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Define \[f(x) = \frac{1}{x + \frac{1}{x}}\] and \[g(x) = \frac{1}{x - \frac{1}{x}}.\] Find the square of the largest real solution to the equation \[ (g(x))^2 - (f(x))^2 = 1.\]

 Sep 9, 2023
 #1
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Find the square of the largest real solution to the equation \((g(x))^2 - (f(x))^2 = 1.\)

 

Hello wiseowl!

 

\((g(x))^2 - (f(x))^2 = 1\\ \dfrac{1}{(x-\dfrac{1}{x})^2}-\dfrac{1}{(x+\dfrac{1}{x})^2}=1\)

 

\(\dfrac{1}{(\dfrac{x^2-1}{x})^2}-\dfrac{1}{(\dfrac{x^2+1}{x})^2}=1\)

 

\(\dfrac{x^2}{(x^2-1)^2}-\dfrac{x^2}{(x^2+1)^2}=1\\ \)

 

\(\dfrac{4x^4}{(x-1)^2(x+1)^2(x^2+1)^2}=1\)

 

Calculated by WolframAlpha.

 

\(x=-\sqrt{\sqrt{2}-1}\\ x=\sqrt{\sqrt{2}-1}\\ x=-\sqrt{1+\sqrt{2}}\\ x=\sqrt{1+\sqrt{2}}\\\)

\(\color{blue}x\in \{-1.55377,-0.64359,0.64359,1.55377\}\)

 

laugh  !

 Sep 15, 2023
edited by asinus  Sep 15, 2023
edited by asinus  Sep 15, 2023

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