Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 05 Oct 2008
Posts: 207

If each expression under the square root is greater than or
[#permalink]
Show Tags
Updated on: 09 Oct 2013, 01:21
Question Stats:
37% (01:47) correct 63% (01:38) wrong based on 1721 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) \(\sqrt{2x}\) (B) \(2x6 + \sqrt{2x}\) (C) \(\sqrt{2x} + x3\) (D) \(2x6 + \sqrt{x2}\) (E) \(x + \sqrt{x2}\) 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 (x3)^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: \(\sqrt{x^2  6x + 9} + \sqrt{2  x} + (x  3) = (x  3) + \sqrt{2  x} + (x  3) = 2x  6 + \sqrt{x  2}\). M1816
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by study on 21 Nov 2009, 00:40.
Last edited by Bunuel on 09 Oct 2013, 01:21, edited 1 time in total.
Renamed the topic, edited the question and added the OA.




Math Expert
Joined: 02 Sep 2009
Posts: 65062

Re: If each expression under the square root is greater than or
[#permalink]
Show Tags
09 Oct 2013, 01:22
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) \(\sqrt{2x}\) (B) \(2x6 + \sqrt{2x}\) (C) \(\sqrt{2x} + x3\) (D) \(2x6 + \sqrt{x2}\) (E) \(x + \sqrt{x2}\) One important note: \(\sqrt{x^2}=x\) \(\sqrt{x^2  6x + 9} + \sqrt{2  x} + x  3=\sqrt{(x3)^2}+\sqrt{2x}+x3=x3+\sqrt{2x}+x3\). Now, as the expressions under the square roots are more than or equal to zero than \(2x\geq{0}\) > \(x\leq{2}\). Next: as \(x\leq{2}\) then \(x3\) becomes \(x3=(x3)=x+3\). \(x3+\sqrt{2x}+x3=x+3+\sqrt{2x}+x3=\sqrt{2x}\). Answer: A.
_________________




Tuck School Moderator
Joined: 20 Aug 2009
Posts: 224
Location: Tbilisi, Georgia
Schools: Stanford (in), Tuck (WL), Wharton (ding), Cornell (in)

Re: Inequalities  M08
[#permalink]
Show Tags
21 Nov 2009, 01:39
studyQuote: Even if x <= 2, rt (x3)^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 (3x). 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 \(\sqrt{(x3)^2}=x3=(x3)=3x\), because (2x)>=0; x<=2; (x3)<0 And so the correct answer is \(3x+\sqrt{2x}+x3=\sqrt{2x}\)




Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10646
Location: Pune, India

Re: Inequalities  M08
[#permalink]
Show Tags
08 Oct 2013, 20:48
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 (x3)^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 \(x3 + \sqrt{2  x} + (x  3)\) Note that \(\sqrt{x^2} = x\) (and not x) Since we know that x < 2, (x  3) must be negative. Hence \(x3 = (x  3)\) Expression becomes: \((x  3) + \sqrt{2  x} + (x  3) = \sqrt{2  x}\)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10646
Location: Pune, India

Re: If each expression under the square root is greater than or
[#permalink]
Show Tags
09 Oct 2013, 19:27
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) \(\sqrt{2x}\) (B) \(2x6 + \sqrt{2x}\) (C) \(\sqrt{2x} + x3\) (D) \(2x6 + \sqrt{x2}\) (E) \(x + \sqrt{x2}\)
One important note: \(\sqrt{x^2}=x\)
\(\sqrt{x^2  6x + 9} + \sqrt{2  x} + x  3=\sqrt{(x3)^2}+\sqrt{2x}+x3=x3+\sqrt{2x}+x3\).
Now, as the expressions under the square roots are more than or equal to zero than \(2x>0\) > \(x<2\). Next: as \(x<2\) then \(x3\) becomes \(x3=(x3)=x+3\).
\(x3+\sqrt{2x}+x3=x+3+\sqrt{2x}+x3=\sqrt{2x}\).
Answer: A. I don't follow the highlighted part at all. So if you have xn, 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, (x3) is negative. So x3 = (x  3) = 3  x
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Math Expert
Joined: 02 Sep 2009
Posts: 65062

