Question 332810
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Let *[tex \Large u\ =\ w^2].


Then solve


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ -\ 12u\ -\ 2\ =\ 0]


for *[tex \Large u_1] and *[tex \Large u_2]


Then set


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ =\ u_1]


and solve for the two roots


Then set


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ =\ u_2]


and solve for the two roots giving you 4 roots in all which is exactly what was expected for a quartic equation.  One of the roots of the equation in *[tex \Large u] will be negative (*[tex \Large -6\ -\ sqrt{38}\ <\ 0], hence one pair of values of *[tex \Large w] will be a complex conjugate pair.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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