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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!

 Sep 19, 2023
 #1
avatar+33616 
+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).

 Sep 20, 2023
 #2
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I acknowledge that the equation $(x² - 4x+2)x²-5x+2 = 1$ has two real roots, approximately equal to 0.2106665 and 3.80167138

 Sep 23, 2023
edited by Guest  Sep 23, 2023

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