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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-

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Question Number 66382 by ajfour last updated on 13/Aug/19
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!
$${Give}\:{me}\:{any}\:{Quintic},\:{i}\:{shall}\:{solve} \\ $$$${it}.\:{For}\:{sure}! \\ $$$${At}^{\mathrm{5}} +{Bt}^{\mathrm{4}} +{Ct}^{\mathrm{3}} +{Dt}^{\mathrm{2}} +{Et}+{F}=\mathrm{0} \\ $$$${wont}\:{even}\:{assume}\:{A}=\mathrm{1},\:{or}\:{B}=\mathrm{0}. \\ $$$${but}\:{if}\:{A}+{C}+{E}={B}+{D}+{F}\: \\ $$$${then}\:{my}\:{formula}\:{dont}\:{work} \\ $$$${but}\:{then}\:{obviously}\:{t}=−\mathrm{1}\:{is}\:{a}\:{root}! \\ $$
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 −((11)/2)x^4 +11x^3 −((77)/8)x^2 +((55)/(16))x−((11)/(32))=0 x^5 −8x^4 −73x^3 +392x^2 +779x−2760=0 please try these with your formula. the first one should be solveable with roots, the second and third ones not. the second one has zeros which can be expressed in a closed form...
$${x}^{\mathrm{5}} +\mathrm{7}{x}^{\mathrm{4}} −\mathrm{22}{x}^{\mathrm{3}} −\mathrm{70}{x}^{\mathrm{2}} +\mathrm{170}{x}−\mathrm{77}=\mathrm{0} \\ $$$${x}^{\mathrm{5}} −\frac{\mathrm{11}}{\mathrm{2}}{x}^{\mathrm{4}} +\mathrm{11}{x}^{\mathrm{3}} −\frac{\mathrm{77}}{\mathrm{8}}{x}^{\mathrm{2}} +\frac{\mathrm{55}}{\mathrm{16}}{x}−\frac{\mathrm{11}}{\mathrm{32}}=\mathrm{0} \\ $$$${x}^{\mathrm{5}} −\mathrm{8}{x}^{\mathrm{4}} −\mathrm{73}{x}^{\mathrm{3}} +\mathrm{392}{x}^{\mathrm{2}} +\mathrm{779}{x}−\mathrm{2760}=\mathrm{0} \\ $$$$\mathrm{please}\:\mathrm{try}\:\mathrm{these}\:\mathrm{with}\:\mathrm{your}\:\mathrm{formula}.\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{one}\:\mathrm{should}\:\mathrm{be}\:\mathrm{solveable}\:\mathrm{with}\:\mathrm{roots},\:\mathrm{the}\:\mathrm{second} \\ $$$$\mathrm{and}\:\mathrm{third}\:\mathrm{ones}\:\mathrm{not}.\:\mathrm{the}\:\mathrm{second}\:\mathrm{one}\:\mathrm{has}\:\mathrm{zeros} \\ $$$$\mathrm{which}\:\mathrm{can}\:\mathrm{be}\:\mathrm{expressed}\:\mathrm{in}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{form}… \\ $$
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 My new formula for one root first: (4def−8cf^2 −e^3 )t^3 +(e^2 f−4df^2 )t^2 −3ef^2 t−5f^3 =0 Please check it MjS Sir!
$${a}=\mathrm{1},\:{b}=\mathrm{7},\:{c}=−\mathrm{22},\:{d}=−\mathrm{70},\: \\ $$$${e}=\mathrm{170},\:{f}=−\mathrm{77} \\ $$$${My}\:{new}\:{formula}\:{for}\:\boldsymbol{{one}}\:{root} \\ $$$${first}: \\ $$$$\left(\mathrm{4}\boldsymbol{{def}}−\mathrm{8}\boldsymbol{{cf}}^{\mathrm{2}} −\boldsymbol{{e}}^{\mathrm{3}} \right)\boldsymbol{{t}}^{\mathrm{3}} +\left(\boldsymbol{{e}}^{\mathrm{2}} \boldsymbol{{f}}−\mathrm{4}\boldsymbol{{df}}^{\mathrm{2}} \right)\boldsymbol{{t}}^{\mathrm{2}} \\ $$$$−\mathrm{3}\boldsymbol{{ef}}^{\mathrm{2}} \boldsymbol{{t}}−\mathrm{5}\boldsymbol{{f}}^{\mathrm{3}} =\mathrm{0} \\ $$$${Please}\:{check}\:{it}\:{MjS}\:{Sir}! \\ $$
Commented by ajfour last updated on 15/Aug/19
i get very little error every time from my formula, i think some coefficient or sign of it needs checking..
