Question


Answer

\because m-\dfrac{1}{m}=3 >0

\therefore m>1

Let A = m^2-\dfrac{1}{m^2}

A^2 = m^4+\dfrac{1}{m^4} -2
(1)
\because m-\dfrac{1}{m}=3

m^2+\dfrac{1}{m^2}-2 = 9

m^2+\dfrac{1}{m^2} = 7

Square the both sides

m^4+\dfrac{1}{m^4}+2=49

\therefore m^4+\dfrac{1}{m^4} = 47

Plug in the expression (1)

A^2 = 47-2 = 45

\therefore A = m^2-\dfrac{1}{m^2} = \sqrt{45}

Steven Zheng posted 2 months ago

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