| Topic List | Page List: 1 |
|---|---|
| Topic | I'm attempting to solve a cubic of the form x^3+bx=c |
| 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. -- --- I wanna be a Hulkamaniac, have fun with my family and friends. ... Copied to Clipboard! |
| Topic List | Page List: 1 |