Question 255806
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Let *[tex \Large u\ =\ x^2]


Substitute:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^4\ -\ 2x^2\ +\ 1\ =\ 0]



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ -\ 2u\ +\ 1\ =\ 0]


Factor:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(u\ -\ 1\right)^2 =\ 0]


Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ 1]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ 1]


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ 1]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 1]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -1]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ 1]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 1]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -1]


In sum, the four roots of your quartic are 1, -1, 1, and -1.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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