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If m>n>0, x=m^2+n^2, and y=2mn, [#permalink]
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30 Nov 2017, 01:31
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[GMAT math practice question] If \(m>n>0, x=m^2+n^2\), and \(y=2mn\), what is the value of \(\sqrt{x^2y^2}\) in terms of \(m\) and \(n\)? A. \(m^2+n^2\) B. \(2m^2+n^2\) C.\(m^2+2n^2\) D. \(n^2m^2\) E. \(m^2n^2\)
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If m>n>0, x=m^2+n^2, and y=2mn, [#permalink]
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30 Nov 2017, 01:51
Formula used: \(x^2  y^2 = (x+y)(xy)\) \(\sqrt{x^2y^2} = \sqrt{(x+y)(xy)}\) In terms of m and n, the equation can be written as \(\sqrt{(m^2+n^2+2mn)(m^2+n^22mn)}\) Simplifying, we get \(\sqrt{(m+n)^2(mn)^2} = (m+n)(mn) = m^2n^2\) (Option E)
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Re: If m>n>0, x=m^2+n^2, and y=2mn, [#permalink]
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30 Nov 2017, 05:11
sqrt (m+n)^2(m−n)^2 m^2  n^2 = Option E
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Re: If m>n>0, x=m^2+n^2, and y=2mn, [#permalink]
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03 Dec 2017, 18:38
=> Plugging in the expressions for \(x\) and \(y\), and expanding gives: \(x^2 – y^2 = (m^2+n^2)^2 – (2mn)^2 = m^4 2m^2n^2 + n^4 = (m^2n^2)^2.\) So, since \(m^2 > n^2\), \(\sqrt{x^2y^2}=\sqrt{(m^2n^2)^2}=m^2n^2=m^2n^2.\) Therefore, the answer is E. Answer: E
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If m>n>0, x=m^2+n^2, and y=2mn, [#permalink]
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27 Feb 2018, 05:38
pushpitkc wrote: Formula used: \(x^2  y^2 = (x+y)(xy)\)
\(\sqrt{x^2y^2} = \sqrt{(x+y)(xy)}\)
In terms of m and n, the equation can be written as \(\sqrt{(m^2+n^2+2mn)(m^2+n^22mn)}\)
Simplifying, we get \(\sqrt{(m+n)^2(mn)^2} = (m+n)(mn) = m^2n^2\)(Option E) I do not understand. I am getting... \(\sqrt{(m^2+n^2)^2(2mn)^2}\) Which then leads me down a long and incorrect path. Thanks in advance for a descriptive solution.




If m>n>0, x=m^2+n^2, and y=2mn,
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27 Feb 2018, 05:38






