all, remember that we cannot cross multiply in this case. The reason is that we do not know the sign of (x - 3). To handle the 2x on the right hand side, bring it to the left hand side to get:

\frac{{x^2 - |x| - 12}}{{x - 3}} - 2x >= 0

\frac{{-x^2 - |x| + 6x - 12}}{{x - 3}} >= 0

Now take 2 cases.

Case 1: If x >= 0, |x| = x
\frac{{-x^2 - x + 6x - 12}}{{x - 3}} >= 0

\frac{{x^2 - 5x + 12}}{{x - 3}} <= 0 (We flipped the sign of left hand side and hence the inequality flipped too.)

Now, x^2 - 5x + 12 has no real roots. It will always be positive. (because b^2 - 4ac is negative so we know that it has no real roots. Put x = 0, you get 12 which is positive. So this quadratic will never touch the x axis and will always remain above it.)

So for the left hand side to be negative, x < 3. (x cannot be 3 because in that case the denominator will be 0.)
Since x has to be greater than 0, the only solution here is 0 <= x < 3.

Case 2: If x <= 0, |x| = -x
\frac{{-x^2 + x + 6x - 12}}{{x - 3}} >= 0

\frac{{x^2 - 7x + 12}}{{x - 3}} <= 0 (We flipped the sign of left hand side and hence the inequality flipped too.)

\frac{{(x - 4)(x - 3)}}{{x - 3}} <= 0
or x <= 4
Since x has to be less than(or equal to) 0, the only solution here is x <= 0.

Combining the solutions above, we get x < 3 as the final solution.

(I have assumed that you are comfortable with quadratic expressions and solving for inequalities once you have factors. If you are not, let me know and I will point you to the related theory.)


+1 to you .
Thats a Good 1 :)
_________________

This time , its my time .

Director
Director
User avatar
Joined: 21 Dec 2010
Posts: 658
Followers: 10

Kudos [?]: 79 [0], given: 51

GMAT Tests User
Re: modulus inequalities [#permalink] New post 31 Jan 2011, 00:16
okies, this is a great explanation , concept cleared , thanks karishma. there is one question : when can one be sure that a quadratic expression is always +ve ?
_________________

What is of supreme importance in war is to attack the enemy's strategy.

Expert Post
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4668
Location: Pune, India
Followers: 1075

Kudos [?]: 4788 [0], given: 163

Re: modulus inequalities [#permalink] New post 31 Jan 2011, 05:43
Expert's post
garimavyas wrote:
okies, this is a great explanation , concept cleared , thanks karishma. there is one question : when can one be sure that a quadratic expression is always +ve ?


When you see x^2 - 5x + 12, think - will it have roots such that their sum is -5 and product is 12? Since product is +ve, and sum -ve, both roots should be negative. But will any two -ve roots add up to give -5 such that their product is 12? (e.g. -1, -4 or -2, -3 etc we do not get 12 as product) No.

To confirm, notice that b^2 - 4ac (where a = 1, b = -5 and c = 12) is -ve here so it has no real roots. Just put x = 0 and you get x^2 - 5x + 12 = 12 which is positive. This means the graph of this expression lies above the x axis. Remember, it will not meet the x axis anywhere because it has no real roots. Hence x^2 - 5x + 12 will always be positive.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Director
Director
User avatar
Joined: 21 Dec 2010
Posts: 658
Followers: 10

Kudos [?]: 79 [0], given: 51

GMAT Tests User
Re: modulus inequalities [#permalink] New post 31 Jan 2011, 07:35
hey ,
clearer now, thanks a ton
_________________

What is of supreme importance in war is to attack the enemy's strategy.

Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 19052
Followers: 3371

Kudos [?]: 24529 [1] , given: 2680

Re: modulus inequalities [#permalink] New post 31 Jan 2011, 08:32
1
This post received
KUDOS
Expert's post
garimavyas wrote:
okies, this is a great explanation , concept cleared , thanks karishma. there is one question : when can one be sure that a quadratic expression is always +ve ?


Quadratic expression ax^2+bx+c is always positive when 1. a, the coefficient of x^2, is positive: a>0 AND 2. the equation ax^2+bx+c=0 has no real roots, which happens when discriminant is negative, so when d=b^2-4ac<0.

In this case the graph (parabola) of the function f(x)=ax^2+bx+c is upward and lies above the X-axis, thus is always positive.

Similarly quadratic expression ax^2+bx+c is always negative when 1. a, the coefficient of x^2, is negative: a<0 AND 2. the equation ax^2+bx+c=0 has no real roots, which happens when discriminant is negative, so when d=b^2-4ac<0.

Check Coordinate Geometry chapter of Math Book for more: math-coordinate-geometry-87652.html

Hope it helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Director
Director
User avatar
Joined: 21 Dec 2010
Posts: 658
Followers: 10

Kudos [?]: 79 [0], given: 51

GMAT Tests User
Re: modulus inequalities [#permalink] New post 31 Jan 2011, 08:54
that solidifies the work of Karishma . now it is formulated .

thanks Karishma and Bunuel
_________________

What is of supreme importance in war is to attack the enemy's strategy.

Re: modulus inequalities   [#permalink] 31 Jan 2011, 08:54
    Similar topics Author Replies Last post
Similar
Topics:
Experts publish their posts in the topic DS inequalities and Modulus pnf619 3 02 Sep 2012, 02:52
4 Experts publish their posts in the topic 2 sums with Modulus and Inequality both rahuljaiswal 16 27 Oct 2010, 01:55
3 Experts publish their posts in the topic Multiple modulus inequalities zaarathelab 7 19 Dec 2009, 01:01
Modulus reply2spg 4 28 Oct 2009, 15:40
Inequalities and modulus subarao 0 13 Aug 2008, 20:25
Display posts from previous: Sort by

modulus inequalities

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.