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If x<0 and y>0 what will be the value of
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Updated on: 09 Jan 2017, 00:05
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54% (00:47) correct 46% (00:55) wrong based on 411 sessions
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If x<0 and y>0 what will be the value of \(\sqrt{x^2*y^2}\) a. xy b. xy c. xy d. yx e. No solution. I put my best on C as it will give the negative value which we are looking for. But he said that because of square root, we want a non negative value and so a xy is the answer. I am still not able to get it Can anyone explain why is it A? He said, its a question from Veritas. Source: selfmade
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Originally posted by skyfarer on 01 Mar 2016, 17:26.
Last edited by Bunuel on 09 Jan 2017, 00:05, edited 5 times in total.
Reformatted the question and added the OA.




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Re: If x<0 and y>0 what will be the value of
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02 Mar 2016, 00:16
skyfarer wrote: If x<0 and y>0 what will be the value of \(\sqrt{x^2*y^2}\) a. xy b. xy c. xy d. yx e. No solution. I put my best on C as it will give the negative value which we are looking for. But he said that because of square root, we want a non negative value and so a xy is the answer. I am still not able to get it Can anyone explain why is it A? He said, its a question from Veritas. Source: selfmade A few definitions will help solve such questions easily: \(\sqrt{x^2} = x\) x = x if x >= 0 (x is 0 or positive) x = x if x < 0 (x is negative) \(\sqrt{x^2*y^2}\) \(= \sqrt{x^2} * \sqrt{y^2}\) \(= x * y\) Since x < 0, x = x Since y > 0, y = y x * y = x * y Answer (A)
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Re: If x<0 and y>0 what will be the value of
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01 Mar 2016, 19:09
skyfarer wrote: If x<0 and y>0 what will be the value of \(\sqrt{x^2*y^2}\) a. xy b. xy c. xy d. yx e. No solution. I put my best on C as it will give the negative value which we are looking for. But he said that because of square root, we want a non negative value and so a xy is the answer. I am still not able to get it Can anyone explain why is it A? He said, its a question from Veritas. Source: selfmade Follow posting guidelines (link in my signatures). Select the proper forum to post your question and put all your analyses either in the next post or under "spoilers". Do format the question properly.An alternate method is to assume particular values based on the information in the question. It is given that x<0 and y>0, so assume x=2 and y=3. Thus \(\sqrt{x^2*y^2} = \sqrt{36} = +6\). Now check which of the options give you the same value when you put x=2 and y=3. a. xy = (2)*3=6. Keep.b. xy = 2*3=6. Eliminate.c. xy = 6 = 6 . Eliminate.d. yx = 3*(2)=6. Eliminate.e. No solution. Not possible as both x^2 and y^2 are positive and hence there will be a solution. Eliminate.Hence A is the correct answer. Hope this helps.




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Re: If x<0 and y>0 what will be the value of
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01 Mar 2016, 18:56
skyfarer wrote: Hi, I have a question which my Friend asked me. IF x<0 and y>0 what will be the value of \sqrt{X^2.Y^2} Choices a. xy b. xy c. xy d. yx e. No solution. I put my best on C as it will give the negative value which we are looking for. But he said that because of square root, we want a non negative value and so a xy is the answer. I am still not able to get it Can anyone explain why is it A? He said, its a question from Veritas. Hi, Firstly, the Q has not been posted in correct format.. Pl follow guidelines
Now on your Q.. \(\sqrt{x^2*y^2}\)... \(\sqrt{x^2}*\sqrt{y^2}\)... \(\sqrt{a^2}\) will always be positive a.. so \(\sqrt{x^2}*\sqrt{y^2}\)= positive x*positive y.... since y<0, we have to add a ive sign so our ansewr becomes x*(y)=xy
Hope it clears your doubt
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Re: If x<0 and y>0 what will be the value of
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06 Jan 2017, 11:29
x < 0 y > 0 assume x = 2 , y = 3 \(substitute \sqrt{x^{2} y^{2}}\) \(\sqrt{36}\) 6 Substitute in options xy = (2x3) = 6 A
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Re: If x<0 and y>0 what will be the value of
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19 Feb 2019, 16:42
chetan2u wrote: skyfarer wrote: Hi, I have a question which my Friend asked me. IF x<0 and y>0 what will be the value of \sqrt{X^2.Y^2} Choices a. xy b. xy c. xy d. yx e. No solution. I put my best on C as it will give the negative value which we are looking for. But he said that because of square root, we want a non negative value and so a xy is the answer. I am still not able to get it Can anyone explain why is it A? He said, its a question from Veritas. Hi, Firstly, the Q has not been posted in correct format.. Pl follow guidelines
Now on your Q.. \(\sqrt{x^2*y^2}\)... \(\sqrt{x^2}*\sqrt{y^2}\)... \(\sqrt{a^2}\) will always be positive a.. so \(\sqrt{x^2}*\sqrt{y^2}\)= positive x*positive y.... since y<0, we have to add a ive sign so our ansewr becomes x*(y)=xy
Hope it clears your doubtHello chetan2u ! Where does the +/ after rooting a number enter in your explanation? \(\sqrt{x^2}\) Why in some exercises you must consider +x and x? Kind regards!



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Re: If x<0 and y>0 what will be the value of
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21 Feb 2019, 00:47
jfranciscocuencag wrote: chetan2u wrote: skyfarer wrote: Hi, I have a question which my Friend asked me. IF x<0 and y>0 what will be the value of \sqrt{X^2.Y^2} Choices a. xy b. xy c. xy d. yx e. No solution. I put my best on C as it will give the negative value which we are looking for. But he said that because of square root, we want a non negative value and so a xy is the answer. I am still not able to get it Can anyone explain why is it A? He said, its a question from Veritas. Hi, Firstly, the Q has not been posted in correct format.. Pl follow guidelines
Now on your Q.. \(\sqrt{x^2*y^2}\)... \(\sqrt{x^2}*\sqrt{y^2}\)... \(\sqrt{a^2}\) will always be positive a.. so \(\sqrt{x^2}*\sqrt{y^2}\)= positive x*positive y.... since y<0, we have to add a ive sign so our ansewr becomes x*(y)=xy
Hope it clears your doubtHello chetan2u ! Where does the +/ after rooting a number enter in your explanation? \(\sqrt{x^2}\) Why in some exercises you must consider +x and x? Kind regards! Check this post: https://www.veritasprep.com/blog/2016/0 ... ootsgmat/It discusses when to take both positive and negative values and when to take only positive values.
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Re: If x<0 and y>0 what will be the value of
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21 Feb 2019, 00:47