$${i}\:{get}\:{very}\:{little}\:{error}\:{every}\:{time} \\ $$$${from}\:{my}\:{formula},\:{i}\:{think}\:{some} \\ $$$${coefficient}\:{or}\:{sign}\:{of}\:{it}\:{needs} \\ $$$${checking}.. \\ $$
Commented by ajfour last updated on 15/Aug/19
But its for sure, i found an awesome method!
$${But}\:{its}\:{for}\:{sure},\:{i}\:{found}\:{an} \\ $$$${awesome}\:{method}! \\ $$
Commented by Rio Michael last updated on 15/Aug/19
i think me too,but it needs proper checking,i′ll check it then i′ll post...
$${i}\:{think}\:{me}\:{too},{but}\:{it}\:{needs}\:{proper}\:{checking},{i}'{ll}\:{check}\:{it}\:{then}\:{i}'{ll}\:{post}… \\ $$
Commented by MJS last updated on 16/Aug/19
does this mean your formula for one root is independent of a and b? this cannot be true...
$$\mathrm{does}\:\mathrm{this}\:\mathrm{mean}\:\mathrm{your}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{one}\:\mathrm{root}\:\mathrm{is} \\ $$$$\mathrm{independent}\:\mathrm{of}\:{a}\:\mathrm{and}\:{b}? \\ $$$$\mathrm{this}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{true}… \\ $$
Commented by ajfour last updated on 16/Aug/19
i′ll try modifying it further sir. however it gives 98% accurate answer for the inevitable real root of the quintic. I have taken, y=(x+R)(x^4 +sx^2 +m) And then x=((pt+q)/t) This may not be the case of a general quintic. In given quintic eq^n A=1 but really B is missing though it isn′t zero. For other roots B appears!
$${i}'{ll}\:{try}\:{modifying}\:{it}\:{further}\:{sir}. \\ $$$${however}\:{it}\:{gives}\:\mathrm{98\%}\:{accurate} \\ $$$${answer}\:{for}\:{the}\:{inevitable}\:{real}\:{root} \\ $$$${of}\:{the}\:{quintic}. \\ $$$${I}\:{have}\:{taken}, \\ $$$$\:\:\:{y}=\left({x}+{R}\right)\left({x}^{\mathrm{4}} +{sx}^{\mathrm{2}} +{m}\right) \\ $$$${And}\:{then}\:\:\:{x}=\frac{{pt}+{q}}{{t}} \\ $$$${This}\:{may}\:{not}\:{be}\:{the}\:{case}\:{of}\:{a} \\ $$$${general}\:{quintic}. \\ $$$${In}\:{given}\:{quintic}\:{eq}^{{n}} \:\:{A}=\mathrm{1} \\ $$$${but}\:{really}\:{B}\:{is}\:{missing}\:{though} \\ $$$${it}\:{isn}'{t}\:{zero}.\:{For}\:{other}\:{roots} \\ $$$${B}\:{appears}! \\ $$
Commented by Sayantan chakraborty last updated on 17/Aug/19
Commented by Sayantan chakraborty last updated on 17/Aug/19
AJFOUR SIR PLEASE HELP ME TO SOLVE THIS.
$$\mathrm{AJFOUR}\:\mathrm{SIR}\:\mathrm{PLEASE}\:\mathrm{HELP}\:\mathrm{ME}\:\mathrm{TO}\:\mathrm{SOLVE}\:\mathrm{THIS}. \\ $$
Commented by Sayantan chakraborty last updated on 17/Aug/19
SOLVE IT .
$$\mathrm{SOLVE}\:\mathrm{IT}\:. \\ $$
Commented by Rasheed.Sindhi last updated on 18/Aug/19
Sir, I think your question & comments are irrelevant here!!! If you want to ask a question go to new question please!
$${Sir},\:{I}\:{think}\:{your}\:{question}\:\&\:{comments} \\ $$$${are}\:{irrelevant}\:{here}!!!\:{If}\:{you}\:{want}\:{to} \\ $$$${ask}\:{a}\:{question}\:{go}\:{to}\:{new}\:{question} \\ $$$${please}! \\ $$
Commented by TawaTawa last updated on 21/Aug/19
weldone sir Ajfour
$$\mathrm{weldone}\:\mathrm{sir}\:\mathrm{Ajfour} \\ $$

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