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.

Inequalities / Quadratic Equation

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 14
Joined: Sun Aug 23, 2009 8:09 am
Thanked: 1 times
by Resurgent » Fri Jan 01, 2010 7:41 am
Which of the following is correct if x is a real number and (x - 11)(x - 3) is negative?

A. x^2 + 5x + 6 < 0
B. x^2 + 5x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0


PS: I do not have the OA for this question.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Wed Jul 22, 2009 1:30 am
Thanked: 1 times

by funx » Fri Jan 01, 2010 8:50 am
I would guess that the answer is E, i.e. 11-x>0

Here is my reasoning:

If (x-11)(x-3)<0 then at least one of them must be negative. Intuitively then, x-11 is the negative part and x-3 is positive because in the former case you subtract a larger number.

x-11<0 | *(-1)
11-x>0

Any other takers?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 39
Joined: Wed Nov 18, 2009 8:41 am
Thanked: 3 times

by sunil_snath » Fri Jan 01, 2010 10:53 am
Even B looks right tho :(
From the question X is greater than 3 and less than 11.

So B, (X+3)(X+2)>0 is true. But E is obvious
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Fri Jan 01, 2010 10:59 am
Resurgent wrote:Which of the following is correct if x is a real number and (x - 11)(x - 3) is negative?

A. x^2 + 5x + 6 < 0
B. x^2 + 5x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0

PS: I do not have the OA for this question.
Here's my solution. It includes a systematic way to solve quadratic inequalities.

Image

The answer is not E since one solution to E is x=0 and x=0 is not a solution to the original inequality.
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 189
Joined: Thu Apr 03, 2008 2:03 pm
Location: USA
Thanked: 21 times

by rohan_vus » Fri Jan 01, 2010 1:07 pm
Brent Hanneson wrote:
Resurgent wrote:Which of the following is correct if x is a real number and (x - 11)(x - 3) is negative?

A. x^2 + 5x + 6 < 0
B. x^2 + 5x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0

PS: I do not have the OA for this question.
Here's my solution. It includes a systematic way to solve quadratic inequalities.

Image

The answer is not E since one solution to E is x=0 and x=0 is not a solution to the original inequality.
With respect to the logic behind bold statement , one solution is x = -4 which satisfies option B but its not a solution to original inequality . This question seems kinda odd to me
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Fri Jan 01, 2010 1:30 pm
rohan_vus wrote:
Brent Hanneson wrote:
Resurgent wrote:Which of the following is correct if x is a real number and (x - 11)(x - 3) is negative?

A. x^2 + 5x + 6 < 0
B. x^2 + 5x + 6 > 0
C. 5 - x < 0
D. x - 5 < 0
E. 11 - x > 0

PS: I do not have the OA for this question.
Here's my solution. It includes a systematic way to solve quadratic inequalities.

Image

The answer is not E since one solution to E is x=0 and x=0 is not a solution to the original inequality.
With respect to the logic behind bold statement , one solution is x = -4 which satisfies option B but its not a solution to original inequality . This question seems kinda odd to me
In our answer choices, we are looking for an inequality such that every possible value for x from the original inequality [ (x-11)(x-3) < 0 ] must also be a solution to the inequality among the answer choices. You are ready the question the other way around.

x = -4 is not a solution to the original inequality (x-11)(x-3) < 0, so we need not consider that value for x when checking the answer choices. We need only consider values of x between 3 and 11.

When we plug these allowable values into the inquality in answer choice B, EVERY value of x is such that x^2 + 5x + 6 > 0

The same cannot be said for the other answer choices.
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 189
Joined: Thu Apr 03, 2008 2:03 pm
Location: USA
Thanked: 21 times

by rohan_vus » Fri Jan 01, 2010 1:51 pm
We need only consider values of x between 3 and 11

This is what i am driving at!.

All possible ranges are 3<x<11. So any of these values even satisfy option E.. So in what way option E is different than option B. Yo mentioned taking x = 0 for kicking out option E , but x = 0 is not in the range of the original inequality .
All values between 3 and 11 satisfy option E also .
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Fri Jan 01, 2010 2:19 pm
rohan_vus wrote:We need only consider values of x between 3 and 11

This is what i am driving at!.

All possible ranges are 3<x<11. So any of these values even satisfy option E.. So in what way option E is different than option B. Yo mentioned taking x = 0 for kicking out option E , but x = 0 is not in the range of the original inequality .
All values between 3 and 11 satisfy option E also .
Sorry, my bad. I was reading something completely different from what you were saying.
All possible values for x (3<x<11) make E valid as well.
Looks like the answer is B and E
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1