Julik
Answered

Find the value of  \(\frac{(m+n)^2}{2k}\)  if   \(m^2-2mn+k^2-2nk+2n^2=0\)  and   \(2mn-k^2=25
\)

Verified Answer :

\(m^{2} - 2mn + k^{2} - 2nk + 2n^{2} = 0 \\ \\ m^{2} - 2mn + n^{2} + k^{2} - 2nk + n^{2} = 0 \\ \\ (m-n)^{2} + (k - n)^{2} = 0 \\ \\ m-n = 0 \ \hbox{and} \ k-n = 0 \\ \\ m = n \ \hbox{and} \ k = n\)

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\(2mn - k^{2} = 25 \\ \\ 2n^{2} - n^{2} = 25 \\ \\ n^{2} = 25 \\ \\ n = 5 \ \vee \ n= -5\)

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\(\dfrac{(m+n)^{2}}{2k} = \dfrac{(n+n)^{2}}{2n} = \dfrac{(2n)^{2}}{2n} = \dfrac{4n^{2}}{2n} = 2n \\ \\ 2n = 10 \ \vee \ 2n = -10\)