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If m<n<0, x=m^2+n^2, and y=2mn, what is the va

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Joined: 16 Aug 2015
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If m<n<0, x=m^2+n^2, and y=2mn, what is the va  [#permalink]

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New post 30 Jan 2018, 03:15
1
2
00:00
A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

73% (01:59) correct 27% (02:19) wrong based on 88 sessions

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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 \(n\)?

\(A. m^2+n^2\)
\(B. 2m^2+n^2\)
\(C. m^2+2n^2\)
\(D. m^2-n^2\)
\(E. n^2-m^2\)

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Re: If m<n<0, x=m^2+n^2, and y=2mn, what is the va  [#permalink]

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New post 30 Jan 2018, 03:46
2
D

x=m²+n², y=2mn

Now rt(\(x²-y²\)) =rt(\((x+y)(x-y)\))

=rt(\((m²+n²+2mn)(m²+n²-2mn)\))
=rt(\((m+n)²(m-n)²\))
=(m+n)(m-n)
=m²-n²
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Math Revolution GMAT Instructor
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Re: If m<n<0, x=m^2+n^2, and y=2mn, what is the va  [#permalink]

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New post 01 Feb 2018, 01:35
=>

\(x^2-y^2\)
\(= (m^2+n^2)^2 – (2mn)^2\)
\(= (m^2+n^2)^2 – 4m^2n^2\)
\(= m^4 + 2m^2n^2 + n^4 – 4m^2n^2\)
\(= m^4 – 2m^2n^2 + n^4\)
\(= (m^2-n^2)^2\)

Therefore,
\(\sqrt{x^2-y^2}\)
\(=\sqrt{(m^2-n^2)^2}\)
\(= |m^2-n^2|\)
\(= m^2-n^2\), since \(m^2 > n^2\).

Therefore, the answer is D.

Answer: D
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Re: If m<n<0, x=m^2+n^2, and y=2mn, what is the va  [#permalink]

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New post 03 Feb 2018, 12:27
1
This is a good, tricky question. If you forget that m and n are negative numbers (based on the given m<n<0) and assign them numbers to work the problem out (in my case, m=2 and n=3), E will work out to be the correct answer.

This just reinforces that you need to remember ALL given prompts. It's not enough to just write them down - I did and still fell into the trap. When testing scenarios, you need to really be sure you are staying in the confines of all the given prompts.
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Re: If m<n<0, x=m^2+n^2, and y=2mn, what is the va   [#permalink] 03 Feb 2018, 12:27
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