Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 59561

Question Stats:
44% (01:31) correct 56% (01:34) wrong based on 162 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. \(x + \sqrt{x  2}\)
Official Answer and Stats are available only to registered users. Register/ Login.



Math Expert
Joined: 02 Sep 2009
Posts: 59561

Re M0816
[#permalink]
Show Tags
16 Sep 2014, 00:37
Official Solution:
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. \(x + \sqrt{x  2}\)
Absolute value properties: When \(x \le 0\) then \(x=x\), or more generally when \(\text{some expression} \le 0\) then \(\text{some expression} = (\text{some expression})\). For example: \(5=5=(5)\); When \(x \ge 0\) then \(x=x\), or more generally when \(\text{some expression} \ge 0\) then \(\text{some expression} = \text{some expression}\). For example: \(5=5\); Another important property to remember: \(\sqrt{x^2}=x\). Back to the original question: \(\sqrt{x^2  6x + 9} + \sqrt{2  x} + x  3=\sqrt{(x3)^2}+\sqrt{2x}+x3=x3+\sqrt{2x}+x3\). Now, as the expressions under the square roots are more than or equal to zero then \(2x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(x3\) becomes \(x3=(x3)=x+3\). \(x3+\sqrt{2x}+x3=x+3+\sqrt{2x}+x3=\sqrt{2x}\).
Answer: A



Manager
Joined: 14 Jul 2014
Posts: 88

Re: M0816
[#permalink]
Show Tags
20 Jan 2015, 06:49
Bunuel wrote: Official Solution:
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. \(x + \sqrt{x  2}\)
Absolute value properties: When \(x \le 0\) then \(x=x\), or more generally when \(\text{some expression} \le 0\) then \(\text{some expression} = (\text{some expression})\). For example: \(5=5=(5)\); When \(x \ge 0\) then \(x=x\), or more generally when \(\text{some expression} \ge 0\) then \(\text{some expression} = \text{some expression}\). For example: \(5=5\); Another important property to remember: \(\sqrt{x^2}=x\). Back to the original question: \(\sqrt{x^2  6x + 9} + \sqrt{2  x} + x  3=\sqrt{(x3)^2}+\sqrt{2x}+x3=x3+\sqrt{2x}+x3\). Now, as the expressions under the square roots are more than or equal to zero than \(2x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(x3\) becomes \(x3=(x3)=x+3\). \(x3+\sqrt{2x}+x3=x+3+\sqrt{2x}+x3=\sqrt{2x}\).
Answer: A Hi Bunuel I understand the properties but unable to comprehend the highlighted part. It says x is less than or equal to 2. But that does not mean that x is less than 0 , then how did we end up with \(x3\) becomes \(x3=(x3)=x+3\) Pls help understand this Thanks



Math Expert
Joined: 02 Sep 2009
Posts: 59561

Re: M0816
[#permalink]
Show Tags
20 Jan 2015, 07:26
buddyisraelgmat wrote: Bunuel wrote: Official Solution:
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. \(x + \sqrt{x  2}\)
Absolute value properties: When \(x \le 0\) then \(x=x\), or more generally when \(\text{some expression} \le 0\) then \(\text{some expression} = (\text{some expression})\). For example: \(5=5=(5)\); When \(x \ge 0\) then \(x=x\), or more generally when \(\text{some expression} \ge 0\) then \(\text{some expression} = \text{some expression}\). For example: \(5=5\); Another important property to remember: \(\sqrt{x^2}=x\). Back to the original question: \(\sqrt{x^2  6x + 9} + \sqrt{2  x} + x  3=\sqrt{(x3)^2}+\sqrt{2x}+x3=x3+\sqrt{2x}+x3\). Now, as the expressions under the square roots are more than or equal to zero than \(2x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(x3\) becomes \(x3=(x3)=x+3\). \(x3+\sqrt{2x}+x3=x+3+\sqrt{2x}+x3=\sqrt{2x}\).
Answer: A Hi Bunuel I understand the properties but unable to comprehend the highlighted part. It says x is less than or equal to 2. But that does not mean that x is less than 0 , then how did we end up with \(x3\) becomes \(x3=(x3)=x+3\) Pls help understand this Thanks When \(x \le 2\) then x3 (expression in modulus) is less than 0, thus \(x3\) becomes \(x3=(x3)=x+3\).



