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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7465
GMAT 1: 760 Q51 V42 GPA: 3.82

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Difficulty:   25% (medium)

Question Stats: 83% (01:07) correct 17% (03:33) wrong based on 6 sessions

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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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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 7465
GMAT 1: 760 Q51 V42 GPA: 3.82

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Official Solution:

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. $$n^2-m^2$$

$$x^2-y^2$$

$$= (m^2+n^2)^2 – (2mn)^2$$

$$= m^4 + 2m^2 n^2 + n^4 – 4m^2 n^2$$

$$= m^4 – 2m^2 n^2 + n^4$$

$$= (m^2-n^2)^2$$

Therefore,

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

$$=\sqrt{(m^2-n^2)^2}$$

$$= |m^2-n^2|$$

$$= m^2-n^2$$, since $$m^2>n^2$$.

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Math Expert V
Joined: 02 Aug 2009
Posts: 7757

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Bunuel wrote:
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. $$n^2-m^2$$

$$X=m^2+n^2.........x^2=(m^2+n^2)^2=m^4+n^4+2(mn)^2=m^4+n^4+2(mn)^2+2(mn)^2-2(mn)^2$$......
$$x^2=m^4+n^4-2(mn)^2+(2mn)^2=(m^2-n^2)^2+y^2........x^2-y^2=(m^2-n^2)^2.......√(x^2-y^2)=m^2-n^2$$

D
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Intern  B
Joined: 24 Feb 2017
Posts: 9

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Why m^2>n^2?

Can anybody explain?
Manager  S
Status: Don't Give Up!
Joined: 15 Aug 2014
Posts: 95
Location: India
Concentration: Operations, General Management
GMAT Date: 04-25-2015
WE: Engineering (Manufacturing)

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Pritam_Pf wrote:
Why m^2>n^2?

Can anybody explain?

m<n<0

That means m and n are negative such as -4 < -2 so m2 > n2
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- Sachin

-If you like my explanation then please click "Kudos" Re: M61-15   [#permalink] 22 May 2019, 08:48
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