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Re: M26-25
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 10 Dec 2016, 04:58
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Re: M26-25
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25 Mar 2017, 19:46
Very Good question
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Re: M26-25
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05 Dec 2017, 22:09
Bunuel wrote: blitheclyde wrote: Hi, Having trouble wrapping my head around why \(\sqrt{x^2}\) is equal to |x| I looked at the post on this page, but did not see any explanation. For example, how is the equation \(\sqrt{4}\) = {-2, 2} any different from \(\sqrt{2^2}\) = ? Is it because the solution of \(\sqrt{x^2}\) must be equal to the solution of \((\sqrt{x})^2\)? if so, why? Thanks in advance for any help! (with regard to bunnel's explanation to xnthic, throughout the explanation bunnel writes "\(\sqrt{25}\) = 5" as though this is given, but my understanding is that \(\sqrt{25}\) could be either positive or negative 5, as either positive or negative 5 squared result in 25. If I am incorrect, then that is most likely the root of my confusion) \(\sqrt{4}=2\) ONLY, not 2 or -2. The square root function cannot give negative result. Check here: http://gmatclub.com/forum/m26-184464.html#p1729559Hi Bunuel, certain application of the sqrt function seem to be contradictory. If sqrt function can't give negative numbers, why do we substitute both -ve and +ve values of \(\sqrt{expression}\) while solving algebraic equations.
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Re: M26-25
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05 Dec 2017, 22:22
abhishek911 wrote: Bunuel wrote: blitheclyde wrote: Hi, Having trouble wrapping my head around why \(\sqrt{x^2}\) is equal to |x| I looked at the post on this page, but did not see any explanation. For example, how is the equation \(\sqrt{4}\) = {-2, 2} any different from \(\sqrt{2^2}\) = ? Is it because the solution of \(\sqrt{x^2}\) must be equal to the solution of \((\sqrt{x})^2\)? if so, why? Thanks in advance for any help! (with regard to bunnel's explanation to xnthic, throughout the explanation bunnel writes "\(\sqrt{25}\) = 5" as though this is given, but my understanding is that \(\sqrt{25}\) could be either positive or negative 5, as either positive or negative 5 squared result in 25. If I am incorrect, then that is most likely the root of my confusion) \(\sqrt{4}=2\) ONLY, not 2 or -2. The square root function cannot give negative result. Check here: http://gmatclub.com/forum/m26-184464.html#p1729559Hi Bunuel, certain application of the sqrt function seem to be contradictory. If sqrt function can't give negative numbers, why do we substitute both -ve and +ve values of \(\sqrt{expression}\) while solving algebraic equations. 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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Re: M26-25
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13 May 2018, 08:02
Bunuel wrote: Official Solution:
If \(x\) and \(y\) are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}\)?
A. \(1+y\)
B. \(1-y\)
C. \(-1-y\)
D. \(y-1\)
E. \(x-y\)
Note that \(\sqrt{a^2}=|a|\). Next, since \(x \lt 0\) and \(y \lt 0\) then \(|x|=-x\) and \(|y|=-y\).
So, \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y\)
Answer: D Hi Brunel, If x <0 . then (Sqrt (X^2))/ X should be equal to 1 Let us take consider X= -2 , then Sqrt ( -2 *-2) should be -2 . so (Sqrt (X^2))/ X = -2/-2 = 1 Can you please explain what mistake am I doing?
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Re: M26-25
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13 May 2018, 08:48
seevishal@gmail.com wrote: Bunuel wrote: Official Solution:
If \(x\) and \(y\) are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}\)?
A. \(1+y\)
B. \(1-y\)
C. \(-1-y\)
D. \(y-1\)
E. \(x-y\)
Note that \(\sqrt{a^2}=|a|\). Next, since \(x \lt 0\) and \(y \lt 0\) then \(|x|=-x\) and \(|y|=-y\).
So, \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y\)
Answer: D Hi Brunel, If x <0 . then (Sqrt (X^2))/ X should be equal to 1 Let us take consider X= -2 , then Sqrt ( -2 *-2) should be -2 . so (Sqrt (X^2))/ X = -2/-2 = 1 Can you please explain what mistake am I doing? I think this is explained several times in this topic. For example, in the post above yours: https://gmatclub.com/forum/m26-184464-20.html#p1973777When 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. \(\sqrt{(-2)^2}=\sqrt{4}=2\), not -2.
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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: M26-25
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11 Jun 2018, 01:11
Hi,
Since x is negative, \sqrt{x^2}/-x - \sqrt{-y*-y}
= \(-x/-x\) - \sqrt{y^2}
= 1-(-y)
= 1+y
Can someone please explain what wrong am I doing here?
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Re: M26-25
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11 Jun 2018, 04:44
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Re: M26-25
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11 Jul 2018, 07:26
Hi,
In one of the other questions, the explanation is provided as follows
If p=−1
then (|p|!)^p=(|−1|!)^−1=1^−1=1
Here |-1| = 1
But in this problem |y| = -y since y is negative
Can you please explain the discrepancy or the misunderstanding in my thought
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Re: M26-25
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11 Jul 2018, 10:26
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Re: M26-25
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21 Aug 2018, 18:15
I used smart numbers, where x = -2 and y = -3. The given formula will get you to -4 and by plugging in numbers, only D ends up with -4.
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Re: M26-25
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22 Aug 2018, 23:04
delm2010 wrote: How does abs(x) = -x, and abs(y) = -y when x<0 and y<0? I am do not understand this concept It's given in the question. If x and y are negative numbers, what is the value.......
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Re: M26-25
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26 Aug 2018, 07:34
as x and y are negative so \(\frac{-x}{x}\) - \(\sqrt{(-y)(-y)}\) \((-1)\) - \(\sqrt{y^2}\) \(-1-(-y)\) \(-1+y\) \(y-1\)
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Re: M26-25
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14 Sep 2018, 07:39
Bunuel wrote: Official Solution:
If \(x\) and \(y\) are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}\)?
A. \(1+y\)
B. \(1-y\)
C. \(-1-y\)
D. \(y-1\)
E. \(x-y\)
Note that \(\sqrt{a^2}=|a|\). Next, since \(x \lt 0\) and \(y \lt 0\) then \(|x|=-x\) and \(|y|=-y\).
So, \(\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y\)
Answer: D For anyone that picked C over D, like I did, plug in numbers: Let X = -2, and Let Y = -3 (also let -X = 2, and Let -Y = 3) You end up with: \(\frac{2}{-2} - \sqrt{3*3}\) = \(-1 - 3\) Now look back - 3 = -Y, so 1- -Y = 1 + Y. D'oh!
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Re M26-25
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15 Sep 2018, 09:12
I think this is a high-quality question.
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