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15 Sep 2014, 23:37
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
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20 Jan 2015, 05:49
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



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20 Jan 2015, 06:26
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 \(x3=(x3)=x+3\).
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20 Jan 2015, 06:42
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



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08 Aug 2015, 22:47
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?
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14 Aug 2015, 11:57
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.



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14 Aug 2015, 23:33
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!
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11 Feb 2016, 08:41
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 ?



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16 Jul 2016, 20:32
I think this is a highquality question and I agree with explanation. Brilliant Question !
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01 Sep 2016, 20:09
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



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26 Sep 2016, 11:42
This is a really great question. Thanks Bunuel



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22 Jan 2017, 04:32
I think this is a highquality question and I agree with explanation.



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02 Jul 2017, 04:44
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.



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02 Jul 2017, 20:22
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.
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04 Jul 2017, 20:48
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.



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24 Oct 2017, 22:53
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?



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24 Oct 2017, 23:00
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.
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24 Oct 2017, 23:05
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







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