Metod Ferrari analitichnij metod rozv yazannya rivnyannya chetvertogo stepenya Nazvanij na chest jogo avtora Lodoviko Ferrari Zmist 1 Kanonizaciya rivnyannya chetvertogo stepenya 2 Rozv yazok Ferrari 2 1 Peretvorennya pohidnogo kubichnogo rivnyannya do kanonichnogo viglyadu 2 2 Rozv yazannya pohidnogo kubichnogo rivnyannya 2 3 Vidobuvannya korenya z oboh chastin i zavershennya rozv yazuvannya 2 4 Pidsumki metodu FerrariKanonizaciya rivnyannya chetvertogo stepenya RedaguvatiRivnyannya chetvertogo stepenya zagalnogo viglyadu A x 4 B x 3 C x 2 D x E 0 1 displaystyle Ax 4 Bx 3 Cx 2 Dx E 0 qquad qquad 1 nbsp Podilimo obidvi chastini na A i pozbavimosya chlena x3 pidstanovkoyu x u B 4 A displaystyle x u B over 4A quad nbsp Zvedemo podibni dodanki i perepoznachimo koeficiyenti pri u a 3 B 2 8 A 2 C A displaystyle alpha 3B 2 over 8A 2 C over A quad nbsp b B 3 8 A 3 B C 2 A 2 D A displaystyle beta B 3 over 8A 3 BC over 2A 2 D over A quad nbsp g 3 B 4 256 A 4 C B 2 16 A 3 B D 4 A 2 E A displaystyle gamma 3B 4 over 256A 4 CB 2 over 16A 3 BD over 4A 2 E over A quad nbsp Otrimayemo kanonichne rivnyannya chetvertogo stepenya u 4 a u 2 b u g 0 1 displaystyle u 4 alpha u 2 beta u gamma 0 qquad qquad 1 quad nbsp Yaksho b 0 displaystyle beta 0 nbsp otrimayemo bikvadratne rivnyannya yake legko rozv yazuyetsya Rozv yazok Ferrari RedaguvatiRozglyanemo sut metodu Ferrari dlya rozv yazannya kanonichnogo rivnyannya chetvertogo stepenya Dlya cogo spochatku zapishemo ochevidnu totozhnist u 2 a 2 u 4 2 a u 2 a 2 displaystyle u 2 alpha 2 u 4 2 alpha u 2 alpha 2 quad nbsp i dodamo yiyi do rivnyannya 1 otrimayemo u 2 a 2 b u g a u 2 a 2 2 displaystyle u 2 alpha 2 beta u gamma alpha u 2 alpha 2 qquad qquad 2 quad nbsp Ce bulo zrobleno dlya togo shob zamist u4 otrimati povnij kvadrat u2 a 2 Drugij dodanok au2 ne znik prote jogo znak zaminivsya na protilezhnij i vin peremistivsya na inshij bik rivnyannya Nastupnim krokom ye vvedennya novoyi zminnoyi y do povnogo kvadrata u rivnyanni 2 i perenesennya 2y razom z koeficiyentom u2 do pravoyi chastini Otrimayemo totozhnu rivnist yaku mi potim dodamo do rivnyannya 2 u 2 a y 2 u 2 a 2 2 y u 2 a y 2 2 y u 2 2 y a y 2 displaystyle begin matrix u 2 alpha y 2 u 2 alpha 2 amp amp 2y u 2 alpha y 2 amp amp 2yu 2 2y alpha y 2 end matrix quad nbsp takozh rozglyanemo ochevidnu rivnist 0 a 2 y u 2 2 y u 2 a u 2 displaystyle 0 alpha 2y u 2 2yu 2 alpha u 2 quad nbsp Dodamo dvi ostanni rivnosti otrimayemo u 2 a y 2 u 2 a 2 a 2 y u 2 a u 2 2 y a y 2 displaystyle u 2 alpha y 2 u 2 alpha 2 alpha 2y u 2 alpha u 2 2y alpha y 2 quad nbsp Dodavshi cyu rivnist do 2 otrimayemo u 2 a y 2 b u g a 2 y u 2 2 y a y 2 a 2 displaystyle u 2 alpha y 2 beta u gamma alpha 2y u 2 2y alpha y 2 alpha 2 quad nbsp Cya rivnist ekvivalentna u 2 a y 2 a 2 y u 2 b u y 2 2 y a a 2 g 3 displaystyle u 2 alpha y 2 alpha 2y u 2 beta u y 2 2y alpha alpha 2 gamma qquad