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skyfarer
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. |y|x
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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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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chetan2u
skyfarer
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. |y|x
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

Hello 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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jfranciscocuencag
chetan2u
skyfarer
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. |y|x
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

Hello 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.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2016/0 ... oots-gmat/
It discusses when to take both positive and negative values and when to take only positive values.
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\(\sqrt{x^2*y^2}\) = |xy| = |x| * |y|
Given x<0, |x| = -x
Given y>0, |y| = y
|x| * |y| = -x * y = -xy
skyfarer
If x<0 and y>0 what will be the value of \(\sqrt{x^2*y^2}\)

a. -xy
b. xy
c. -|xy|
d. |y|x
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.

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