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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8001
GMAT 1: 760 Q51 V42 GPA: 3.82
If m<n<0, x=m^2+n^2, and y=2mn, what is the va  [#permalink]

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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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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 V
Joined: 16 Aug 2015
Posts: 8001
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: If m<n<0, x=m^2+n^2, and y=2mn, what is the va  [#permalink]

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=>

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

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.

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GMAT 1: 700 Q43 V42 Re: If m<n<0, x=m^2+n^2, and y=2mn, what is the va  [#permalink]

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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. 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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