Question

Find the value of  
`((m+n)^2)/(2k)`  if  
`m^2-2mn+k^2-2nk+2n^2=0`  and   [tex]2mn-k^2=25

[/tex]

Answer 1

`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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`((m+n)^(2))/(2k) = ((n+n)^(2))/(2n) = ((2n)^(2))/(2n) = (4n^(2))/(2n) = 2n \\ \\ 2n = 10 \ \vee \ 2n = -10`