BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Doubt in Inequality

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Data Sufficiency |
User avatar
Legendary Member
Posts: 588
Joined: Sun Oct 16, 2011 9:42 amFollowed by: 9 members
Location: New Delhi, India
Thanked: 130 times
GMAT Score:720

by rijul007 » Sat Nov 19, 2011 10:42 am
1. If x^2 > 1, then what are the limits of that inequality?

x^2 > 1
x^2 - 1 > 0
(x+1)(x-1) > 0

Take cases

Case 1: x > 1
(x+1)(x-1) will be positive

Case 2 -1<x<1
(x+1)(x-1) will be ngative

Case 3 x<-1
(x+1)(x-1) will be positive

Hence
x^2 > 1 for x<-1 and x>1

2. If (5X - 2)(3X + 1) > 0, then what are the limits of that inequality?
similarly take cases in these too..

you'll get the ans
x<-1/2 and x>2/5
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 99
Joined: Thu Jun 09, 2011 2:09 pm
Thanked: 11 times

by kanwar86 » Sat Nov 19, 2011 10:46 am
IMO
1) x<-1 or x>1
2) x<-1/3 or x>2/5
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 29
Joined: Fri Oct 14, 2011 10:15 amFollowed by: 1 member
Thanked: 4 times

by zooki » Sat Nov 19, 2011 11:14 am
for Q 2: x=-1/3 does not satisfy the inequality.
neither does x=2/5
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 176
Joined: Thu Sep 22, 2011 5:32 am
Thanked: 5 times

by vishal.pathak » Sat Nov 19, 2011 11:36 am
seema19 wrote:1. If x^2 > 1, then what are the limits of that inequality?
2. If (5X - 2)(3X + 1) > 0, then what are the limits of that inequality?
1.if x^2 > 1 then -infinity < x < -1 and 1 < x < infinity
2. if (5x - 2) (3x + 1) > 0
then
either both (5x-2) and (3x+1) have to be +ve or both have to be -ve for the final product to b +ve (>0)
when both +ve x > 2/5 (5x-2 > 0), x>-1/3(3x+1 > 0)
so we see that x > 2/5 satisfies both x > 2/5 and x>-1/3
when both -ve x < 2/5, x < -1/3
so we see that x < -1/3 satisfies both x<2/5 and x<-1/3
so the range is x < -1/3 , x > 2/5
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 23
Joined: Wed Sep 14, 2011 12:34 am

by seema19 » Sat Nov 19, 2011 11:42 am
X^2 > 1
X^2 - 1 > 0
(X + 1)(X - 1) > 0
Thus, X > -1 and X > +1.
However, according to the answer,this is wrong. Can someone plz explain why this is wrong.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 99
Joined: Thu Jun 09, 2011 2:09 pm
Thanked: 11 times

by kanwar86 » Sat Nov 19, 2011 11:57 am
seema19 wrote:X^2 > 1
X^2 - 1 > 0
(X + 1)(X - 1) > 0
Thus, X > -1 and X > +1.
However, according to the answer,this is wrong. Can someone plz explain why this is wrong.
For expression (X-1)(X+1) to be greater than zero either X should be >1 or <-1
In your case for 1>X>-1, the term (X+1)>0 but (X-1)<0, which implies the product, X^2-1, is less than zero. Hence, -1<x<1 can't be the solution.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 176
Joined: Thu Sep 22, 2011 5:32 am
Thanked: 5 times

by vishal.pathak » Sat Nov 19, 2011 12:06 pm
seema19 wrote:X^2 > 1
X^2 - 1 > 0
(X + 1)(X - 1) > 0
Thus, X > -1 and X > +1.
However, according to the answer,this is wrong. Can someone plz explain why this is wrong.
for (X + 1)(X - 1) > 0 either both ((X + 1) & (X - 1)) have to be +ve or both have to be -ve so that the final product is +ve
+ve*+ve = +ve
-ve*-ve = -ve
if both are +ve then x+1>0 and x-1>0
or x > -1 and x>1
So u can see that x>1 satisfies both x > -1 and x>1 (so range from this is x>1, lets call it eqn (1))
Now if both ((X + 1) & (X - 1)) are -ve
then x<-1 and x<1
So u can see that x<-1 satisfies both x<-1 and x<1 (so range from this is x<-1, lets call it eqn (2))
From eqn (1) and eqn(2)
x<-1 and x >1
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1084
Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members
Thanked: 158 times

by pemdas » Sat Nov 19, 2011 12:09 pm
likewise |x|>|1| you solve by squaring both sides you can consider the roots here
for |x|>|1| the only solutions are mode must be greater than the absolute value of 1 regardless of signs, hence x<-1 and x>1
seema19 wrote:X^2 > 1
X^2 - 1 > 0
(X + 1)(X - 1) > 0
Thus, X > -1 and X > +1.
However, according to the answer,this is wrong. Can someone plz explain why this is wrong.
Success doesn't come overnight!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 99
Joined: Thu Jun 09, 2011 2:09 pm
Thanked: 11 times

by kanwar86 » Sat Nov 19, 2011 12:20 pm
Please note that the solution to 1)is X<-1 OR X>1 instead of X<-1 AND X>1 (which is not possible as both sets are mutually exclusive)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Sun Nov 20, 2011 5:32 am
One approach is to determine the CRITICAL POINTS. Check my posts here:

https://www.beatthegmat.com/inequality-c ... 89518.html

https://www.beatthegmat.com/inequalities ... 89788.html
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1