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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

solution of √ (x + 7) < x

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 » 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
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
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
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
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
Log in to postLogin with Googleuse password
QUOTE IN REPLY
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
Log in to postLogin with Googleuse password
QUOTE IN REPLY
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
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
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
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1