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01 Nov 2017, 07:23
I think this is a highquality question and I agree with explanation.



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09 Nov 2017, 00:51
Using the plugin method is a faster way to get correct answer.
As we know that expression under square root is <=0, x must be <=2.
Plug in any value of x that is <=2, say x=1, in given expressionand each answer choice.
Only Choice A will match.
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23 Nov 2017, 03:28
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. \(x + \sqrt{x  2}\) Hie Bunuel, Can you please provide link to similar kind of question



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23 Nov 2017, 04:28



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20 Feb 2018, 01:00
I think this is a highquality question and I agree with explanation.



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02 Jun 2018, 05:27
Here as per condition we can get both x<=2 and x>=3 wondering how only x<=2 condition was considered initially
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Updated on: 22 Jun 2018, 22:02
Hey Bunuel!
I'm a bit confused over here. I do understand that \(\sqrt{x^2} = x\). However, we are given that each expression under the square root is greater than or equal to 0. Hence \(\sqrt{x^2−6x+9} = \sqrt{(x3)^2} = x3\).
So the entity, \(x3 ≥ 0\) Hence, isn't the range of x becomes \(x≥3\) from this term?
Let me know where am I going wrong.
Thanks a lot PrayasT
Originally posted by PrayasT on 22 Jun 2018, 13:46.
Last edited by PrayasT on 22 Jun 2018, 22:02, edited 1 time in total.



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22 Jun 2018, 21:34
PrayasT wrote: Hey Bunnel!
I'm a bit confused over here. I do understand that \(\sqrt{x^2} = x\). However, we are given that each expression under the square root is greater than or equal to 0. Hence \(\sqrt{x^2−6x+9} = \sqrt{(x3)^2} = x3\).
So the entity, \(x3 ≥ 0\) Hence, isn't the range of x becomes \(x≥3\) from this term?
Let me know where am I going wrong.
Thanks a lot PrayasT Absolute value is always nonnegative so \(x3 ≥ 0\) is true for any value of x. Or in other way, since the square of a number is nonnegative then \(\sqrt{(x3)^2}\) is defined for any value of x. We get the range of x from \(\sqrt{2  x}\). The square root to be defined the expression under the square root must be greater than or equal to 0 > \(2x \ge 0\) \(\rightarrow\) \(x \le 2\).
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22 Jun 2018, 22:05
Bunuel wrote: PrayasT wrote: Hey Bunnel!
I'm a bit confused over here. I do understand that \(\sqrt{x^2} = x\). However, we are given that each expression under the square root is greater than or equal to 0. Hence \(\sqrt{x^2−6x+9} = \sqrt{(x3)^2} = x3\).
So the entity, \(x3 ≥ 0\) Hence, isn't the range of x becomes \(x≥3\) from this term?
Let me know where am I going wrong.
Thanks a lot PrayasT Absolute value is always nonnegative so \(x3 ≥ 0\) is true for any value of x. Or in other way, since the square of a number is nonnegative then \(\sqrt{(x3)^2}\) is defined for any value of x. We get the range of x from \(\sqrt{2  x}\). The square root to be defined the expression under the square root must be greater than or equal to 0 > \(2x \ge 0\) \(\rightarrow\) \(x \le 2\). That makes perfect sense! Thanks a lot! Regards, PrayasT



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20 Jul 2018, 20:36
I solved in the following manner:
Once you realize x<=2, plug in x=2 in the prompt, you get 0. Now check answer choices by plugging in x=2 to get the value 0.
Only answer choice A fits.



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12 Aug 2018, 10:25
What a great question.
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17 Nov 2018, 03:52
Hi Bunuel, In ques  Link: https://gmatclub.com/forum/d01183474.html#p1571995You have made a point that: "When the GMAT provides the square root sign for an even root, such as a square root, fourth root, erc. then the only accepted answer is the positive root. That is: √9=3, NOT +3 or 3; Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and 3." Why same is not valid here? Why are we considering the value of sqrt as positive and negative?



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17 Nov 2018, 07:14
SSwami wrote: Hi Bunuel, In ques  Link: https://gmatclub.com/forum/d01183474.html#p1571995You have made a point that: "When the GMAT provides the square root sign for an even root, such as a square root, fourth root, erc. then the only accepted answer is the positive root. That is: √9=3, NOT +3 or 3; Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and 3." Why same is not valid here? Why are we considering the value of sqrt as positive and negative? Where do you see a contradiction? When we have that \(\sqrt{x^2}=x\), still the square root is giving a nonnegative result because the absolute value of a number is always nonnegative. Please read the previous pages, I think several posts there explain this clearly.
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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.
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23 Dec 2018, 07:58
Why is the expression >=0 but becomes <=0 in the explanation? If according to the setting of the question, shouldn't the answer be D?



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23 Dec 2018, 10:43
Because we know X<=2, (X3) instead of (X3). But how do we make sure if (X3) satisfy X<=2?



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07 Jan 2019, 20:10
Fab Question!







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