Manager
Joined: 14 Jul 2014
Posts: 88

Re: M0816
[#permalink]
Show Tags
20 Jan 2015, 07:42
Bunuel wrote: buddyisraelgmat wrote: Bunuel wrote: Official Solution:
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. \(x + \sqrt{x  2}\)
Absolute value properties: When \(x \le 0\) then \(x=x\), or more generally when \(\text{some expression} \le 0\) then \(\text{some expression} = (\text{some expression})\). For example: \(5=5=(5)\); When \(x \ge 0\) then \(x=x\), or more generally when \(\text{some expression} \ge 0\) then \(\text{some expression} = \text{some expression}\). For example: \(5=5\); Another important property to remember: \(\sqrt{x^2}=x\). Back to the original question: \(\sqrt{x^2  6x + 9} + \sqrt{2  x} + x  3=\sqrt{(x3)^2}+\sqrt{2x}+x3=x3+\sqrt{2x}+x3\). Now, as the expressions under the square roots are more than or equal to zero than \(2x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(x3\) becomes \(x3=(x3)=x+3\). \(x3+\sqrt{2x}+x3=x+3+\sqrt{2x}+x3=\sqrt{2x}\).
Answer: A Hi Bunuel I understand the properties but unable to comprehend the highlighted part. It says x is less than or equal to 2. But that does not mean that x is less than 0 , then how did we end up with \(x3\) becomes \(x3=(x3)=x+3\) Pls help understand this Thanks When \(x \le 2\) then x3 (expression in modulus) is less than 0, thus \(x3\) becomes \(x3=(x3)=x+3\). Oh Ok. Logically makes sense. Thanks



Manager
Joined: 08 Sep 2012
Posts: 59
Location: India
Concentration: Social Entrepreneurship, General Management
WE: Engineering (Investment Banking)

Re: M0816
[#permalink]
Show Tags
08 Aug 2015, 23:47
buddyisraelgmat wrote: Bunuel wrote: Official Solution:
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. \(x + \sqrt{x  2}\)
Absolute value properties: When \(x \le 0\) then \(x=x\), or more generally when \(\text{some expression} \le 0\) then \(\text{some expression} = (\text{some expression})\). For example: \(5=5=(5)\); When \(x \ge 0\) then \(x=x\), or more generally when \(\text{some expression} \ge 0\) then \(\text{some expression} = \text{some expression}\). For example: \(5=5\); Another important property to remember: \(\sqrt{x^2}=x\). Back to the original question: \(\sqrt{x^2  6x + 9} + \sqrt{2  x} + x  3=\sqrt{(x3)^2}+\sqrt{2x}+x3=x3+\sqrt{2x}+x3\). Now, as the expressions under the square roots are more than or equal to zero than \(2x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(x3\) becomes \(x3=(x3)=x+3\). \(x3+\sqrt{2x}+x3=x+3+\sqrt{2x}+x3=\sqrt{2x}\).
Answer: A Hi Bunuel I understand the properties but unable to comprehend the highlighted part. It says x is less than or equal to 2. But that does not mean that x is less than 0 , then how did we end up with \(x3\) becomes \(x3=(x3)=x+3\) Pls help understand this Thanks Bunuel, I still not do not understand that when the question stem states that expressions under the root are positive, then why have you considered x3 ? We should then only consider the positive option (x3) I chose (x3) for the first term and ended up as option B Am i missing anything?



Intern
Joined: 30 Jul 2014
Posts: 11
Location: Brazil
Concentration: Other, Strategy
GPA: 3.95
WE: Consulting (Education)

Re: M0816
[#permalink]
Show Tags
14 Aug 2015, 12:57
sagar2911 wrote: buddyisraelgmat wrote: Bunuel wrote: Official Solution:
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. \(x + \sqrt{x  2}\)
Absolute value properties: When \(x \le 0\) then \(x=x\), or more generally when \(\text{some expression} \le 0\) then \(\text{some expression} = (\text{some expression})\). For example: \(5=5=(5)\); When \(x \ge 0\) then \(x=x\), or more generally when \(\text{some expression} \ge 0\) then \(\text{some expression} = \text{some expression}\). For example: \(5=5\); Another important property to remember: \(\sqrt{x^2}=x\). Back to the original question: \(\sqrt{x^2  6x + 9} + \sqrt{2  x} + x  3=\sqrt{(x3)^2}+\sqrt{2x}+x3=x3+\sqrt{2x}+x3\). Now, as the expressions under the square roots are more than or equal to zero than \(2x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(x3\) becomes \(x3=(x3)=x+3\). \(x3+\sqrt{2x}+x3=x+3+\sqrt{2x}+x3=\sqrt{2x}\).
Answer: A Hi Bunuel I understand the properties but unable to comprehend the highlighted part. It says x is less than or equal to 2. But that does not mean that x is less than 0 , then how did we end up with \(x3\) becomes \(x3=(x3)=x+3\) Pls help understand this Thanks Bunuel, I still not do not understand that when the question stem states that expressions under the root are positive, then why have you considered x3 ? We should then only consider the positive option (x3) I chose (x3) for the first term and ended up as option B Am i missing anything? Because the question says "each expression under the square root is greater than or equal to 0", we need to understand that if x3 is 0, then 2x is a negative number, so the root of 2  x is root of 1 what breaks the rule stated in the question.