Re: If each expression under the square root is greater than or
[#permalink]
Show Tags
27 May 2019, 00:52
Asad wrote: 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) \(\sqrt{2x}\) (B) \(2x6 + \sqrt{2x}\) (C) \(\sqrt{2x} + x3\) (D) \(2x6 + \sqrt{x2}\) (E) \(x + \sqrt{x2}\)
Hi Bunuel, IanStewartIf the question stem is changed with below, the correct choice still be A, right? If each expression under the square root is greater than 0, what is \(\sqrt{x^2  6x + 9} + \sqrt{3  x} + x  3\)? (A) \(\sqrt{2x}\) (B) \(2x6 + \sqrt{2x}\) (C) \(\sqrt{2x} + x3\) (D) \(2x6 + \sqrt{x2}\) (E) \(x + \sqrt{x2}\) In this case the correct answer would be \(\sqrt{3x}\). \(\sqrt{x^2  6x + 9} + \sqrt{3  x} + x  3=\sqrt{(x3)^2}+\sqrt{3x}+x3=x3+\sqrt{3x}+x3\). Now, as the expressions under the square roots are more than or equal to zero than \(3x\geq{0}\) > \(x\leq{3}\). Next: as \(x\leq{3}\) then \(x3\) becomes \(x3=(x3)=x+3\). \(x3+\sqrt{3x}+x3=x+3+\sqrt{3x}+x3=\sqrt{3x}\).
_________________



VP
Joined: 06 Sep 2013
Posts: 1491
Concentration: Finance

Re: Inequalities  M08
[#permalink]
Show Tags
08 Oct 2013, 15:16
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 (x3)^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



Intern
Joined: 09 Jan 2015
Posts: 1

Re: If each expression under the square root is greater than or
[#permalink]
Show Tags
22 Jul 2015, 00:22
If we just plug in the number 1 we can get the answer in 30 seconds max.



Manager
Joined: 06 Jun 2013
Posts: 140
Location: India
Concentration: Finance, Economics
GPA: 3.6
WE: Engineering (Computer Software)

Re: If each expression under the square root is greater than or
[#permalink]
Show Tags
23 Sep 2015, 10:41
i assumed x=1 and compared the answer with the options and A is only matching.



Manager
Joined: 26 Sep 2013
Posts: 180
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
09 Oct 2013, 17:29
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) \(\sqrt{2x}\) (B) \(2x6 + \sqrt{2x}\) (C) \(\sqrt{2x} + x3\) (D) \(2x6 + \sqrt{x2}\) (E) \(x + \sqrt{x2}\)
One important note: \(\sqrt{x^2}=x\)
\(\sqrt{x^2  6x + 9} + \sqrt{2  x} + x  3=\sqrt{(x3)^2}+\sqrt{2x}+x3=x3+\sqrt{2x}+x3\).
Now, as the expressions under the square roots are more than or equal to zero than \(2x>0\) > \(x<2\). Next: as \(x<2\) then \(x3\) becomes \(x3=(x3)=x+3\).
\(x3+\sqrt{2x}+x3=x+3+\sqrt{2x}+x3=\sqrt{2x}\).
Answer: A. I don't follow the highlighted part at all. So if you have xn, 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?



Manager
Joined: 26 Sep 2013
Posts: 180
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
10 Oct 2013, 05:57
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) \(\sqrt{2x}\) (B) \(2x6 + \sqrt{2x}\) (C) \(\sqrt{2x} + x3\) (D) \(2x6 + \sqrt{x2}\) (E) \(x + \sqrt{x2}\)
One important note: \(\sqrt{x^2}=x\)
\(\sqrt{x^2  6x + 9} + \sqrt{2  x} + x  3=\sqrt{(x3)^2}+\sqrt{2x}+x3=x3+\sqrt{2x}+x3\).
Now, as the expressions under the square roots are more than or equal to zero than \(2x>0\) > \(x<2\). Next: as \(x<2\) then \(x3\) becomes \(x3=(x3)=x+3\).
\(x3+\sqrt{2x}+x3=x+3+\sqrt{2x}+x3=\sqrt{2x}\).
Answer: A. I don't follow the highlighted part at all. So if you have xn, 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, (x3) is negative. So x3 = (x  3) = 3  x OOOOH I see now, thank you,



Manager
Joined: 03 Jan 2013
Posts: 184
Location: United States
Concentration: Finance, Entrepreneurship
GPA: 3.02
WE: Engineering (Other)

Re: If each expression under the square root is greater than or
[#permalink]
Show Tags
05 Nov 2013, 08:03
WOW! That is a really well concealed trap.



Intern
Joined: 24 Sep 2012
Posts: 8

Re: If each expression under the square root is greater than or
[#permalink]
Show Tags
06 Jul 2014, 20:55
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) \(\sqrt{2x}\) (B) \(2x6 + \sqrt{2x}\) (C) \(\sqrt{2x} + x3\) (D) \(2x6 + \sqrt{x2}\) (E) \(x + \sqrt{x2}\)
One important note: \(\sqrt{x^2}=x\)
\(\sqrt{x^2  6x + 9} + \sqrt{2  x} + x  3=\sqrt{(x3)^2}+\sqrt{2x}+x3=x3+\sqrt{2x}+x3\).
Now, as the expressions under the square roots are more than or equal to zero than \(2x\geq{0}\) > \(x\leq{2}\). Next: as \(x\leq{2}\) then \(x3\) becomes \(x3=(x3)=x+3\).
\(x3+\sqrt{2x}+x3=x+3+\sqrt{2x}+x3=\sqrt{2x}\).
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 x3 as the square root? The square root can only be +ve (x3) Please explain. Thanks



