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Re M1920 [#permalink]
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16 Sep 2014, 00:06



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Hi Bunuel, I got the roots for the equation as (x3)^2 (reading the equation as x^2  6x +9  We've flipped the values of a and b in the equation a^2  2ab + b^2). Based on my approach x  3 would be the correct answer and on yours 3  x is the correct answer. Does my above para make any sense? If it does, and if the question has both of these options, which would be the correct answer? If it doesn't please help me understand my mistake.
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Re: M1920 [#permalink]
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21 Nov 2014, 03:51
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Re: M1920 [#permalink]
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21 Nov 2014, 04:05
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Bunuel wrote: The point is that x  3 = 3  x. Both indicate the distance between x and 3. So, I understand that both of our answers are correct and that both these values won't appear as options in a single question.
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21 Nov 2014, 04:07



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Re: M1920 [#permalink]
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24 Jul 2015, 06:44
Bunuel wrote: Official Solution:
Which of the following always equals \(\sqrt{9 + x^2  6x}\) ?
A. \(x  3\) B. \(3 + x\) C. \(3  x\) D. \(3 + x\) E. \(3  x\)
\(\sqrt{9 + x^2  6x} = \sqrt{(3  x)^2} = 3  x\) (by the definition of a square root).
Answer: C i answered choice a : x3 i took the the expression as x^26x+9 > (x3)^2. no my understanding is that on GMAT when some expression is under root sign, only positive root is considered...? is it correct. if it is so, the we need not take mod value...? pls help



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Re: M1920 [#permalink]
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24 Jul 2015, 06:51
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riyazgilani wrote: Bunuel wrote: Official Solution:
Which of the following always equals \(\sqrt{9 + x^2  6x}\) ?
A. \(x  3\) B. \(3 + x\) C. \(3  x\) D. \(3 + x\) E. \(3  x\)
\(\sqrt{9 + x^2  6x} = \sqrt{(3  x)^2} = 3  x\) (by the definition of a square root).
Answer: C i answered choice a : x3 i took the the expression as x^26x+9 > (x3)^2. no my understanding is that on GMAT when some expression is under root sign, only positive root is considered...? is it correct. if it is so, the we need not take mod value...? pls help Exactly because the square root function cannot give negative result the answer is 3x, an absolute value of 3x, which also cannot be negative. MUST KNOW: \(\sqrt{x^2}=x\):The point here is that since square root function can not 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 \(\sqrt{x^2}=x\). I'd advice to go through basics and only after to practice questions:
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Re M1920 [#permalink]
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22 Aug 2016, 13:02
I think this is a highquality question and I agree with explanation.



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Re: M1920 [#permalink]
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12 Sep 2016, 14:23
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After solving this question I got the result as X3 and I answered it as X3, And got this wrong. Thanks for giving the link which clears the basic of Absolute value including the result:
AB = BA
Which is clearly applicable in this question provided options.



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Bunuel wrote: riyazgilani wrote: Bunuel wrote: Official Solution:
Which of the following always equals \(\sqrt{9 + x^2  6x}\) ?
A. \(x  3\) B. \(3 + x\) C. \(3  x\) D. \(3 + x\) E. \(3  x\)
\(\sqrt{9 + x^2  6x} = \sqrt{(3  x)^2} = 3  x\) (by the definition of a square root).
Answer: C i answered choice a : x3 i took the the expression as x^26x+9 > (x3)^2. no my understanding is that on GMAT when some expression is under root sign, only positive root is considered...? is it correct. if it is so, the we need not take mod value...? pls help Exactly because the square root function cannot give negative result the answer is 3x, an absolute value of 3x, which also cannot be negative. MUST KNOW: \(\sqrt{x^2}=x\):The point here is that since square root function can not 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 \(\sqrt{x^2}=x\). I'd advice to go through basics and only after to practice questions: Hello Bunuel, VyshakI thought the equation was as follows: \(\sqrt{x^2}=x\), if \(x>{0}\); \(\sqrt{x^2}=x\), if \(x<={0}\). Am i wrong? Please correct me if so. Thanks



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Re: M1920 [#permalink]
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30 Mar 2017, 21:01



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Re: M1920 [#permalink]
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31 Mar 2017, 09:00
Dear Bunuel,
The answer choices CAN"T have the following answers in their list: \(3  x\) and \(x  3\) Because the following rule:
\(3  x\) = \(x  3\)
Am I right?
Thanks in advance



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Re: M1920 [#permalink]
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31 Mar 2017, 14:07
Bunuel wrote: manishtank1988 wrote: Hello Bunuel, VyshakI thought the equation was as follows: \(\sqrt{x^2}=x\), if \(x>{0}\); \(\sqrt{x^2}=x\), if \(x<={0}\). Am i wrong? Please correct me if so. Thanks You can put = sign in any of the two because \(\sqrt{0^2}=0=0\). Understood thanks...



