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Intern
Joined: 04 May 2017
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Re: M08-16
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 01 Nov 2017, 07:23
I think this is a high-quality question and I agree with explanation.
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Manager
Joined: 01 Feb 2017
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Re: M08-16
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09 Nov 2017, 00:51
Using the plug-in 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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Re: M08-16
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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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Re: M08-16
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23 Nov 2017, 04:28
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Re M08-16
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20 Feb 2018, 01:00
I think this is a high-quality question and I agree with explanation.
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Joined: 26 Dec 2017
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Re: M08-16
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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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Re: M08-16
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02 Jun 2018, 08:27
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Manager
Joined: 26 Feb 2018
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M08-16
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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 \(\sqrt{x^2−6x+9} = \sqrt{(x-3)^2} = |x-3|\).
So the entity, \(|x-3| ≥ 0\) 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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Re: M08-16
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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 \(\sqrt{x^2−6x+9} = \sqrt{(x-3)^2} = |x-3|\).
So the entity, \(|x-3| ≥ 0\) 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 non-negative so \(|x-3| ≥ 0\) is true for any value of x. Or in other way, since the square of a number is non-negative then \(\sqrt{(x-3)^2}\) 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 --> \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\).
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Joined: 26 Feb 2018
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Re: M08-16
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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 \(\sqrt{x^2−6x+9} = \sqrt{(x-3)^2} = |x-3|\).
So the entity, \(|x-3| ≥ 0\) 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 non-negative so \(|x-3| ≥ 0\) is true for any value of x. Or in other way, since the square of a number is non-negative then \(\sqrt{(x-3)^2}\) 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 --> \(2-x \ge 0\) \(\rightarrow\) \(x \le 2\). That makes perfect sense! Thanks a lot! Regards, PrayasT
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Re: M08-16
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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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Re: M08-16
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12 Aug 2018, 10:25
What a great question.
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Re: M08-16
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17 Nov 2018, 03:52
Hi Bunuel, In ques - Link: https://gmatclub.com/forum/d01-183474.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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Joined: 02 Sep 2009
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Re: M08-16
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17 Nov 2018, 07:14
SSwami wrote: Hi Bunuel, In ques - Link: https://gmatclub.com/forum/d01-183474.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 non-negative result because the absolute value of a number is always non-negative. Please read the previous pages, I think several posts there explain this clearly.
_________________
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
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Intern
Joined: 18 Dec 2018
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Re M08-16
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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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Joined: 18 Dec 2018
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Re M08-16
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23 Dec 2018, 10:43
Because we know X<=2, -(X-3) instead of (X-3).
But how do we make sure if -(X-3) satisfy X<=2?
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Joined: 22 Oct 2018
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Re M08-16
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07 Jan 2019, 20:10
Fab Question!
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