Math Expert
Joined: 02 Sep 2009
Posts: 65062

Re: If each expression under the square root is greater than or
[#permalink]
Show Tags
07 Jul 2014, 00:27
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) \(\sqrt{2x}\) (B) \(2x6 + \sqrt{2x}\) (C) \(\sqrt{2x} + x3\) (D) \(2x6 + \sqrt{x2}\) (E) \(x + \sqrt{x2}\)
One important note: \(\sqrt{x^2}=x\)
\(\sqrt{x^2  6x + 9} + \sqrt{2  x} + x  3=\sqrt{(x3)^2}+\sqrt{2x}+x3=x3+\sqrt{2x}+x3\).
Now, as the expressions under the square roots are more than or equal to zero than \(2x\geq{0}\) > \(x\leq{2}\). Next: as \(x\leq{2}\) then \(x3\) becomes \(x3=(x3)=x+3\).
\(x3+\sqrt{2x}+x3=x+3+\sqrt{2x}+x3=\sqrt{2x}\).
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 x3 as the square root? The square root can only be +ve (x3) Please explain. Thanks An absolute value is always more than or equal to zero. So, x  3 >=0.
_________________



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10646
Location: Pune, India

Re: If each expression under the square root is greater than or
[#permalink]
Show Tags
07 Jul 2014, 04:12
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) \(\sqrt{2x}\) (B) \(2x6 + \sqrt{2x}\) (C) \(\sqrt{2x} + x3\) (D) \(2x6 + \sqrt{x2}\) (E) \(x + \sqrt{x2}\)
One important note: \(\sqrt{x^2}=x\)
\(\sqrt{x^2  6x + 9} + \sqrt{2  x} + x  3=\sqrt{(x3)^2}+\sqrt{2x}+x3=x3+\sqrt{2x}+x3\).
Now, as the expressions under the square roots are more than or equal to zero than \(2x\geq{0}\) > \(x\leq{2}\). Next: as \(x\leq{2}\) then \(x3\) becomes \(x3=(x3)=x+3\).
\(x3+\sqrt{2x}+x3=x+3+\sqrt{2x}+x3=\sqrt{2x}\).
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 x3 as the square root? The square root can only be +ve (x3) 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 x3 is 7, then x3 is 7. When x3 is 7, x3 is 7 only.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 01 Aug 2014
Posts: 50

Re: If each expression under the square root is greater than or
[#permalink]
Show Tags
09 Jan 2016, 17:16
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?



Math Expert
Joined: 02 Sep 2009
Posts: 65062

Re: If each expression under the square root is greater than or
[#permalink]
Show Tags
10 Jan 2016, 05:27
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 reread the solutions above. We derived that x≤2 not x≤0.
_________________



Manager
Joined: 01 Aug 2014
Posts: 50

Re: If each expression under the square root is greater than or
[#permalink]
Show Tags
10 Jan 2016, 10:23
Ah, thank you, I see how the condition on x was derived now.



Director
Joined: 17 Dec 2012
Posts: 635
Location: India

If each expression under the square root is greater than or
[#permalink]
Show Tags
08 Mar 2018, 16:16
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) \(\sqrt{2x}\) (B) \(2x6 + \sqrt{2x}\) (C) \(\sqrt{2x} + x3\) (D) \(2x6 + \sqrt{x2}\) (E) \(x + \sqrt{x2}\)
Main Idea: We are likely looking for cancellation of terms Details: We see the first term is sqrt(x3)^2 and looks like the right term for cancellation with (x3) sqrt(x3)^2 = x3 which can be x3 or (x3) = x+3 We can rule out x3 because x< 2 and so it is negative We now have the expressions as : X+3 + sqrt(2x) +x3 = sqrt(2x) Hence A
_________________
Srinivasan Vaidyaraman Magical LogiciansHolistic and Holy Approach



Manager
Joined: 31 Jul 2017
Posts: 195
Location: Tajikistan

Re: If each expression under the square root is greater than or
[#permalink]
Show Tags
30 Jun 2018, 08:12
Hi Bunuel, chetan2u, VeritasPrepKarishmaI used the following approach, not sure if its the right reasoning though, can you please check it out x^26x+9=0 D=(6^2)364*9=0 x=(6+0)=3, so x is 3. substitute x with 0 in the expression and we get A as correct answer only.




Re: If each expression under the square root is greater than or
[#permalink]
30 Jun 2018, 08:12



Go to page
1 2
Next
[ 28 posts ]

