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

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
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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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Difficulty:

15% (low)

Question Stats:

80% (01:20) correct 20% (02:07) wrong based on 173 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. $$n^2-m^2$$
E. $$m^2-n^2$$

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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" BSchool Forum Moderator Joined: 26 Feb 2016 Posts: 2931 Location: India GPA: 3.12 If m>n>0, x=m^2+n^2, and y=2mn, [#permalink] ### Show Tags 30 Nov 2017, 01:51 1 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) _________________ You've got what it takes, but it will take everything you've got Manager Joined: 07 Jun 2017 Posts: 177 Location: India Concentration: Technology, General Management GMAT 1: 660 Q46 V38 GPA: 3.6 WE: Information Technology (Computer Software) Re: If m>n>0, x=m^2+n^2, and y=2mn, [#permalink] ### Show Tags 30 Nov 2017, 05:11 sqrt (m+n)^2(m−n)^2 m^2 - n^2 = Option E _________________ Regards, Naveen email: nkmungila@gmail.com Please press kudos if you like this post Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 5825 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: If m>n>0, x=m^2+n^2, and y=2mn, [#permalink] ### Show Tags 03 Dec 2017, 18:38 1 => 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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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: $$x^2 - y^2 = (x+y)(x-y)$$

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

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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