Manager
Joined: 08 Sep 2012
Posts: 59
Location: India
Concentration: Social Entrepreneurship, General Management
WE: Engineering (Investment Banking)

Re: M0816
[#permalink]
Show Tags
15 Aug 2015, 00:33
roneileonel wrote: sagar2911 wrote: Bunuel,
I still not do not understand that when the question stem states that expressions under the root are positive, then why have you considered x3 ? We should then only consider the positive option (x3) I chose (x3) for the first term and ended up as option B
Am i missing anything?
Because the question says "each expression under the square root is greater than or equal to 0", we need to understand that if x3 is 0, then 2x is a negative number, so the root of 2  x is root of 1 what breaks the rule stated in the question. Thank you dude! This helps a lot!



Intern
Joined: 29 Oct 2014
Posts: 24
GMAT Date: 12062014

Re: M0816
[#permalink]
Show Tags
11 Feb 2016, 09:41
schak2rhyme wrote: I think this is a highquality question and the explanation isn't clear enough, please elaborate. Next: as x≤2 then x−3 becomes x−3=−(x−3)=−x+3.
In the excerpt from the solution above why is x−3 not >=0 instead of <=.Please explain. The question says the expression cant be negative .. not roots .. x=3 because the whole expression root of 2x should be positive ..even if x is negative the first expression remains positive .. am i correct ?



Manager
Joined: 21 Sep 2015
Posts: 73
Location: India
GMAT 1: 730 Q48 V42 GMAT 2: 750 Q50 V41

Re: M0816
[#permalink]
Show Tags
16 Jul 2016, 21:32
I think this is a highquality question and I agree with explanation. Brilliant Question !



Intern
Joined: 13 Jun 2016
Posts: 18

Re: M0816
[#permalink]
Show Tags
01 Sep 2016, 21:09
Please let me know if I did this correctly alternatively.
\(\sqrt{x^2  6x + 9}\) , solving (x3)*(x3) , x = 3,
plug in 3 for all x, becomes answer A



Math Expert
Joined: 02 Sep 2009
Posts: 59561

Re: M0816
[#permalink]
Show Tags
02 Sep 2016, 03:58
xnthic wrote: Please let me know if I did this correctly alternatively.
\(\sqrt{x^2  6x + 9}\) , solving (x3)*(x3) , x = 3,
plug in 3 for all x, becomes answer A No, that's not correct. Why are you equating \(\sqrt{x^2  6x + 9}\) to 0? Why should x be 3? As shown in the solution x must be less than or equal to 2 for \(\sqrt{2x}\) to be defined.



Intern
Joined: 26 Apr 2016
Posts: 5

Re: M0816
[#permalink]
Show Tags
26 Sep 2016, 12:42
This is a really great question. Thanks Bunuel



Intern
Joined: 17 Aug 2016
Posts: 46

Re: M0816
[#permalink]
Show Tags
22 Jan 2017, 05:32
I think this is a highquality question and I agree with explanation.



Intern
Joined: 12 Jan 2017
Posts: 1

Re: M0816
[#permalink]
Show Tags
02 Jul 2017, 05:44
Bunuel wrote: Official Solution:
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. \(x + \sqrt{x  2}\)
Absolute value properties: When \(x \le 0\) then \(x=x\), or more generally when \(\text{some expression} \le 0\) then \(\text{some expression} = (\text{some expression})\). For example: \(5=5=(5)\); When \(x \ge 0\) then \(x=x\), or more generally when \(\text{some expression} \ge 0\) then \(\text{some expression} = \text{some expression}\). For example: \(5=5\); Another important property to remember: \(\sqrt{x^2}=x\). Back to the original question: \(\sqrt{x^2  6x + 9} + \sqrt{2  x} + x  3=\sqrt{(x3)^2}+\sqrt{2x}+x3=x3+\sqrt{2x}+x3\). Now, as the expressions under the square roots are more than or equal to zero than \(2x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(x3\) becomes \(x3=(x3)=x+3\). \(x3+\sqrt{2x}+x3=x+3+\sqrt{2x}+x3=\sqrt{2x}\).
Answer: A Hey Bunuel, i have a doubt, it will be of great help if you can reply. Since the question tells you that the quantity under the root should be positive, why are you considering x3 to be positive. the quantity under the root is (x3)^2, which is always positie irrespective of what value x takes.



