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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. \(xy\)
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16 Sep 2014, 01:25
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
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Re: M2625
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25 Oct 2014, 05:44
How does abs(x) = x, and abs(y) = y when x<0 and y<0? I am do not understand this concept



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25 Oct 2014, 05:47
delm2010 wrote: How does abs(x) = x, and abs(y) = y when x<0 and y<0? I am do not understand this concept Below links on absolute value should help you:
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Re: M2625
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25 Oct 2014, 23:33
Hi Bunuel, I have a query.... wouldn't \sqrt{x^2}/ x would be equal to +1?
I say so since it's already given in the question that x is a ve number, so by \sqrt{x^2} we know that we need to consider only the ve Root.
But in the denominator the variable x would be a ve number. Hence after the resolution of \sqrt{x^2} one part of the question would be x/x, where x would give us x(since the ve root needs to be considered) and the denominator in isolation is again a ve number... so the fraction becomes \frac{x}{x} = 1.
Kindly clarify



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Re: M2625
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26 Oct 2014, 06:08
avaneeshvyas wrote: Hi Bunuel, I have a query.... wouldn't \sqrt{x^2}/ x would be equal to +1?
I say so since it's already given in the question that x is a ve number, so by \sqrt{x^2} we know that we need to consider only the ve Root.
But in the denominator the variable x would be a ve number. Hence after the resolution of \sqrt{x^2} one part of the question would be x/x, where x would give us x(since the ve root needs to be considered) and the denominator in isolation is again a ve number... so the fraction becomes \frac{x}{x} = 1.
Kindly clarify First of all, read this: rulesforpostingpleasereadthisbeforeposting133935.html#p1096628 (Writing Mathematical Formulas on the Forum) Next, the point is that the square root function cannot give negative result (\(\sqrt{some \ expression}\geq{0}\)) and \(\sqrt{x^2}=x\), which means that \(\sqrt{x^2}/ x=\frac{x}{x}=\frac{x}{x}=1\). Please follow the links in my previous post for more on absolute values.
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[quote="Bunuel"]Official Solution:
1y=1+y
I have not understood the above portion. Could you please tell me how to get this?
I have done it by taking numbers.
Let x=1 and y=1 Here i have taken integer value thinking that I will get an answer which can be in option also.
Now putting the value of x and y I am getting the value of question is 2.
Only option D gives me the value. Is my approach correct?



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Re: M2625
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28 Nov 2014, 05:38
Raihanuddin wrote: Bunuel wrote: Official Solution:
1y=1+y
I have not understood the above portion. Could you please tell me how to get this?
I have done it by taking numbers.
Let x=1 and y=1 Here i have taken integer value thinking that I will get an answer which can be in option also.
Now putting the value of x and y I am getting the value of question is 2.
Only option D gives me the value. Is my approach correct? 1  y = 1 + y because we are given that y is negative. When y is negative y = y. Hence 1  y = 1  (y) = 1 + y.
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Re: M2625
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16 Oct 2015, 07:57
There is a step that is off in this question. If you substitute negative y from the get go (y) = +y * (y) leaves you with a negative inside the square root and this cannot happen. Should be written as square root of y*lyl



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18 Oct 2015, 13:39
danjbon wrote: There is a step that is off in this question. If you substitute negative y from the get go (y) = +y * (y) leaves you with a negative inside the square root and this cannot happen. Should be written as square root of y*lyl That's wrong. \(\sqrt{y*y}=\sqrt{(negative)*positive}=\sqrt{positive*positive}\)
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having a hard time understanding this basic rule: \(\sqrt{x^2}=x\)
Assuming it's an integer, shouldn't this be the positive integer of x? why x



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xnthic wrote: having a hard time understanding this basic rule: \(\sqrt{x^2}=x\)
Assuming it's an integer, shouldn't this be the positive integer of x? why x MUST KNOW: \(\sqrt{x^2}=x\):The point here is that since square root function cannot 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 \(\sqrt{x^2}=x\). For example if x = 5, then \(\sqrt{x^2}=\sqrt{25}=5=5=x\)
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Re: M2625
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01 Sep 2016, 00:12
Bunuel wrote: 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. \(xy\) Alternate approach is to testing some easy values: As per question stem, x & y are negative number.. Let x= 1 & y=1 Substituting the in equation \(\frac{\sqrt{1}}{1}\sqrt{(1)*1}\) = 2 Checking answer choices y1=11=2 only answer that matches what we get above Answer: D



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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)



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Re: M2625
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08 Dec 2016, 13:53
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: m26184464.html#p1729559
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Re: M2625
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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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Re: M2625
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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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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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Re: M2625
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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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Re: M2625
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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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