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Re: M2625
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10 Dec 2016, 05:58
cuhmoon wrote: Doubt for this q:
\(\sqrt{y * y}\)
= \(\sqrt{y^2}\)
and Square root of Y has to be the positive root.. Thus = y
Thus the answer must be 1y.
I do understand that \(\sqrt{x^2}\)= x ... and if X is ve then mod x = x, but we don't have to simplify\(\sqrt{y^2}\) to mod y. We can just take the positive root.. since per GMAT anything under the square root sign is the positive root ?
What am I missing here?
I guess my q is on the GMAT, square root of numbers are ONLY positive.. Please check the discussion above.
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25 Mar 2017, 20:46
Very Good question
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05 Dec 2017, 23:09
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/m26184464.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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05 Dec 2017, 23:22
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/m26184464.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. 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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13 May 2018, 09:02
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. \(1y\) C. \(1y\) D. \(y1\) E. \(xy\)
Note that \(\sqrt{a^2}=a\). Next, since \(x \lt 0\) and \(y \lt 0\) then \(x=x\) and \(y=y\). So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\)
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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13 May 2018, 09:48
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. \(1y\) C. \(1y\) D. \(y1\) E. \(xy\)
Note that \(\sqrt{a^2}=a\). Next, since \(x \lt 0\) and \(y \lt 0\) then \(x=x\) and \(y=y\). So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\)
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/m2618446420.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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Re: M2625
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11 Jun 2018, 02:11
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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11 Jun 2018, 05:44
dishadesai wrote: 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? Why are you changing x to x there? The stem just says that x is negative, so x stands for some negative number, changing x to x does not make sense.
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11 Jul 2018, 08:26
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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11 Jul 2018, 11:26
Fervids77 wrote: 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 When x < 0, then x = x. Say x = 1, then 1 = (1) = 1.
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21 Aug 2018, 19:15
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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23 Aug 2018, 00:04
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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26 Aug 2018, 08:34
as x and y are negative so \(\frac{x}{x}\)  \(\sqrt{(y)(y)}\) \((1)\)  \(\sqrt{y^2}\) \(1(y)\) \(1+y\) \(y1\)
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Re: M2625
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14 Sep 2018, 08:39
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. \(1y\) C. \(1y\) D. \(y1\) E. \(xy\)
Note that \(\sqrt{a^2}=a\). Next, since \(x \lt 0\) and \(y \lt 0\) then \(x=x\) and \(y=y\). So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\)
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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15 Sep 2018, 10:12
I think this is a highquality question.



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03 Jan 2019, 11:31
The answer can be calculate very easily Do 1 thing : substitute x = a and y = b (as question says both x and y are negative numbers) and solve with basic math and reverse back the result in terms of "y". Answer will be "1+y". Try your hand once.



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19 Mar 2019, 20:47
I agree It should be 1 y since the expression after the ve sign should be positive and would be the absolute value of y



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19 Mar 2019, 21:11
Bokaage wrote: I agree It should be 1 y since the expression after the ve sign should be positive and would be the absolute value of y I suggest to read the hole thread. The answer is correct as it is.
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