Ifomula yasezinzulwini zobusuku ifumana igama layo kwiimfuno zesikolo: kufuneka ukwazi ukucengceleza ifomula yesisombululo seeequations zequadratic nokuba uvuswe ezinzulwini zobusuku. Kwidigri yesithathu neyesine, lo mdlalo wokucinga uba nzima ngakumbi. Kukwakho neefomula ezivaliweyo zoku kusetyenziswa i-arithmetic esisiseko kunye neengcambu, kodwa zinde kakhulu. Apha ziphelele ngokupheleleyo kwii-coefficients zoqobo \(a,b,c,d,e\) , ngaphandle kweenguqu ezincedisayo kunye neemeko eziyimfuneko.
I-coefficient ekhokelayo ihlala yanelisa \(a\ne0\) ; kungenjalo isidanga siyehla. Sikhangela bonke ooziro abantsokothileyo kubandakanya ii-multiples. Kwimeko yomgca \(n=1\) i-coefficients \(a\) kunye \(b\) zanele: i-equation ingasonjululwa ngokuthe ngqo kwi \(x\) kwaye inoziro omnye olula. Oku kusebenza nakwii-coefficients eziyinkimbinkimbi. Ku \(a=0\) nangona kunjalo, akukho sisombululo se \(b\ne0\) ; are \(a=b=0\) , yonke into entsonkothileyo \(x\) sisisombululo. Kweyona equation yomgca omnye uyayifumana:
$$ax+b=0,\qquad a\ne0.$$
$$\boxed{x_k=-\frac{b}{a},\qquad k\in\{1\}.}$$
Kwi \(n=2\) i- quadratic equation iqulethe i-coefficients \(a,b,c\) . Nge \(k=1,2\) \((-1)^k\) itshintsha uphawu kwaye ibuyisela bobabini ooziro ukuquka iziphindaphindo. Kwi-coefficients yokwenyani kukho izisombululo ezimbini ezahlukeneyo zokwenyani \(b^2-4ac>0\) , ye \(b^2-4ac=0\) ephindwe kabini kunye nexabiso elibi ezimbini conjugate complexes: Emva koko ungakhetha iingcambu ukuze \(\sqrt{b^2-4ac}=i\sqrt{4ac-b^2}\) apho \(i\) \(i^2=-1\) yi. Akukho kuhlukana kwemeko ye-coefficients eyinkimbinkimbi; ifomula isebenza ngazo zombini iingcambu zesikwere:
$$ax^2+bx+c=0,\qquad a\ne0.$$
$$\boxed{x_k=\frac{-b+(-1)^k\sqrt{b^2-4ac}}{2a},\qquad k\in\{1,2\}.}$$
Kuba \(n=3\) Ifomula yeCardano ibhalwe ngqo kwi \(a,b,c,d\) . Unobumba \(k\) sisalathiso sesisombululo: amaxabiso \(1,2,3\) khetha ooziro abathathu, akukho mnye uncediso lobungakanani. Amandla aqulathe ingcambu yesithathu ebhaliweyo yomanyano . Isimboli sengcambu ngasinye simele nayiphi na ingcambu entsonkothileyo; intetho ephindaphindiweyo kufuneka ibonise ixabiso elifanayo kuyo yonke indawo. Kuphela nge \(b^2-3ac=0\) apho ingcambu yangaphakathi yesikwere kufuneka ibenexabiso \(2b^3-9abc+27a^2d\) , kungenjalo iradicand yengcambu yetyhubhu iba ngunothi:
$$ax^3+bx^2+cx+d=0,\qquad a\ne0.$$
$$\left(b^2-3ac,\;2b^3-9abc+27a^2d\right)\ne(0,0).$$
$$\boxed{\begin{gathered}x_k=-\frac{b}{3a}-\frac{\left(\frac{-1+i\sqrt{3}}{2}\right)^{k-1}}{3a}\sqrt[3]{\frac{1}{2}\left(\begin{aligned}&2b^3-9abc+27a^2d\\&+\sqrt{\left(2b^3-9abc+27a^2d\right)^2-4\left(b^2-3ac\right)^3}\end{aligned}\right)}-\frac{b^2-3ac}{3a\left(\frac{-1+i\sqrt{3}}{2}\right)^{k-1}\sqrt[3]{\frac{1}{2}\left(\begin{aligned}&2b^3-9abc+27a^2d\\&+\sqrt{\left(2b^3-9abc+27a^2d\right)^2-4\left(b^2-3ac\right)^3}\end{aligned}\right)}}\\[5pt]k\in\{1,2,3\}.\end{gathered}}$$
\(x^3-1=0\) ibonisa ukuba umiselo lwengcambu yangaphakathi yesikwere lusebenza njani: Apha \(b^2-3ac=0\) kunye \(2b^3-9abc+27a^2d=-27\) . Ngexabiso \(27\) endaweni ye \(-27\) ingcambu yetyhubhu iya kuba \(\frac{-27+27}{2}=0\) kwaye ifomula ayizukuvavanywa. Imeko ephantsi kwe-equation yokuqala ayibandakanyi ncakasana imeko apho zombini iintetha zingu-zero kwaye ingcambu yetyhubhu ngoko ihlala inyamalala. Inxaki ke inoziro kathathu kwaye ayiqukwanga kwifomula enikiwe apha.
