Question #b3a3e

1 Answer
NickTheTurtle
Dec 17, 2017

I assume that you want to solve for #x/(x^2+1)=(x^2+1)/x#.

Cross-multiplying gives #x^2=(x^2+1)^2#.

Take the square root of both sides: #+-x=x^2+1#. The plus or minus is important as #(+-x)^2=x^2#.

Then, #x^2+-x+1=0#. Use the quadratic formula to find that the solutions for #x# are #(-1+-sqrt(1^2-4*1*1))/(2*1)=(-1+-isqrt(3))/2# and #(1+-sqrt((-1)^2-4*1*1))/(2*1)=(1+-isqrt(3))/2#.

Thus, the solutions are #(isqrt(3)-1)/2#, #(1+isqrt(3))/2#, #(1-isqrt(3))/2#, and #-(1+isqrt(3))/2#.

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