M08 #16 - squares : Retired Discussions [Locked]
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 07 Dec 2016, 05:02

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M08 #16 - squares

Author Message
Manager
Joined: 17 Dec 2008
Posts: 177
Followers: 2

Kudos [?]: 121 [7] , given: 0

### Show Tags

01 Feb 2009, 07:09
7
KUDOS
13
This post was
BOOKMARKED
If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

(A) $$\sqrt{2-x}$$
(B) $$2x-6 + \sqrt{2-x}$$
(C) $$\sqrt{2-x} + x-3$$
(D) $$2x-6 + \sqrt{x-2}$$
(E) $$x + \sqrt{x-2}$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions
SVP
Joined: 17 Jun 2008
Posts: 1569
Followers: 11

Kudos [?]: 245 [7] , given: 0

### Show Tags

05 Feb 2009, 03:25
7
KUDOS
A.

Since, each expression under the square root is positive, 2-x > 0 or, x < 2.

Thus, x-3 < 0.

Now, first expression = sqrt(x-3)^2 = |x-3| = 3-x

Thus, the whole expression will reduce to sqrt(2-x)
Manager
Joined: 17 Dec 2008
Posts: 177
Followers: 2

Kudos [?]: 121 [0], given: 0

### Show Tags

09 Feb 2009, 05:05
>>Now, first expression = sqrt(x-3)^2 = |x-3| = 3-x

Can you explain the above? Isn't the sqrt of a number always its positive root?

sqrt(x-3)^2 = x-3 ? NOT |x-3| ?
SVP
Joined: 17 Jun 2008
Posts: 1569
Followers: 11

Kudos [?]: 245 [0], given: 0

### Show Tags

09 Feb 2009, 09:38
ConkergMat wrote:
>>Now, first expression = sqrt(x-3)^2 = |x-3| = 3-x

Can you explain the above? Isn't the sqrt of a number always its positive root?

sqrt(x-3)^2 = x-3 ? NOT |x-3| ?

You are right. That is why 3-x as (x-3) is negative.
Senior Manager
Joined: 01 Mar 2009
Posts: 372
Location: PDX
Followers: 6

Kudos [?]: 87 [4] , given: 24

### Show Tags

17 Oct 2009, 17:43
4
KUDOS
1
This post was
BOOKMARKED
xALIx wrote:
scthakur wrote:
ConkergMat wrote:
>>Now, first expression = sqrt(x-3)^2 = |x-3| = 3-x

Can you explain the above? Isn't the sqrt of a number always its positive root?

sqrt(x-3)^2 = x-3 ? NOT |x-3| ?

You are right. That is why 3-x as (x-3) is negative.

I still don't get how you got 3-x.

It's way too late for the reply but here it goes

x< 2 and sqrt ( x-3 ) ^2 >=0

now for sqrt(x-3)^2 there are two values : x-3 or -(x-3)
x-3>0 and x>3 - This is impossible because we have x<2

So sqrt(x-3)^2 = 3-x.
The rest is pretty straightforward.
_________________

In the land of the night, the chariot of the sun is drawn by the grateful dead

Intern
Joined: 04 Sep 2008
Posts: 1
Followers: 0

Kudos [?]: 1 [1] , given: 0

Re: M08 #16 - squares [#permalink]

### Show Tags

16 Aug 2010, 06:14
1
KUDOS
I think it should be B.

sqrt((x-3) (x-3)) should give (x-3) because it is mentioned that each expression under the square root is greater than or equal to 0.

Rest is simple --> x-3 +sqrt(x-2) +x -3

2x -6 +sqrt(x-2)
Intern
Joined: 07 Jul 2010
Posts: 13
Followers: 0

Kudos [?]: 8 [0], given: 3

Re: M08 #16 - squares [#permalink]

### Show Tags

16 Aug 2010, 07:49
2-x> o then x<2 or x is negative

(x-3)^2 >o then x is different from 3

if x is negative then squareroot((x-3)^2 )=|x-3|=3-x then answer a

if 0<x<2 with x different from 3 then squareroot((x-3)^2 )=|x-3|=3-x then answer a too

in both cases x-3 is negative hence the squareroot of it's square is equal to |x-3|

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 35909
Followers: 6850

Kudos [?]: 90010 [8] , given: 10402

Re: M08 #16 - squares [#permalink]

### Show Tags

16 Aug 2010, 08:46
8
KUDOS
Expert's post
5
This post was
BOOKMARKED
ConkergMat wrote:
If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

(A) $$\sqrt{2-x}$$
(B) $$2x-6 + \sqrt{2-x}$$
(C) $$\sqrt{2-x} + x-3$$
(D) $$2x-6 + \sqrt{x-2}$$
(E) $$x + \sqrt{x-2}$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

One important note: $$\sqrt{x^2}=|x|$$

$$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3=\sqrt{(x-3)^2}+\sqrt{2-x}+x-3=|x-3|+\sqrt{2-x}+x-3$$.