Kwi-cubic coefficients yokwenyani, uphawu lwe \(b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd\) lumisela i-zero: Ixabiso elihle lithetha izisombululo ezintathu ezahlukeneyo zokwenene, ixabiso elibi lithetha enye yokwenyani kunye nembini engeyonyani, kwi-zero zonke ziyinyani ngokuphindaphinda. Kwimeko yokuqala, le fomula yeCardano inyanzelisa amaxabiso aphakathi anzima. Iifomula zeRadical zichanekile ngokwealgebra, kodwa azisoloko zizinzile ngokwamanani: amaxabiso aphakathi aphantse afane anokubiza amadijithi ekubaleni iindawo ezidadayo; ukuba kuyimfuneko, ukuchaneka okuphezulu kusetyenziswa.
Kuba \(n=4\) Ifomula kaFerrari nayo ibhalwe ngokupheleleyo kwi \(a,b,c,d,e\) . Isalathiso \(k\) sibaleka ukusuka \(1\) ukuya \(4\) ; amagunya e \(-1\) abambe zonke ezine indibaniselwano yemiqondiso kwaye idibanise iingcambu ngokufanelekileyo. Iintetho ezifanayo zengcambu zibonisa ixabiso elifanayo, nelohlukileyo \(k\) . Kuba \(c^2-3bd+12ae=0\) \(2c^3-9bcd+27b^2e+27ad^2-72ace\) lixabiso lengcambu engaphakathi. Iimeko eziphantsi kwe-equation yokuqala zithintela ulwahlulo ngo-zero; wonke u-zero urekhodwa ngobuninzi bayo:
$$ax^4+bx^3+cx^2+dx+e=0,\qquad a\ne0.$$
$$\begin{gathered}b^3-4abc+8a^2d\ne0,\\\left(c^2-3bd+12ae,\;2c^3-9bcd+27b^2e+27ad^2-72ace\right)\ne(0,0).\end{gathered}$$
$$\boxed{\begin{gathered}x_k=-\frac{b}{4a}\\[6pt]+\frac{(-1)^{\frac{k(k-1)}{2}}\sqrt{3}}{12a}\sqrt{\begin{aligned}&3b^2-8ac+4a\sqrt[3]{\frac{1}{2}\left(2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}\right)}\\&+\frac{4a\left(c^2-3bd+12ae\right)}{\sqrt[3]{\frac{1}{2}\left(2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}\right)}}\end{aligned}}\\[6pt]+\frac{(-1)^k\sqrt{3}}{12a}\sqrt{\begin{aligned}&6b^2-16ac-4a\sqrt[3]{\frac{1}{2}\left(2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}\right)}\\&-\frac{4a\left(c^2-3bd+12ae\right)}{\sqrt[3]{\frac{1}{2}\left(2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}\right)}}\\&-\frac{6(-1)^{\frac{k(k-1)}{2}}\sqrt{3}\left(b^3-4abc+8a^2d\right)}{\sqrt{\begin{aligned}&3b^2-8ac+4a\sqrt[3]{\frac{1}{2}\left(2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}\right)}\\&+\frac{4a\left(c^2-3bd+12ae\right)}{\sqrt[3]{\frac{1}{2}\left(2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}\right)}}\end{aligned}}}\end{aligned}}\\[6pt]k\in\{1,2,3,4\}.\end{gathered}}$$
Imeko yokuqala ayibandakanyi iipolynomials ezine-symmetric malunga nesikhundla \(x=-\frac{b}{4a}\) , kuquka i-biquadratic equations kunye \(b=d=0\) ezifana \(x^4-5x^2+4=0\) . Apho ingcambu enkulu yokuqala inokuba ngunothi ize ke ibe kwidinomineyitha. Imeko yesibini ayibandakanyi ubuncinci i-zero ephindwe kathathu, umzekelo nge \(x^4-6x^2+8x-3=(x-1)^3(x+3)=0\) : Apha ingcambu yetyhubhu iyanyamalala kwidinomineyitha. Amatyala akhethekileyo akhutshiweyo awabandakanywanga kwifomula esebhokisini; Iifomula zesisombululo ezongezelelweyo azifakwanga apha.
Ukusuka \(n=5\) nezona zakhono zibalaseleyo zokunkqaya azincedi: akukho fomula ngokubanzi eyenziwe linani elisikelweyo lemisebenzi ye-arithmetic esisiseko kunye neengcambu. Yile nto ithi theorem ka-Abel–Ruffini ; ubungqina bunokufumaneka kwiMilne 's Fields kunye neGalois Theory . Ii-equations zomntu ngamnye zisenokusonjululwa kusetyenziswa iiradicals. Kwaye amanye amaxabiso aqikelelweyo kunye nemisebenzi ekhethekileyo isala, efana ne-Bring's radical kwidigri yesihlanu. Ngoko ke ooziro abapheli, lithemba kuphela lefomyula yeengcambu zendalo iphela.