qquad 3 quad nbsp Viberemo zminnu y tak shob u pravij chastini rivnosti 3 utvorivsya povnij kvadrat Ce stanetsya yaksho diskriminant pravoyi chastini dorivnyuvatime 0 Dlya poyasnennya cogo yavisha rozglyanemo povnij kvadrat yak deyaku kvadratichnu funkciyu s u t 2 s 2 u 2 2 s t u t 2 displaystyle su t 2 s 2 u 2 2st u t 2 quad nbsp Kvadratichna funkciya z pravogo boku nerivnosti maye tri koeficiyenti Mozhna perekonatisya sho kvadrat drugogo z nih minus pochetverenij dobutok pershogo na tretogo dast nul 2 s t 2 4 s 2 t 2 0 displaystyle 2st 2 4 s 2 t 2 0 quad nbsp Tomu dlya togo shob peretvoriti pravu chastinu rivnyannya 3 na povnij kvadrat potribno rozv yazati shodo parametra y take rivnyannya b 2 4 2 y a y 2 2 y a a 2 g 0 displaystyle beta 2 4 2y alpha y 2 2y alpha alpha 2 gamma 0 quad nbsp Vikonayemo mnozhennya i zvedemo podibni dodanki pri y b 2 4 2 y 3 5 a y 2 4 a 2 2 g y a 3 a g 0 displaystyle beta 2 4 2y 3 5 alpha y 2 4 alpha 2 2 gamma y alpha 3 alpha gamma 0 quad nbsp Podilimo obidvi chastini na 4 i perenesemo b2 4 u pravu chastinu 2 y 3 5 a y 2 4 a 2 2 g y a 3 a g b 2 4 0 displaystyle 2y 3 5 alpha y 2 4 alpha 2 2 gamma y left alpha 3 alpha gamma beta 2 over 4 right 0 qquad qquad quad nbsp Mayemo kubichne rivnyannya shodo y Podilimo obidvi chastini na 2 y 3 5 2 a y 2 2 a 2 g y a 3 2 a g 2 b 2 8 0 4 displaystyle y 3 5 over 2 alpha y 2 2 alpha 2 gamma y left alpha 3 over 2 alpha gamma over 2 beta 2 over 8 right 0 qquad qquad 4 quad nbsp Peretvorennya pohidnogo kubichnogo rivnyannya do kanonichnogo viglyadu Redaguvati Rivnyannya 4 ye pohidnim kubichnim rivnyannyam vid rivnyannya chetvertogo stepenya Shob jogo rozv yazati potribno privesti jogo do kanonichnogo viglyadu Zrobimo zaminu y v 5 6 a displaystyle y v 5 over 6 alpha quad nbsp Rivnyannya 4 nabuvaye viglyadu v 5 6 a 3 5 2 a v 5 6 a 2 2 a 2 g v 5 6 a a 3 2 a g 2 b 2 8 0 displaystyle left v 5 over 6 alpha right 3 5 over 2 alpha left v 5 over 6 alpha right 2 2 alpha 2 gamma left v 5 over 6 alpha right left alpha 3 over 2 alpha gamma over 2 beta 2 over 8 right 0 quad nbsp Rozkriyemo duzhki v 3 5 2 a v 2 25 12 a 2 v 125 216 a 3 5 2 a v 2 5 3 a v 25 36 a 2 2 a 2 g v 5 6 a 2 a 2 g a 3 2 a g 2 b 2 8 0 displaystyle left v 3 5 over 2 alpha v 2 25 over 12 alpha 2 v 125 over 216 alpha 3 right 5 over 2 alpha left v 2 5 over 3 alpha v 25 over 36 alpha 2 right 2 alpha 2 gamma v 5 over 6 alpha 2 alpha 2 gamma left alpha 3 over 2 alpha gamma over 2 beta 2 over 8 right 0 quad nbsp Zvedemo podibni dodanki pri stepenyah v vrahuvavshi sho koeficiyent pri v2 dorivnyuye nulyu i cej dodanok znishuyetsya v 3 a 2 12 g v a 3 108 a g 3 b 2 8 0 displaystyle v 3 left alpha 2 over 12 gamma right v left alpha 3 over 108 alpha gamma over 3 beta 2 over 8 right 0 quad nbsp Mi otrimali kanonichne kubichne rivnyannya Perepoznachimo jogo koeficiyenti P a 2 12 g displaystyle P alpha 2 over 12 gamma quad nbsp Q a 3 108 