Julik
contestada

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

Respuesta :

[tex]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[/tex]

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[tex]2mn - k^{2} = 25 \\ \\ 2n^{2} - n^{2} = 25 \\ \\ n^{2} = 25 \\ \\ n = 5 \ \vee \ n= -5[/tex]

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