Question
Given positive real number m, if m-\cfrac{1}{m}=3 , find the value of m^2-\cfrac{1}{m^2}
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)
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}