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

Redeem

Tricky roots problem

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 158
Joined: Mon Nov 02, 2009 5:49 pm
Thanked: 2 times
Followed by:3 members

Tricky roots problem

by tonebeeze » Wed Apr 20, 2011 1:48 pm
If each expression under the square root is greater than or equal to 0, what is √(x^2 - 6x + 9) + √(2 - x) + x - 3?

a. √(2-x)
b. 2x - 6 + √(2-x)
c. √(2-x) + x - 3
d. 2x - 6 + √(x-2)
e. x + √(x-2)

OA = A
Top
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 965
Joined: Thu Jan 28, 2010 12:52 am
Thanked: 156 times
Followed by:34 members
GMAT Score:720

by vineeshp » Wed Apr 20, 2011 5:16 pm
I am not sure there is enough info in the question.

√(x^2 - 6x + 9) + √(2 - x) + x - 3.
√(x - 3)^2 + √(2 - x) + x - 3.

Now (x - 3)^2 >= 0 . This is already proven as it is a square.
(2 - x) >= 0 --> x <=2.

Now if the root of
√(x^2 - 6x + 9) is +- (x-3) > Both values are valid, as the square of this still satisfies the condition.

If root is - (x-3) . Then eqn comes down to - (x-3) + √(2 - x) + x - 3 = √(2 - x) . Answer will be A.

If root is (x-3), Then eqn comes down to - (x-3) + √(2 - x) + x - 3 = 2x - 6 + √(2 - x) . Answer will be B.

So what am I missin?
Vineesh,
Just telling you what I know and think. I am not the expert. :)
Top
User avatar
Senior | Next Rank: 100 Posts
Posts: 40
Joined: Tue Jan 18, 2011 5:14 am
Location: Bangalore
Thanked: 3 times
Followed by:2 members

by bajwa2307 » Wed Apr 20, 2011 5:17 pm
you sure all the signs in the equation are correct? I think there should be a minus in there instead of a plus

x^2 - 6x + 9 can be written as (x -3)^2

so, √(x^2 - 6x + 9) can be written as √(x -3)^2 = x - 3


now is where the question might be wrong, if there is minus infront of the other x-3 such that it is written as -(x+3), the terms cancel out and we are left with the desired value √(2 - x)

can someone verify this please.
Verbal is testing my patience
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 243
Joined: Sun Jul 12, 2009 7:12 am
Location: Dominican Republic
Thanked: 31 times
Followed by:2 members
GMAT Score:480

by MAAJ » Wed Apr 20, 2011 5:48 pm
I get (A) and (B) :/.... there most be something wrong with this problem. What's the source?

sqrt(x^2 - 6x + 9) + sqrt(2 - x) + x - 3
sqrt((x - 3)^2) + sqrt(2 - x) + x - 3
|x - 3| + sqrt(2 - x) + x - 3

Positive abs
x - 3 + sqrt(2 - x) + x - 3
2x - 6 + sqrt(2 - x)

Negative abs
-x + 3 + sqrt(2 - x) + x - 3
sqrt(2 - x)
"There's a difference between interest and commitment. When you're interested in doing something, you do it only when circumstance permit. When you're committed to something, you accept no excuses, only results."
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Thu Apr 21, 2011 9:38 pm
tonebeeze wrote:If each expression under the square root is greater than or equal to 0, what is √(x^2 - 6x + 9) + √(2 - x) + x - 3?

a. √(2-x)
b. 2x - 6 + √(2-x)
c. √(2-x) + x - 3
d. 2x - 6 + √(x-2)
e. x + √(x-2)

OA = A
Solution:
Now all expressions under square root sign are positive or zero.
So, 2-x >= 0, or x <= 2.
Now x^2 - 6x + 9 = (x-3)^2.
So, sqrt(x^2 -6x+9) = (x-3) or -(x-3) = 3-x.
Now, in GMAT we just consider positive square root value.
Now, for x <= 2, x-3 < 0. So, just take 3-x as the square root.
Hence, the expression becomes 3-x +sqrt(2-x) + x-3 = sqrt(2-x).

The correct answer is a.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Fri Apr 22, 2011 3:50 am
tonebeeze wrote:If each expression under the square root is greater than or equal to 0, what is √(x^2 - 6x + 9) + √(2 - x) + x - 3?

a. √(2-x)
b. 2x - 6 + √(2-x)
c. √(2-x) + x - 3
d. 2x - 6 + √(x-2)
e. x + √(x-2)

OA = A
We can plug in a value for x.

Let x=1:
√(x² - 6x + 9) + √(2 - x) + x - 3
= √(1² - 6*1 + 9) + √(2-1) + 1 - 3
= √4 - 1
= 1. This is our target.

Now we plug x=1 into all the answers to which yields our target of 1.

Only answer choice A works:
√(2-x) = √(2-1) = 1.

The correct answer is A.
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
Post new topic Post Reply
• Page 1 of 1