Determine the sum of all real numbers \(x\) satisfying

\((x^2-4x+2)^{x^2-5x+2} = 1 \) I tried solving and tried the answers: 5, 2, 4, 3, 9 But they were all wrong. Please help!

WaterPheonix Sep 19, 2023

#1**+3 **

The left hand side of the expression is 1 if (i) \(x^2 -5x+2=0\) or (ii) \(x^2 - 4x + 2 = 1\)

These are quadratics that can be solved using the standard quadratic formula. They will give values of x in the form \(x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\)

Adding these two values will cancel the discriminant part leaving you with a sum of \(\frac{-b}{a}\).

Do this for each quadratic and then add the results.

Also if \(x^2 - 4x + 2 = -1\) and \(x^2 -5x+2\) is even then the RHS will be 1 (This will give you two more x values).

Alan Sep 20, 2023