a g 3 b 2 8 displaystyle Q alpha 3 over 108 alpha gamma over 3 beta 2 over 8 quad nbsp Otrimayemo rivnyannya v 3 P v Q 0 5 displaystyle v 3 Pv Q 0 qquad qquad 5 quad nbsp Rozv yazannya pohidnogo kubichnogo rivnyannya Redaguvati Rozglyanemo pitannya pro rozv yazannya nas zadovolnit bud yakij rozv yazok rivnyannya 5 Poznachimo U Q 2 Q 2 4 P 3 27 3 displaystyle U sqrt 3 Q over 2 pm sqrt Q 2 over 4 P 3 over 27 quad nbsp vzyato z kubichne rivnyannya dd otrimayetsya takij rozv yazok kubichnogo rivnyannya 4 ye y 5 6 a P 3 U U 6 displaystyle y 5 over 6 alpha P over 3U U qquad qquad 6 quad nbsp Mozhna pokazati sho mayut misce zalezhnosti 1 P 0 Q 2 Q 2 4 P 3 27 0 displaystyle P 0 Longleftarrow Q over 2 sqrt Q 2 over 4 P 3 over 27 0 quad nbsp 2 lim P 0 P Q 2 Q 2 4 P 3 27 3 0 displaystyle lim P to 0 P over sqrt 3 Q over 2 sqrt Q 2 over 4 P 3 over 27 0 quad nbsp Vidobuvannya korenya z oboh chastin i zavershennya rozv yazuvannya Redaguvati Rozglyanemo shemu zgortannya povnogo kvadrata s 2 u 2 2 s t u t 2 s 2 u 2 s t 2 s 2 2 displaystyle s 2 u 2 2st u t 2 left left sqrt s 2 right u 2st over 2 sqrt s 2 right 2 quad nbsp Vona ye virnoyu dlya oboh znakiv kvadratnih koreniv yaksho yih brati odnakovimi Mi ne budemo pisati vlasne znak oskilki ce viklikatime pevni trudnoshi zvazhayuchi na te sho dali vzhivatimutsya inshi znaki yaki viniknut potim Natomist poryad z cim znakom mi budemo staviti indeks sho yavlyatime soboyu zminnu znak yakoyi beretsya do uvagi dd Zvazhayuchi na ce mi otrimayemo a 2 y u 2 b u y 2 2 y a a 2 g a 2 y u b 2 a 2 y 2 displaystyle alpha 2y u 2 beta u y 2 2y alpha alpha 2 gamma left left sqrt alpha 2y right u beta over 2 sqrt alpha 2y right 2 quad nbsp Zauvazhennya Yaksho b 0 todi a 2y 0 A yaksho b 0 to mi otrimayemo bikvadratne rivnyannya sho bulo rozglyanute vishe dd Zvazhayuchi na ce 3 peretvoryuyetsya na u 2 a y 2 a 2 y u b 2 a 2 y 2 7 displaystyle u 2 alpha y 2 left left sqrt alpha 2y right u beta over 2 sqrt alpha 2y right 2 qquad qquad 7 quad nbsp Rivnist 7 mistit lishe povni kvadrati odin u livij chastini i odin u pravij Yaksho kvadrati dvoh viraziv rivni to i sami virazi rivni abo vidriznyayutsya lishe znakom tobto u 2 a y a 2 y u b 2 a 2 y 7 displaystyle u 2 alpha y pm left left sqrt alpha 2y right u beta over 2 sqrt alpha 2y right qquad qquad 7 quad nbsp Zvedemo podibni dodanki pri u u 2 s a 2 y u a y s b 2 a 2 y 0 8 displaystyle u 2 left mp s sqrt alpha 2y right u left alpha y pm s beta over 2 sqrt alpha 2y right 0 qquad qquad 8 quad nbsp Zauvazhennya Znaki s sho figuruyut u fomuli yak s displaystyle pm s quad nbsp i s displaystyle mp s quad nbsp ye velichinami zalezhnimi dd Rivnyannya 8 ye kvadratnim rivnyannyam shodo u Jogo rozv yazok maye viglyad u s a 2 y t a 2 y 4 a y s b 2 a 2 y 2 displaystyle u pm s sqrt alpha 2y pm t sqrt alpha 2y 4 alpha y pm s beta over 2 sqrt alpha 2y over 2 quad nbsp Abo pislya sproshennya u