Math Expert
Joined: 02 Sep 2009
Posts: 59561

Re: M0816
[#permalink]
Show Tags
02 Jul 2017, 21:22
prateeksaxena55 wrote: Bunuel wrote: Official Solution:
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. \(x + \sqrt{x  2}\)
Absolute value properties: When \(x \le 0\) then \(x=x\), or more generally when \(\text{some expression} \le 0\) then \(\text{some expression} = (\text{some expression})\). For example: \(5=5=(5)\); When \(x \ge 0\) then \(x=x\), or more generally when \(\text{some expression} \ge 0\) then \(\text{some expression} = \text{some expression}\). For example: \(5=5\); Another important property to remember: \(\sqrt{x^2}=x\). Back to the original question: \(\sqrt{x^2  6x + 9} + \sqrt{2  x} + x  3=\sqrt{(x3)^2}+\sqrt{2x}+x3=x3+\sqrt{2x}+x3\). Now, as the expressions under the square roots are more than or equal to zero than \(2x \ge 0\) \(\rightarrow\) \(x \le 2\). Next: as \(x \le 2\) then \(x3\) becomes \(x3=(x3)=x+3\). \(x3+\sqrt{2x}+x3=x+3+\sqrt{2x}+x3=\sqrt{2x}\).
Answer: A Hey Bunuel, i have a doubt, it will be of great help if you can reply. Since the question tells you that the quantity under the root should be positive, why are you considering x3 to be positive. the quantity under the root is (x3)^2, which is always positie irrespective of what value x takes. The point is that \(\sqrt{x^2}=x\). Next, \(x = x\), when \(x \geq 0\) and \(x = x\), when \(x \leq 0\). We have \(\sqrt{x^2  6x + 9}=\sqrt{(x3)^2}\). Thus according to the above property \(\sqrt{x^2  6x + 9}=\sqrt{(x3)^2}=x3\). Next, since \(x \le 2\), then \(x  3 < 0\) and therefore \(x3 = (x3) = 3x\). Notice here that since \(x \le 2\), then 3  x will be positive, so the condition that the square root must give a nonnegative result is satisfied.



Intern
Joined: 30 Jun 2017
Posts: 7
Location: India

Re: M0816
[#permalink]
Show Tags
04 Jul 2017, 21:48
The question says " each expression under the square root is greater than or equal to 0" Hence, It comes out that x is less than or equal to 2 by second expression and for the first expression why at all we are thinking it to be square of (x3) as this will result in values greater or equal to 3 not valid as it is less than 2. but if we take it as square of (3x) then it will less than 3 satisfying outcome from other equation.
Therefore, the shorter method of solving will be as follows.
Step 1. from second expression find x is less than or equal to 2 Step 2. solve first expression as sqaure of (3x) not square of (x3) { as this gives x is greater or equal to 3} and then you get the answer.
It avoids complexity of modulus.



Manager
Joined: 31 Oct 2016
Posts: 105

Re: M0816
[#permalink]
Show Tags
24 Oct 2017, 23:53
In the GMAT Club math book: "When the GMAT provides the square root sign for an even root, then the only accepted answer is the positive root". That is, sqrt(25) = 5, NOT +5.
So, why sqrt(x^2) must be x, not just +x?



Math Expert
Joined: 02 Sep 2009
Posts: 59561

Re: M0816
[#permalink]
Show Tags
25 Oct 2017, 00:00
Terabyte wrote: In the GMAT Club math book: "When the GMAT provides the square root sign for an even root, then the only accepted answer is the positive root". That is, sqrt(25) = 5, NOT +5.
So, why sqrt(x^2) must be x, not just +x? \(\sqrt{x^2}=x\):[/b] The point here is that since square root function cannot give negative result then \(\sqrt{some \ expression}\geq{0}\). So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to? Let's consider following examples: If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\). So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=x\), if \(x<0\). What function does exactly the same thing? The absolute value function: \(x=x\), if \(x\geq{0}\) and \(x=x\), if \(x<0\). That is why \(\sqrt{x^2}=x\). Hope it's clear.



Manager
Joined: 31 Oct 2016
Posts: 105

Re: M0816
[#permalink]
Show Tags
25 Oct 2017, 00:05
Bunuel wrote: \(\sqrt{x^2}=x\):[/b]
The point here is that since square root function cannot give negative result then \(\sqrt{some \ expression}\geq{0}\).
So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?
Let's consider following examples: If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\); If \(x=5\) > \(\sqrt{x^2}=\sqrt{25}=5=x=positive\).
So we got that: \(\sqrt{x^2}=x\), if \(x\geq{0}\); \(\sqrt{x^2}=x\), if \(x<0\).
What function does exactly the same thing? The absolute value function: \(x=x\), if \(x\geq{0}\) and \(x=x\), if \(x<0\). That is why \(\sqrt{x^2}=x\).
Hope it's clear.
Yes, it's clear. Thank you very much, Bunuel. Learning a lot from your posts. Aiming for Q50







Go to page
1 2
Next
[ 38 posts ]