Now, as the expressions under the square roots are more than or equal to zero than $$2-x>0$$ --> $$x<2$$. Next: as $$x<2$$ then $$|x-3|$$ becomes $$|x-3|=-(x-3)=-x+3$$.

$$|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}$$.

_________________
Intern
Joined: 19 Jul 2010
Posts: 3
Followers: 0

Kudos [?]: 2 [1] , given: 0

Re: M08 #16 - squares [#permalink]

### Show Tags

16 Aug 2010, 09:57
1
KUDOS
I have done it as follows..

Plugin a value for x where the value of the expression under the squareroot will be greater than or equal to zero, i.e., x^2-6x+9 >=0 and 2-x >=0.

I took x=1, substituing that gives 4 for 1st expression and 1 for 2nd expression. Now, Substituing x=1 in the answer choices. Except option A, all other choices give a negative answer. So, answer is A.
Manager
Joined: 12 Jul 2010
Posts: 67
Followers: 1

Kudos [?]: 2 [0], given: 3

Re: M08 #16 - squares [#permalink]

### Show Tags

17 Aug 2010, 00:41
Its A.

Since each expression under the square root is greater than or equal to 0, x<2. For all values of x< 1, the equation is true.

Lets suppose x = 1! Putting x=1 in sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3 gives 1 as solution.

Intern
Joined: 10 Jun 2010
Posts: 20
Followers: 0

Kudos [?]: 1 [0], given: 0

Re: M08 #16 - squares [#permalink]

### Show Tags

17 Aug 2010, 07:14
a is for me

\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3 = |x-3|+ \sqrt{2 - x} + x - 3

because 2-x >=0, x<= 2, |x-3| can be expressed as 3-x

the result is 3-x + \sqrt{2 - x} + x - 3 = \sqrt{2 - x}
Intern
Joined: 16 Nov 2010
Posts: 10
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: M08 #16 - squares [#permalink]

### Show Tags

13 May 2011, 02:25
ssing1 wrote:
I think it should be B.

sqrt((x-3) (x-3)) should give (x-3) because it is mentioned that each expression under the square root is greater than or equal to 0.

Rest is simple --> x-3 +sqrt(x-2) +x -3

2x -6 +sqrt(x-2)

nope
it is mentioned that the expression under the square root > 0.
Also square root of 16 is 4 and not 4,-4.
There4, sgrt [(x-3)(x-3) ] becomes 3-x and not x-3 to keep the sqrt positive since x<2.
Intern
Joined: 15 May 2011
Posts: 40
WE 1: IT Consulting - 5 Years
Followers: 0

Kudos [?]: 3 [0], given: 4

Re: M08 #16 - squares [#permalink]

### Show Tags

01 Jul 2011, 07:24
Awesome question! I fell for B like a mortal! Every piece of information on the GMAT counts, if the question stem mentions that every expression under the square is greater than or equal to zero, we HAVE to use it somewhere!
Manager
Joined: 23 Jan 2011
Posts: 127
Followers: 1

Kudos [?]: 67 [0], given: 13

Re: M08 #16 - squares [#permalink]

### Show Tags

07 Jul 2011, 10:23
AriBenCanaan wrote:
Awesome question! I fell for B like a mortal! Every piece of information on the GMAT counts, if the question stem mentions that every expression under the square is greater than or equal to zero, we HAVE to use it somewhere!

agree, probably questions on gmat may not be so straight forward after all
Senior Manager
Joined: 25 Nov 2011
Posts: 261
Location: India
Concentration: Technology, General Management
GPA: 3.95
WE: Information Technology (Computer Software)
Followers: 4

Kudos [?]: 160 [1] , given: 20

Re: M08 #16 - squares [#permalink]

### Show Tags

12 Feb 2012, 03:45
1
KUDOS
Pretty tough....its not the question itself tough, but remembering underlying algebra principles...

whenever I see the expression $$a^2 -2.a.b + b^2$$ I immediately think it as $$(a-b)^2$$, as we learned in the school. But the fact is, it depends on how one perceives.
It can be either $$(a-b)^2$$ or $$(b-a)^2$$ as there is no info about a and b. That is why we should remember the value of \sqrt{$$a^2 -2.a.b + b^2$$} as |a-b| and it should be evaluated based on the a & b variables info. This is what actually tested in the question - a tough question, but very good question.
Thanks
_________________

