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M61-15

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MathRevolution
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M61-15 [#permalink] New post  18 Jun 2018, 04:12
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Question Stats:

81% (01:39) correct 19% (02:40) wrong based on 27 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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MathRevolution
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Re M61-15 [#permalink] New post  18 Jun 2018, 04:12
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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 \(m^2>n^2\).


Answer: D
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chetan2u
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M61-15 [#permalink] New post  18 Jun 2018, 04:41
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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. \(n^2-m^2\)


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

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Pritam_Pf
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Re: M61-15 [#permalink] New post  11 Jan 2019, 16:39
Why m^2>n^2?

Can anybody explain?
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sach24x7
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Re: M61-15 [#permalink] New post  22 May 2019, 08:48
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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Bambi2021
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Re: M61-15 [#permalink] New post  16 Jul 2021, 17:07
\(\sqrt{x^2-y^2} = \sqrt{(x+y)(x-y)}\)

Squaring the expression gives:

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

Now take the square root gives:

\((m+n)(m-n) = m^2 - n^2\)
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Re: M61-15 [#permalink]
16 Jul 2021, 17:07
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