GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 19 Jun 2018, 09:17

LIVE NOW:

Tips for Your Best Possible Application - Live Chat with Tuck Admissions | Click Here to JOIN!


Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If each expression under the square root is greater than or

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

14 KUDOS received
Senior Manager
Senior Manager
avatar
Joined: 05 Oct 2008
Posts: 258
If each expression under the square root is greater than or [#permalink]

Show Tags

New post Updated on: 09 Oct 2013, 02:21
14
66
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

36% (01:12) correct 64% (01:07) wrong based on 1392 sessions

HideShow timer Statistics

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) \(x + \sqrt{x-2}\)

First, \((2 - x)\) has to be \(\ge 0\) for the expression \(\sqrt{2 - x}\) to be defined (GMAT deals only with real numbers). Therefore, \(x \le 2\) .

Even if x <= 2, rt (x-3)^2 is still a valid explanation. Since there is a square below the rt sign the number is obviously positive. How can one conveniently change from (x - 3) to (3 - x)?

\(\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = 3 - x\) (because \(x \le 2\) and thus \(x - 3 \lt 0\) ).

IMO:
\(\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = x - 3\) is valid


Summing up: \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + (x - 3) = (x - 3) + \sqrt{2 - x} + (x - 3) = 2x - 6 + \sqrt{x - 2}\).


M18-16

Originally posted by study on 21 Nov 2009, 01:40.
Last edited by Bunuel on 09 Oct 2013, 02:21, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
6 KUDOS received
Tuck Thread Master
User avatar
Joined: 20 Aug 2009
Posts: 289
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)
Re: Inequalities - M08 [#permalink]

Show Tags

New post 21 Nov 2009, 02:39
6
7
study

Quote:
Even if x <= 2, rt (x-3)^2 is still a valid explanation. Since there is a square below the rt sign the number is obviously positive. How can one conveniently change from (x - 3) to (3 - x)?


yes, you can and in fact should change it to (3-x). Here's explanation:

http://en.wikipedia.org/wiki/Square_root

\(sqrt(x^2) = |x|\)

|x| = x, for all x>=0
|x| = -x, for all x<0

In our case we have \(\sqrt{(x-3)^2}=|x-3|=-(x-3)=3-x\), because (2-x)>=0; x<=2; (x-3)<0

And so the correct answer is

\(3-x+\sqrt{2-x}+x-3=\sqrt{2-x}\)
1 KUDOS received
SVP
SVP
User avatar
Joined: 06 Sep 2013
Posts: 1885
Concentration: Finance
GMAT ToolKit User
Re: Inequalities - M08 [#permalink]

Show Tags

New post 08 Oct 2013, 16:16
1
study wrote:
Can someone please explain the logic here. This problem is from M08.


If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\) ?


* \(\sqrt{2 - x}\)
* \(2x - 6 + \sqrt{2 - x}\)
* \(\sqrt{2 - x} + {x - 3}\)
* \(2x - 6 + \sqrt{x - 2}\)
* \(x + \sqrt{x - 2}\)

First, \((2 - x)\) has to be \(\ge 0\) for the expression \(\sqrt{2 - x}\) to be defined (GMAT deals only with real numbers). Therefore, \(x \le 2\) .

Even if x <= 2, rt (x-3)^2 is still a valid explanation. Since there is a square below the rt sign the number is obviously positive. How can one conveniently change from (x - 3) to (3 - x)?

\(\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = 3 - x\) (because \(x \le 2\) and thus \(x - 3 \lt 0\) ).

IMO:
\(\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = x - 3\) is valid


Summing up: \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + (x - 3) = (x - 3) + \sqrt{2 - x} + (x - 3) = 2x - 6 + \sqrt{x - 2}\).


What's the OA? I chose (B) I'm afraid I might be wrong though
Expert Post
19 KUDOS received
Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8100
Location: Pune, India
Re: Inequalities - M08 [#permalink]

Show Tags

New post 08 Oct 2013, 21:48
19
9
jlgdr wrote:
study wrote:
Can someone please explain the logic here. This problem is from M08.


If each expression under the square root is greater than or equal to 0, what is \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3\) ?


* \(\sqrt{2 - x}\)
* \(2x - 6 + \sqrt{2 - x}\)
* \(\sqrt{2 - x} + {x - 3}\)
* \(2x - 6 + \sqrt{x - 2}\)
* \(x + \sqrt{x - 2}\)

First, \((2 - x)\) has to be \(\ge 0\) for the expression \(\sqrt{2 - x}\) to be defined (GMAT deals only with real numbers). Therefore, \(x \le 2\) .

Even if x <= 2, rt (x-3)^2 is still a valid explanation. Since there is a square below the rt sign the number is obviously positive. How can one conveniently change from (x - 3) to (3 - x)?

\(\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = 3 - x\) (because \(x \le 2\) and thus \(x - 3 \lt 0\) ).

