Current Events > I'm attempting to solve a cubic of the form x^3+bx=c

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FLOUR
09/15/18 9:14:29 AM
#1:


Where b > 0. That's a huge requirement. There's only one real solution as a result. Why?

Anyways, define k to be the square root of (81c^2+12b^3)

Then x = the cube root of [(c/2)+(k/18)] + the cube root of [(c/2)-(k/18)]

Is this right? Kind of an ugly solution but it appears to work.

If b < 0 then it gets even more ugly as k could be an imaginary number as a result and the computation becomes a disaster.

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SageHarpuia
09/15/18 9:15:29 AM
#2:


24
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"You will pay dearly for your futile resistance!"
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clearaflagrantj
09/15/18 9:16:46 AM
#3:


Man I completed Calc 1-3 and ODE/PDE and I remember none of this shit
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frozenshock
09/15/18 9:17:40 AM
#4:


Yep. Semester started.
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