Question 322668
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What you wrote is 2x^-96 = 0, which means *[tex \Large 2x^{-96}\ =\ 0] which renders as *[tex \Large \frac{2}{x^{96}}] -- the graph of which is asymptotic to the *[tex \Large x]-axis, and hence has no zeros.


But I don't think that is what you meant.


I suspect you meant *[tex \Large 2x^2\ -\ 96\ =\ 0], which is VASTLY different than what you wrote.  I suspect you are laboring under the misapprehension that the caret mark means to square the variable that precedes it.  It does not mean that -- it means to raise the variable that precedes it to the power that follows it. I.e. x^2 means x squared, x^3 means x cubed, x^6 means x to the sixth power...and x^-96 means to raise x to the -96th power.


Going on my presumption, proceed as follows:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ -\ 96\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ =\ 96]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ 48]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \pm\sqrt{48}\ =\ \pm\sqrt{16\,\cdot\,3}\ =\ \pm4\sqrt{3}]


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