Question 358694
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ =\ 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \pm\sqrt{3}]


The *[tex \Large x]-intercepts of


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ 2x^2\ -\ 6]


are the points *[tex \Large \left(\alpha_1,0\right)] and *[tex \Large \left(\alpha_2,0\right)] such that *[tex \Large f\left(\alpha_1\right)\ =\ 0] and *[tex \Large f\left(\alpha_2\right)\ =\ 0], meaning *[tex \Large \alpha_1] and *[tex \Large \alpha_2] are the roots of the equation that we found in part a).


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