s a 2 y t 3 a 2 y s 2 b a 2 y 2 displaystyle u pm s sqrt alpha 2y pm t sqrt left 3 alpha 2y pm s 2 beta over sqrt alpha 2y right over 2 quad nbsp Ce rozv yazok kanonichnogo kvadratnogo rivnyannya Rozv yazok vihidnogo rivnyannya mozhna podati u viglyadi x B 4 A s a 2 y t 3 a 2 y s 2 b a 2 y 2 8 displaystyle x B over 4A pm s sqrt alpha 2y pm t sqrt left 3 alpha 2y pm s 2 beta over sqrt alpha 2y right over 2 qquad qquad 8 quad nbsp Vazhlivo Dva znaki s displaystyle pm s quad nbsp otrimani z rivnyannya 7 ye zalezhnimi tomu yavlyayut soboyu odin i toj samij znak a znak t displaystyle pm t quad nbsp nezalezhnij dd Pidsumki metodu Ferrari Redaguvati Rozv yazok rivnyannya chetvertogo stepenya A x 4 B x 3 C x 2 D x E 0 displaystyle Ax 4 Bx 3 Cx 2 Dx E 0 quad nbsp znahoditsya pislya provedennya obchislen a 3 B 2 8 A 2 C A displaystyle alpha 3B 2 over 8A 2 C over A quad nbsp b B 3 8 A 3 B C 2 A 2 D A displaystyle beta B 3 over 8A 3 BC over 2A 2 D over A quad nbsp g 3 B 4 256 A 4 C B 2 16 A 3 B D 4 A 2 E A displaystyle gamma 3B 4 over 256A 4 CB 2 over 16A 3 BD over 4A 2 E over A quad nbsp Yaksho b 0 displaystyle beta 0 quad nbsp to dorechno rozv yazuvati u 4 a u 2 g 0 displaystyle u 4 alpha u 2 gamma 0 quad nbsp i pidstanovkoyu x u B 4 A displaystyle x u B over 4A quad nbsp znahoditi koreniNemozhlivo rozibrati viraz SVG MathML mozhna vvimknuti cherez plagin brauzera Nedijsna vidpovid Math extension cannot connect to Restbase vid servera http localhost 6011 uk wikipedia org v1 displaystyle x B over 4A pm s sqrt alpha pm t sqrt alpha 2 4 gamma over 2 qquad beta 0 quad dd dd P a 2 12 g displaystyle P alpha 2 over 12 gamma quad nbsp Q a 3 108 a g 3 b 2 8 displaystyle Q alpha 3 over 108 alpha gamma over 3 beta 2 over 8 quad nbsp R Q 2 Q 2 4 P 3 27 displaystyle R Q over 2 pm sqrt Q 2 over 4 P 3 over 27 quad nbsp pidhodyat obidva znaki kvadratnogo korenya U R 3 displaystyle U sqrt 3 R quad nbsp v cogo rivnyannya isnuyut tri kompleksni koreni bud yakij z nih nas zadovolnit y 5 6 a U U 0 0 U 0 P 3 U displaystyle y 5 over 6 alpha U begin cases U 0 amp to 0 U neq 0 amp to P over 3U end cases quad nbsp W a 2 y displaystyle W sqrt alpha 2y quad nbsp x B 4 A s W t 3 a 2 y s 2 b W 2 displaystyle x B over 4A pm s W pm t sqrt left 3 alpha 2y pm s 2 beta over W right over 2 quad nbsp Dva simvoli s povinni mati odnakovi znaki a simvol t nezalezhnij Shob znajti vsi koreni treba znajti znachennya x dlya vsih kombinacij simvoliv s t spochatku toreba rozv yazati dlya vipadku potim dlya dali dlya i naostanok dlya Korin podvijnoyi kratnosti mi otrimayemo dvichi potrijnoyi trichi a korin kratnosti chotiri chotiri razi shopravda u comu vipadku u nas buv bi vipadok koli b 0 yakij ne ye zagalnim a prizvodit do bikvadratnogo rivnyannya Poryadok koreniv viznachayetsya tim yake U bulo obrano dd Otrimano z https uk wikipedia org w index php title Metod Ferrari amp oldid 33892518
