Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
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.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 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
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
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
GMATOnDemandStart 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 563 reviewsTarget Test Prep

solution of √ (x + 7) < x

Expert replies
Post new topic Post Reply
Subscribe topic
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 » Thu Oct 14, 2010 12:11 am
What is the solution of √ (x + 7) < x?
(A) x > 2
(B) x > ½ √30
(C) x > ½ (1 + √29)
(D) x > 1 + ½ √29
(E) x > 1 + ½ √30


[spoiler]Made up[/spoiler]
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
Source: — Problem Solving |
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members
Location: Bangalore, India
Thanked: 146 times

by shovan85 » Thu Oct 14, 2010 1:00 am
IMO C

Square both sides and get the eqn of the form x^2 - x - 7 > 0.

Solving for this we can get either

x> [1+sqrt(29)]/2 or x> [1-sqrt(29)]/2
Top
Join the discussion
Master | Next Rank: 500 Posts
Posts: 270
Joined: Wed Apr 07, 2010 9:00 amFollowed by: 2 members
Thanked: 24 times

by neerajkumar1_1 » Thu Oct 14, 2010 1:06 am
root(x+7)<x

x will have to be +ve since root(x+7) will always be +ve...

now squaring both sides we get

x+7 < x^2
=> x^2 - x - 7 > 0
solve for roots of x using formula [-b +- root(b^2 - 4ac)]/2a

we get x to be 1/2(1 + root(29)) and 1/2(1 - root(29))

so
(x - 1/2(1 + root(29)))(x- 1/2(1 - root(29))) > 0

hence x > 1/2(1 + root(29))
or x> 1/2(1 - root(29))

first one is among the answer choices...

PIck C
Top
Join the discussion
Legendary Member
Posts: 1119
Joined: Fri May 07, 2010 8:50 amFollowed by: 3 members
Thanked: 29 times

by diebeatsthegmat » Sat Oct 16, 2010 7:59 pm
sanju09 wrote:What is the solution of √ (x + 7) < x?
(A) x > 2
(B) x > ½ √30
(C) x > ½ (1 + √29)
(D) x > 1 + ½ √29
(E) x > 1 + ½ √30


[spoiler]Made up[/spoiler]
only C.,..
hey, where is the source? i saw you posted so many nice maths but some Ps is weird... :)
Top
Join the discussion
Master | Next Rank: 500 Posts
Posts: 155
Joined: Wed Sep 22, 2010 3:54 am
Thanked: 6 times

by pzazz12 » Tue Oct 19, 2010 2:33 am
neerajkumar1_1 wrote:root(x+7)<x

x will have to be +ve since root(x+7) will always be +ve...

now squaring both sides we get

x+7 < x^2
=> x^2 - x - 7 > 0
solve for roots of x using formula [-b +- root(b^2 - 4ac)]/2a

we get x to be 1/2(1 + root(29)) and 1/2(1 - root(29))

so
(x - 1/2(1 + root(29)))(x- 1/2(1 - root(29))) > 0

hence x > 1/2(1 + root(29))
or x> 1/2(1 - root(29))

first one is among the answer choices...

PIck C
thank u.......is there any other way to solve these type of problems....
Top
Join the discussion

GMAT/MBA Expert

GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pmFollowed by: 88 members
Location: Milpitas, CA
Thanked: 447 times

by Rahul@gurome » Tue Oct 19, 2010 11:07 pm
We get that x^2 – x – 7 > 0.
Instead of using the formula for roots we can write that x^2 – x =(x – ½)^2 – ¼.
Or x^2 – x – 7 > 0 can be written as (x – ½)^2 – ¼ -7 > 0.
Or we have that (x – ½)^2 > 29/4.
So either (x-1/2) > sqrt(29)/2 or (x-1/2) < -sqrt(29)/2.
So either x >1/2(1 + sqrt29) or x < ½(1 – sqrt29) {this is negative}.Now x cannot be negative because sqrt(x+7) is positive and positive can never be less than negative.
So the only possible range for x is x >1/2(1 + sqrt29).
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
Join the discussion