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Determine the sum of all real numbers \(x\) satisfying \((x^2-4x+2)^{x^2-5x+2} = 1\)

 Jul 18, 2021
 #1
avatar+78 
+2

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. smiley

 Jul 18, 2021
 #2
avatar+126 
+2

thank you for the hint! i originally tried solving the problem like this, but i got stuck, haha. i'm probably just not good enough 

uvacowdo  Jul 18, 2021
 #3
avatar+115426 
+3

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

 

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

 Jul 19, 2021
 #4
avatar+126 
+1

thanks!

uvacowdo  Jul 20, 2021

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