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If , m+(1/m) = 8 then what is the value of m^2 + (1/m^2)+4 ?

 Jun 12, 2021

Best Answer 

 #1
avatar+517 
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\(m+{1\over m}=8\)

⇒ \((m+ {1\over m})^2=64\)

\(m^2+{1\over m^2}+2=64\)

\(m^2+{1\over m^2}=62\)

 

\(m^2+{1\over m^2}+4 = 66\)

 

 

~Hope you got it. 

 Jun 12, 2021
 #1
avatar+517 
+2
Best Answer

\(m+{1\over m}=8\)

⇒ \((m+ {1\over m})^2=64\)

\(m^2+{1\over m^2}+2=64\)

\(m^2+{1\over m^2}=62\)

 

\(m^2+{1\over m^2}+4 = 66\)

 

 

~Hope you got it. 

amygdaleon305 Jun 12, 2021

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