-------------------------
-Aravind Chembeti

Intern
Status: gmat fresher
Joined: 07 Jun 2012
Posts: 25
GPA: 3.87
Followers: 9

Kudos [?]: 209 [1] , given: 12

Re: M08 #16 - squares [#permalink]

### Show Tags

20 Aug 2012, 09:09
1
KUDOS
ConkergMat wrote:
If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

(A) $$\sqrt{2-x}$$
(B) $$2x-6 + \sqrt{2-x}$$
(C) $$\sqrt{2-x} + x-3$$
(D) $$2x-6 + \sqrt{x-2}$$
(E) $$x + \sqrt{x-2}$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

Suppose choose the value x=1 which satisfies the condition
Hence the given becomes 2+1+1-3=1
1.1
2.3.4.5.not equal
Hence 1 is correct
Manager
Joined: 14 Jun 2012
Posts: 66
Followers: 0

Kudos [?]: 13 [1] , given: 1

Re: M08 #16 - squares [#permalink]

### Show Tags

29 Aug 2012, 22:40
1
KUDOS
Really good question. There are a few parts to this question and I did all but one correct. I could make out that the first expression is (x-3)^2 and that there are two options i.e. +(x-3) and -(x-3). Using these two values I got A and B as the possible answer choices.

At this point the question stem that each expression under the square root is greater than or equal to zero.

This is where I made the mistake and chose B. Looking back through the answer choice right now it was really a stupid mistake to consider the option as -(x-3). I misinterpreted the question stem of value under the square root as non-negative for the value of x to be non-negative. A real stupid blunder.

Really good question though. Nice learning.
_________________

My attempt to capture my B-School Journey in a Blog : tranquilnomadgmat.blogspot.com

There are no shortcuts to any place worth going.

Intern
Joined: 04 Apr 2013
Posts: 16
Concentration: Finance, Economics
GPA: 3
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: M08 #16 - squares [#permalink]

### Show Tags

22 Aug 2013, 03:38

I spent 3.41 on this one, after looking at it for about a minute... easy solution is to substitute a value of '1' into the expression (obtaining '1' as a result), then confirm which answer option provides the same value.

Only option A also produces '1': SQRT(2-1) = 1.

Current Student
Joined: 19 Dec 2011
Posts: 12
Concentration: Healthcare, Operations
Schools: Olin '17 (M)
GPA: 3.24
WE: Supply Chain Management (Military & Defense)
Followers: 0

Kudos [?]: 7 [0], given: 54

Re: M08 #16 - squares [#permalink]

### Show Tags

20 Oct 2013, 13:30
Hi Bunuel,
You posted: as x<2, then |x-3|=-(x-3)=-x+3

But |x-3| equals either x-3 or -(x-3)
If x-3>0, then x>3
If x-3<0, then x-3<-0, x<3

How did you conclude to use -(x-3) for |x-3|?

Thanks,
Francis
Math Expert
Joined: 02 Sep 2009
Posts: 35909
Followers: 6850

Kudos [?]: 90010 [2] , given: 10402

Re: M08 #16 - squares [#permalink]

### Show Tags

20 Oct 2013, 13:39
2
KUDOS
Expert's post
jrawls wrote:
Hi Bunuel,
You posted: as x<2, then |x-3|=-(x-3)=-x+3

But |x-3| equals either x-3 or -(x-3)
If x-3>0, then x>3
If x-3<0, then x-3<-0, x<3

How did you conclude to use -(x-3) for |x-3|?

Thanks,
Francis

ABSOLUTE VALUE PROPERTIES:

When $$x\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \ expression\leq{0}$$ then $$|some \ expression|={-(some \ expression)}$$. For example: $$|-5|=5=-(-5)$$;

When $$x\geq{0}$$ then $$|x|=x$$, or more generally when $$some \ expression\geq{0}$$ then $$|some \ expression|={some \ expression}$$. For example: $$|5|=5$$.

BACK TO THE QUESTION:
|x-3|=x-3, when x>3.
|x-3|=-(x-3), when x<3.

Since we know that x<2, then we have the second case (when x<2, x-3<0, thus |x-3|=-(x-3)).

Hope it's clear.
_________________
Re: M08 #16 - squares   [#permalink] 20 Oct 2013, 13:39

Go to page    1   2    Next  [ 25 posts ]

Similar topics Replies Last post
Similar
Topics:
9 m08 q#6 11 25 Nov 2008, 10:19
18 m08, #11 20 06 Nov 2008, 21:41
20 m08#27 18 02 Nov 2008, 14:48
4 m08#23 16 02 Nov 2008, 14:43
8 m08 Q#7 16 01 Aug 2008, 09:56
Display posts from previous: Sort by

# M08 #16 - squares

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.