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If m>n>0, x=m^2+n^2, and y=2mn,

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If m>n>0, x=m^2+n^2, and y=2mn, [#permalink]

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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 \(\sqrt{x^2-y^2}\) in terms of \(m\) and \(n\)?

A. \(m^2+n^2\)
B. \(2m^2+n^2\)
C.\(m^2+2n^2\)
D. \(n^2-m^2\)
E. \(m^2-n^2\)
[Reveal] Spoiler: OA

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If m>n>0, x=m^2+n^2, and y=2mn, [#permalink]

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Formula used: \(x^2 - y^2 = (x+y)(x-y)\)

\(\sqrt{x^2-y^2} = \sqrt{(x+y)(x-y)}\)

In terms of m and n, the equation can be written as \(\sqrt{(m^2+n^2+2mn)(m^2+n^2-2mn)}\)

Simplifying, we get \(\sqrt{(m+n)^2(m-n)^2} = (m+n)(m-n) = m^2-n^2\)(Option E)
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Re: If m>n>0, x=m^2+n^2, and y=2mn, [#permalink]

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New post 30 Nov 2017, 05:11
sqrt (m+n)^2(m−n)^2
m^2 - n^2 = Option E
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Plugging in the expressions for \(x\) and \(y\), and expanding gives:

\(x^2 – y^2 = (m^2+n^2)^2 – (2mn)^2 = m^4 -2m^2n^2 + n^4 = (m^2-n^2)^2.\)
So, since \(m^2 > n^2\),
\(\sqrt{x^2-y^2}=\sqrt{(m^2-n^2)^2}=|m^2-n^2|=m^2-n^2.\)

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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New post 27 Feb 2018, 05:38
pushpitkc wrote:
Formula used: \(x^2 - y^2 = (x+y)(x-y)\)

\(\sqrt{x^2-y^2} = \sqrt{(x+y)(x-y)}\)

In terms of m and n, the equation can be written as \(\sqrt{(m^2+n^2+2mn)(m^2+n^2-2mn)}\)

Simplifying, we get \(\sqrt{(m+n)^2(m-n)^2} = (m+n)(m-n) = m^2-n^2\)(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,   [#permalink] 27 Feb 2018, 05:38
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