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18 Jun 2018, 04:12
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D
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Difficulty:
25%
(medium)
Question Stats:
76% (01:43) correct
24%
(02:39)
wrong
based on 33
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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. \(n^2-m^2\)
A. \(m^2+n^2\)
B. \(2m^2+n^2\)
C. \(m^2+2n^2\)
D. \(m^2-n^2\)
E. \(n^2-m^2\)
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
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
18 Jun 2018, 04:41
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Bunuel
\(X=m^2+n^2.........x^2=(m^2+n^2)^2=m^4+n^4+2(mn)^2=m^4+n^4+2(mn)^2+2(mn)^2-2(mn)^2\)......
\(x^2=m^4+n^4-2(mn)^2+(2mn)^2=(m^2-n^2)^2+y^2........x^2-y^2=(m^2-n^2)^2.......√(x^2-y^2)=m^2-n^2\)
D