IMO:
\(\sqrt{x^2 - 6x + 9} = \sqrt{(x - 3)^2} = x - 3\) is valid


Summing up: \(\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + (x - 3) = (x - 3) + \sqrt{2 - x} + (x - 3) = 2x - 6 + \sqrt{x - 2}\).


What's the OA? I chose (B) I'm afraid I might be wrong though


The correct answer will be (A)
The expression becomes

\(|x-3| + \sqrt{2 - x} + (x - 3)\)

Note that \(\sqrt{x^2} = |x|\) (and not x)

Since we know that x < 2, (x - 3) must be negative. Hence \(|x-3| = -(x - 3)\)

Expression becomes: \(-(x - 3) + \sqrt{2 - x} + (x - 3) = \sqrt{2 - x}\)
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

Expert Post
10 KUDOS received
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 46164
Re: If each expression under the square root is greater than or [#permalink]

Show Tags

New post 09 Oct 2013, 02:22
10
17
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) \(x + \sqrt{x-2}\)

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 \(2-x\geq{0}\) --> \(x\leq{2}\). Next: as \(x\leq{2}\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

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

Answer: A.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Manager
Manager
avatar
Joined: 26 Sep 2013
Posts: 204
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41
GMAT 2: 730 Q49 V41
Re: If each expression under the square root is greater than or [#permalink]

Show Tags

New post 09 Oct 2013, 18:29
Bunuel 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) \(x + \sqrt{x-2}\)

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 \(|x-3|=-(x-3)=-x+3\).

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

Answer: A.


I don't follow the highlighted part at all. So if you have x-n, and it is supposed to be positive you can just change around the order? Shouldn't the expression just be undefined, since the facts we are given, and the expression we are given can't coexist?
Expert Post
1 KUDOS received
Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8100
Location: Pune, India
Re: If each expression under the square root is greater than or [#permalink]

Show Tags

New post 09 Oct 2013, 20:27
1
1
AccipiterQ wrote:
Bunuel 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) \(x + \sqrt{x-2}\)

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 \(|x-3|=-(x-3)=-x+3\).

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

Answer: A.


I don't follow the highlighted part at all. So if you have x-n, and it is supposed to be positive you can just change around the order? Shouldn't the expression just be undefined, since the facts we are given, and the expression we are given can't coexist?


Absolutely not! Ordinarily, if you change the order, it changes everything. But this situation is not ordinary. It is a mod situation i.e. there is modulus around x - n.

Recall how mods are defined:
|x| = x if x is positive
= -x if x is negative

So if you want to get rid of the mod sign, you need to know whether x is positive or negative.

Here we know that x <= 2. For all such values of x, (x-3) is negative.
So |x-3| = -(x - 3) = 3 - x
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

Manager
Manager
avatar
Joined: 26 Sep 2013
Posts: 204
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41
GMAT 2: 730 Q49 V41
Re: If each expression under the square root is greater than or [#permalink]

Show Tags

New post 10 Oct 2013, 06:57
VeritasPrepKarishma wrote:
AccipiterQ wrote:
Bunuel 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) \(x + \sqrt{x-2}\)

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 \(|x-3|=-(x-3)=-x+3\).

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

Answer: A.


I don't follow the highlighted part at all. So if you have x-n, and it is supposed to be positive you can just change around the order? Shouldn't the expression just be undefined, since the facts we are given, and the expression we are given can't coexist?


Absolutely not! Ordinarily, if you change the order, it changes everything. But this situation is not ordinary. It is a mod situation i.e. there is modulus around x - n.

Recall how mods are defined:
|x| = x if x is positive
= -x if x is negative

So if you want to get rid of the mod sign, you need to know whether x is positive or negative.

Here we know that x <= 2. For all such values of x, (x-3) is negative.
So |x-3| = -(x - 3) = 3 - x



OOOOH I see now, thank you,
Manager
Manager
avatar
Joined: 03 Jan 2013
Posts: 185
Location: United States
Concentration: Finance, Entrepreneurship
GMAT 1: 750 Q48 V46
GPA: 3.02
WE: Engineering (Other)
GMAT ToolKit User
Re: If each expression under the square root is greater than or [#permalink]

Show Tags

New post 05 Nov 2013, 09:03
WOW! That is a really well concealed trap.
Intern
Intern
avatar
Joined: 24 Sep 2012
Posts: 8
Re: If each expression under the square root is greater than or [#permalink]

Show Tags

New post 06 Jul 2014, 21:55
Bunuel 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) \(x + \sqrt{x-2}\)

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 \(2-x\geq{0}\) --> \(x\leq{2}\). Next: as \(x\leq{2}\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

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

Answer: A.




Hi,
As per GMAT Club Math book v3 page 14 - Roots... It is mentioned " • When the GMAT provides the square root sign for an even root, such as or , then the only accepted
answer is the positive root."

