Timeline for How to avoid Vieta's formulas when simplifying?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 17, 2024 at 10:09 | comment | added | user64494 | OK with Solve[{w==(2*x1^2+3*x1*x2+2*x2^2+3 x2*x3+2 x3^2+3 x1*x3)/(4 x1^2*x2+4 x1^2*x3+4*x2^2*x1+4 x2^2*x3+4 x3^2*x1+4 x3^2*x2),2 x1^3+5 x1+8==0,2 x2^3+5 x2+8==0,2 x3^3+5 x3+8==0,x1!=x2,x1!=x3,x2!=x3},w,{x1,x2,x3}]. | |
| Oct 17, 2024 at 5:58 | comment | added | user64494 | Something to improve in your code, since the above command produces extraneous results {{w->-(5/96)},{w->(\[InvisiblePrefixScriptBase]^3)\[NegativeVeryThinSpace] -0.575\[Ellipsis]},{w->(\[InvisiblePrefixScriptBase]^3)\[NegativeVeryThinSpace] 0.0923\[Ellipsis]-0.312\[Ellipsis] I},{w->(\[InvisiblePrefixScriptBase]^3)\[NegativeVeryThinSpace] 0.0923\[Ellipsis]+0.312\[Ellipsis] I},...} too | |
| Oct 17, 2024 at 4:43 | comment | added | user64494 | Even Solve[{w == (2*x1^2 + 3*x1*x2 + 2*x2^2 + 3 x2*x3 + 2 x3^2 + 3 x1*x3)/(4 x1^2*x2 + 4 x1^2*x3 + 4*x2^2*x1 + 4 x2^2*x3 + 4 x3^2*x1 + 4 x3^2*x2), 2 x1^3 + 5 x1 + 8 == 0, 2 x2^3 + 5 x2 + 8 == 0, 2 x3^3 + 5 x3 + 8 == 0}, w, {x1, x2, x3}] works well. | |
| Oct 17, 2024 at 4:43 | vote | accept | user64494 | ||
| Oct 17, 2024 at 2:29 | history | answered | cvgmt | CC BY-SA 4.0 |