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 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point 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.

solution to x ^2 + 7 x + 12 > 0

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members
Location: India
Thanked: 267 times
GMAT Score:760
by sanju09 » Tue May 26, 2009 5:04 am
The solution to x ^2 + 7 x + 12 > 0 is

A. -4 < x < -3
B. x < -4 and x > -3
C. x > -4
D. x < -3
E. 7 < x < 12



MBM
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 487
Joined: Fri Mar 27, 2009 5:49 am
Thanked: 36 times

Re: solution to x ^2 + 7 x + 12 > 0

by dtweah » Tue May 26, 2009 5:15 am
sanju09 wrote:The solution to x ^2 + 7 x + 12 > 0 is

A. -4 < x < -3
B. x < -4 and x > -3
C. x > -4
D. x < -3
E. 7 < x < 12



MBM
Find the zero's of the inequality which will be excluded and plot them on a number line, or imagine them on a number line. You will have 3 intervals
(x+4)(x+3)>0
zero's are -4, -3.

Intervals

x< -4, -4<x<-3, x>-3

The middle interval fails-- the product is negative in this region for any x- and the union of the first and third gives the answer.

Hence B.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members
Location: Toronto
Thanked: 1710 times
GMAT Score:800

Re: solution to x ^2 + 7 x + 12 > 0

by Stuart@KaplanGMAT » Tue May 26, 2009 9:57 am
dtweah wrote:
sanju09 wrote:The solution to x ^2 + 7 x + 12 > 0 is

A. -4 < x < -3
B. x < -4 and x > -3
C. x > -4
D. x < -3
E. 7 < x < 12



MBM
Find the zero's of the inequality which will be excluded and plot them on a number line, or imagine them on a number line. You will have 3 intervals
(x+4)(x+3)>0
zero's are -4, -3.

Intervals

x< -4, -4<x<-3, x>-3

The middle interval fails-- the product is negative in this region for any x- and the union of the first and third gives the answer.

Hence B.
Slightly different approach (but very similar):

To get a positive product, both brackets must have the same sign.

Therefore, either:

x + 4 > 0 and x + 3 > 0

or

x + 4 < 0 and x + 3 < 0

In the first case, we get:

x > -4 AND x > -3. We always take the more restrictive condition:

x > -3

In the second case, we get

x < -4 and x < -3

Again, taking the more restrictive condition:

x < -4

So, x < -4 OR x > -3.. choose (b).
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 487
Joined: Fri Mar 27, 2009 5:49 am
Thanked: 36 times

Re: solution to x ^2 + 7 x + 12 > 0

by dtweah » Tue May 26, 2009 12:26 pm
Stuart Kovinsky wrote:
dtweah wrote:
sanju09 wrote:The solution to x ^2 + 7 x + 12 > 0 is

A. -4 < x < -3
B. x < -4 and x > -3
C. x > -4
D. x < -3
E. 7 < x < 12



MBM
Find the zero's of the inequality which will be excluded and plot them on a number line, or imagine them on a number line. You will have 3 intervals
(x+4)(x+3)>0
zero's are -4, -3.

Intervals

x< -4, -4<x<-3, x>-3

The middle interval fails-- the product is negative in this region for any x- and the union of the first and third gives the answer.

Hence B.
Slightly different approach (but very similar):

To get a positive product, both brackets must have the same sign.

Therefore, either:

x + 4 > 0 and x + 3 > 0

or

x + 4 < 0 and x + 3 < 0

In the first case, we get:

x > -4 AND x > -3. We always take the more restrictive condition:

x > -3

In the second case, we get

x < -4 and x < -3

Again, taking the more restrictive condition:

x < -4

So, x < -4 OR x > -3.. choose (b).
This used to be my favorite way of tackling these but I switched to the sign method.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1