Question 676177
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Your problem makes little sense.  Are you completely misusing the term "square root" and trying to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ 3x^4\ +\ x^2\ -\ 1]


If so, let *[tex \LARGE u\ =\ x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(u)\ =\ 3u^2\ +\ u\ -\ 1]


Solve for *[tex \LARGE u], discard the negative value of *[tex \LARGE u], then solve *[tex \LARGE \pm\sqrt{u}], and evaluate both using your calculator.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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