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Re: M1920 [#permalink]
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21 Aug 2017, 04:28
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This question is based on the principle of \(\sqrt{x^2}=x\) Hence its \(x3=3x\) Straight away C. Press Kudos id this helps!



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Re: M1920 [#permalink]
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09 Feb 2018, 06:42
Hi Bunuel,
I used the plugin method...but still a little confused as to why C was correct...I did the following:
made x=2...then...√9+412 = √1 as my goal.
I then plugged in the two in place of x in the answer choices...to see if I got a figure that corresponded with my goal of √1.
A) 23= 1 NO B) 3+2= 5 NO C) [32] = [1] MAYBE D) [3+2] = [5] NO E) 32 = 1 MAYBE
I originally chose E because the goal of √1, seems to result in 1 and 1 only when one finds the square root. However, upon a second look...is [1] another way of saying √1, thus making C the correct answer?



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Re: M1920 [#permalink]
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09 Feb 2018, 06:49
ttaiwo wrote: Hi Bunuel,
I used the plugin method...but still a little confused as to why C was correct...I did the following:
made x=2...then...√9+412 = √1 as my goal.
I then plugged in the two in place of x in the answer choices...to see if I got a figure that corresponded with my goal of √1.
A) 23= 1 NO B) 3+2= 5 NO C) [32] = [1] MAYBE D) [3+2] = [5] NO E) 32 = 1 MAYBE
I originally chose E because the goal of √1, seems to result in 1 and 1 only when one finds the square root. However, upon a second look...is [1] another way of saying √1, thus making C the correct answer? C is correct because it's generally true, always true for ALL values of x. As for your approach: for plugin method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only. For example, check x = 4.
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Re: M1920 [#permalink]
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09 Feb 2018, 07:46
Hi Bunuel,
Yup, I've plugged in x=4 and got √1 as the goal (so the goal remained unchanged)
A) 1, NO B) 7, NO C) [1], MAYBE D) [7], NO E) 1, NO
So in this case, C is conclusively the case because it is always true for all values of x...
My question remains though...is √1 always [1], and is that also the case for √1?
Thanks again,
Tosin



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Re: M1920 [#permalink]
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09 Feb 2018, 07:51
ttaiwo wrote: Hi Bunuel,
Yup, I've plugged in x=4 and got √1 as the goal (so the goal remained unchanged)
A) 1, NO B) 7, NO C) [1], MAYBE D) [7], NO E) 1, NO
So in this case, C is conclusively the case because it is always true for all values of x...
My question remains though...is √1 always [1], and is that also the case for √1?
Thanks again,
Tosin \(\sqrt{1}=1\) only. \(\sqrt{1}\) is an imaginary number and since all numbers on the GMAT are real numbers you don't need to worry about that at all.. When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is: \(\sqrt{9} = 3\), NOT +3 or 3; \(\sqrt[4]{16} = 2\), NOT +2 or 2; Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and 3. Because \(x^2 = 9\) means that \(x =\sqrt{9}=3\) or \(x=\sqrt{9}=3\).
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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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Re: M1920 [#permalink]
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09 Feb 2018, 08:18
Bunuel wrote: ttaiwo wrote: Hi Bunuel,
Yup, I've plugged in x=4 and got √1 as the goal (so the goal remained unchanged)
A) 1, NO B) 7, NO C) [1], MAYBE D) [7], NO E) 1, NO
So in this case, C is conclusively the case because it is always true for all values of x...
My question remains though...is √1 always [1], and is that also the case for √1?
Thanks again,
Tosin \(\sqrt{1}=1\) only. \(\sqrt{1}\) is an imaginary number and since all numbers on the GMAT are real numbers you don't need to worry about that at all.. When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is: \(\sqrt{9} = 3\), NOT +3 or 3; \(\sqrt[4]{16} = 2\), NOT +2 or 2; Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and 3. Because \(x^2 = 9\) means that \(x =\sqrt{9}=3\) or \(x=\sqrt{9}=3\).Thanks