So how can one take |x-3| as the square root? The square root can only be +ve (x-3)
Please explain.
Thanks
Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 46164
Re: If each expression under the square root is greater than or [#permalink]

Show Tags

New post 07 Jul 2014, 01:27
nehamodak wrote:
Bunuel 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) \(x + \sqrt{x-2}\)

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 \(2-x\geq{0}\) --> \(x\leq{2}\). Next: as \(x\leq{2}\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

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

Answer: A.




Hi,
As per GMAT Club Math book v3 page 14 - Roots... It is mentioned " • When the GMAT provides the square root sign for an even root, such as or , then the only accepted
answer is the positive root."

So how can one take |x-3| as the square root? The square root can only be +ve (x-3)
Please explain.
Thanks


An absolute value is always more than or equal to zero. So, |x - 3| >=0.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Expert Post
Veritas Prep GMAT Instructor
User avatar
P
Joined: 16 Oct 2010
Posts: 8100
Location: Pune, India
Re: If each expression under the square root is greater than or [#permalink]

Show Tags

New post 07 Jul 2014, 05:12
nehamodak wrote:
Bunuel 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) \(x + \sqrt{x-2}\)

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 \(2-x\geq{0}\) --> \(x\leq{2}\). Next: as \(x\leq{2}\) then \(|x-3|\) becomes \(|x-3|=-(x-3)=-x+3\).

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

Answer: A.




Hi,
As per GMAT Club Math book v3 page 14 - Roots... It is mentioned " • When the GMAT provides the square root sign for an even root, such as or , then the only accepted
answer is the positive root."

So how can one take |x-3| as the square root? The square root can only be +ve (x-3)
Please explain.
Thanks


Let me also add here that when we take absolute value of a number/expression, we are implying that it doesn't matter whether the number is positive or negative - we will take it to be positive only.
Say, if x-3 is 7, then |x-3| is 7. When x-3 is -7, |x-3| is 7 only.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

1 KUDOS received
Intern
Intern
avatar
B
Joined: 09 Jan 2015
Posts: 1
Re: If each expression under the square root is greater than or [#permalink]

Show Tags

New post 22 Jul 2015, 01:22
1
If we just plug in the number 1 we can get the answer in 30 seconds max.
1 KUDOS received
Manager
Manager
avatar
S
Joined: 06 Jun 2013
Posts: 179
Location: India
Concentration: Finance, Economics
Schools: Tuck
GMAT 1: 640 Q49 V30
GPA: 3.6
WE: Engineering (Computer Software)
Premium Member
Re: If each expression under the square root is greater than or [#permalink]

Show Tags

New post 23 Sep 2015, 11:41
1
i assumed x=1 and compared the answer with the options and A is only matching.
Manager
Manager
avatar
Joined: 01 Aug 2014
Posts: 57
Schools: Rotman '17 (A)
GMAT 1: 710 Q44 V42
Re: If each expression under the square root is greater than or [#permalink]

Show Tags

New post 09 Jan 2016, 18:16
How should we intuitively know that x is = or < 0?

Would a correct approach include setting x to 0 and plugging in to derive an answer, then setting x to a negative number say -2 and plugging in to derive an answer, and then picking the corresponding answer from the answer choices? Or am I way off here?
Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 46164
Re: If each expression under the square root is greater than or [#permalink]

Show Tags

New post 10 Jan 2016, 06:27
Anonamy wrote:
How should we intuitively know that x is = or < 0?

Would a correct approach include setting x to 0 and plugging in to derive an answer, then setting x to a negative number say -2 and plugging in to derive an answer, and then picking the corresponding answer from the answer choices? Or am I way off here?


Please re-read the solutions above. We derived that x≤2 not x≤0.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Manager
Manager
avatar
Joined: 01 Aug 2014
Posts: 57
Schools: Rotman '17 (A)
GMAT 1: 710 Q44 V42
Re: If each expression under the square root is greater than or [#permalink]

Show Tags

New post 10 Jan 2016, 11:23
Ah, thank you, I see how the condition on x was derived now.
Expert Post
Director
Director
User avatar
B
Joined: 17 Dec 2012
Posts: 635
Location: India
If each expression under the square root is greater than or [#permalink]

Show Tags

New post 08 Mar 2018, 17:16
study 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) \(x + \sqrt{x-2}\)


Main Idea: We are likely looking for cancellation of terms

Details: We see the first term is sqrt(x-3)^2 and looks like the right term for cancellation with (x-3)

sqrt(x-3)^2 = |x-3| which can be x-3 or -(x-3) = -x+3

We can rule out x-3 because x< 2 and so it is negative

We now have the expressions as : -X+3 + sqrt(2-x) +x-3 = sqrt(2-x)

Hence A
_________________

Srinivasan Vaidyaraman
Sravna
http://www.sravnatestprep.com/best-online-gre-preparation.php

Improve Intuition and Your Score
Systematic Approaches

If each expression under the square root is greater than or   [#permalink] 08 Mar 2018, 17:16
Display posts from previous: Sort by

If each expression under the square root is greater than or

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

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®.