an interesting problem i need help on

I would either try to get (x²-4x+2) to 1, which would maintain any power applied to it to have an output of one, or getting the exponent (x²-5x+2) to 0, so that it would constantly be an output of one. Of course, I don't know exactly how to solve this problem, but hopefully this helps. 

\(either\\\quad x^2-5x+2 = 0 \quad and \quad x^2-4x+2\ne0\\ \quad x=\frac{5\pm\sqrt{17}}{2} \quad and \quad x\ne\frac{4\pm \sqrt{8}}{2}\\ \quad x=\frac{5\pm\sqrt{17}}{2}\\ or\\ \quad x^2-4x+2=1\\ \quad x^2-4x+1=0\\ \quad x=\frac{4\pm\sqrt{12}}{2}=2\pm\sqrt{3}\\ sum=\frac{5+\sqrt{17}}{2}+\frac{5-\sqrt{17}}{2}+2+\sqrt{3}+2-\sqrt{3}\\ sum=\frac{5}{2}+\frac{5}{2}+2+2\\ sum=9 \)

Note. I have not checked this answer, there could easily be careless errors.

LaTex:

either\\\quad x^2-5x+2 = 0   \quad and \quad x^2-4x+2\ne0\\
\quad x=\frac{5\pm\sqrt{17}}{2} \quad and \quad x\ne\frac{4\pm \sqrt{8}}{2}\\
\quad x=\frac{5\pm\sqrt{17}}{2}\\
or\\
\quad x^2-4x+2=1\\
\quad x^2-4x+1=0\\
\quad x=\frac{4\pm\sqrt{12}}{2}=2\pm\sqrt{3}\\
sum=\frac{5+\sqrt{17}}{2}+\frac{5-\sqrt{17}}{2}+2+\sqrt{3}+2-\sqrt{3}\\
sum=\frac{5}{2}+\frac{5}{2}+2+2\\
sum=9