Give-me-any-Quintic-i-shall-solve-it-For-sure-At-5-Bt-4-Ct-3-Dt-2-Et-F-0-wont-even-assume-A-1-or-B-0-but-if-A-C-E-B-D-F-then-my-formula-dont-work-but-then-obviously-t-1-is-a-root-

Question Number 66382 by ajfour last updated on 13/Aug/19

$$GivemeanyQuintic,ishallsolve$$$$it.Forsure!$$$${At}^5+{Bt}^4+{Ct}^3+{Dt}^2+Et+F=0$$$$wontevenassumeA=1,orB=0.$$$$butifA+C+E=B+D+F$$$$thenmyformuladontwork$$$$butthenobviouslyt=-1isaroot!$$

Commented by alphaprime last updated on 13/Aug/19

Sir , it clearly means that you've created a new research profile which yields solutions of quintics , It would be helpful in resolving higher curves , please get it published in any maths journals. it would be helpful and I wish & pray that it brings you capital and happiness simultaneously.

Commented by MJS last updated on 14/Aug/19

$$x^5+7x^4-22x^3-70x^2+170x-77=0$$$$x^5-\frac{11}{2}{x}^4+11x^3-\frac{77}{8}{x}^2+\frac{55}{16}x-\frac{11}{32}=0$$$$x^5-8x^4-73x^3+392x^2+779x-2760=0$$$$\mathrm{please}\mathrm{try}\mathrm{these}\mathrm{with}\mathrm{your}\mathrm{formula}.\mathrm{the}\mathrm{first}$$$$\mathrm{one}\mathrm{should}\mathrm{be}\mathrm{solveable}\mathrm{with}\mathrm{roots},\mathrm{the}\mathrm{second}$$$$\mathrm{and}\mathrm{third}\mathrm{ones}\mathrm{not}.\mathrm{the}\mathrm{second}\mathrm{one}\mathrm{has}\mathrm{zeros}$$$$\mathrm{which}\mathrm{can}\mathrm{be}\mathrm{expressed}\mathrm{in}\mathrm{a}\mathrm{closed}\mathrm{form}\dots$$

Commented by Tanmay chaudhury last updated on 14/Aug/19

Commented by ajfour last updated on 15/Aug/19

$$a=1,b=7,c=-22,d=-70,$$$$e=170,f=-77$$$$Mynewformulafor\mathit{o}\mathit{n}\mathit{e}root$$$$first:$$$$(4\mathit{d}\mathit{e}\mathit{f}-8{\mathit{c}\mathit{f}}^2-{\mathit{e}}^3){\mathit{t}}^3+({\mathit{e}}^2\mathit{f}-4{\mathit{d}\mathit{f}}^2){\mathit{t}}^2$$$$-3{\mathit{e}\mathit{f}}^2\mathit{t}-5{\mathit{f}}^3=0$$$$PleasecheckitMjSSir!$$

Commented by ajfour last updated on 15/Aug/19

$$igetverylittleerroreverytime$$$$frommyformula,ithinksome$$$$coefficientorsignofitneeds$$$$checking..$$

Commented by ajfour last updated on 15/Aug/19

$$Butitsforsure,ifoundan$$$$awesomemethod!$$

Commented by Rio Michael last updated on 15/Aug/19

$$ithinkmetoo,butitneedsproperchecking,i^{\prime }llcheckitthen{i}^{\prime }llpost\dots$$

Commented by MJS last updated on 16/Aug/19

$$\mathrm{does}\mathrm{this}\mathrm{mean}\mathrm{your}\mathrm{formula}\mathrm{for}\mathrm{one}\mathrm{root}\mathrm{is}$$$$\mathrm{independent}\mathrm{of}a\mathrm{and}b?$$$$\mathrm{this}\mathrm{cannot}\mathrm{be}\mathrm{true}\dots$$

Commented by ajfour last updated on 16/Aug/19

$$i^{\prime }lltrymodifyingitfurthersir.$$$$howeveritgives98\mathrm{\%}accurate$$$$answerfortheinevitablerealroot$$$$ofthequintic.$$$$Ihavetaken,$$$$y=(x+R)(x^4+{sx}^2+m)$$$$Andthenx=\frac{pt+q}{t}$$$$Thismaynotbethecaseofa$$$$generalquintic.$$$$Ingivenquintic{eq}^{n}A=1$$$$butreallyBismissingthough$$$$it{isn}^{\prime }tzero.Forotherroots$$$$Bappears!$$

Commented by Sayantan chakraborty last updated on 17/Aug/19

Commented by Sayantan chakraborty last updated on 17/Aug/19

$$\mathrm{AJFOUR}\mathrm{SIR}\mathrm{PLEASE}\mathrm{HELP}\mathrm{ME}\mathrm{TO}\mathrm{SOLVE}\mathrm{THIS}.$$

Commented by Sayantan chakraborty last updated on 17/Aug/19

$$\mathrm{SOLVE}\mathrm{IT}.$$

Commented by Rasheed.Sindhi last updated on 18/Aug/19

$$Sir,Ithinkyourquestion\mathrm{\&}comments$$$$areirrelevanthere!!!Ifyouwantto$$$$askaquestiongotonewquestion$$$$please!$$

Commented by TawaTawa last updated on 21/Aug/19

$$\mathrm{weldone}\mathrm{sir}\mathrm{Ajfour}$$

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