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$If $m+\frac{1}{m}=8$, then what is the value of $m^{2}+\frac{1}{m^{2}}+4$?$

Guest Feb 1, 2015

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Best Answer

+128407

+5

m + 1/m = 8 square both sides

m^2 + 2(m)(1/m) + 1/m^2 = 64

m^2 + 2 + 1/m^2 = 64 subtract 2 from both sides

m^2 + 1/m^2 = 62 ....therefore

m^2 + 1/m^2 + 4 = 62 + 4 = 66

CPhill Feb 1, 2015

1⁺⁰ Answers

+128407

+5

Best Answer

m + 1/m = 8 square both sides

m^2 + 2(m)(1/m) + 1/m^2 = 64

m^2 + 2 + 1/m^2 = 64 subtract 2 from both sides

m^2 + 1/m^2 = 62 ....therefore

m^2 + 1/m^2 + 4 = 62 + 4 = 66

CPhill Feb 1, 2015

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