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Intern
Joined: 04 May 2017
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Re: M08-16
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 01 Nov 2017, 08:23
I think this is a high-quality question and I agree with explanation.
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Joined: 01 Feb 2017
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Re: M08-16
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09 Nov 2017, 01:51
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, 04:28
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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Joined: 02 Sep 2009
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Re: M08-16
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23 Nov 2017, 05:28
prk0789 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. \(\sqrt{2 - x}\)
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 8. Exponents and Roots of Numbers
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Re M08-16
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20 Feb 2018, 02:00
I think this is a high-quality question and I agree with explanation.
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Re: M08-16
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02 Jun 2018, 06:27
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, 09:27
tejyr wrote: Here as per condition we can get both x<=2 and x>=3
wondering how only x<=2 condition was considered initially The square of a number is 0 or positive, so from \(\sqrt{(x-3)^2}\) we don't have x >= 3 because (x - 3)^2 is non-negative for any value of x, not only for x >= 3.
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Joined: 26 Feb 2018
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M08-16
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Updated on: 22 Jun 2018, 23:02
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, 14:46.
Last edited by PrayasT on 22 Jun 2018, 23:02, edited 1 time in total.
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Joined: 02 Sep 2009
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Re: M08-16
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22 Jun 2018, 22:34
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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Re: M08-16
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22 Jun 2018, 23:05
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, 21:36
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, 11:25
What a great question.
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Re: M08-16
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17 Nov 2018, 04:52
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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Re: M08-16
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17 Nov 2018, 08:14
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.
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Re M08-16
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23 Dec 2018, 08:58
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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Re M08-16
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23 Dec 2018, 11:43
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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Re M08-16
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07 Jan 2019, 21:10
Fab Question!
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