Ifomula yaphakathi kwamabili ithola igama layo esimisweni sesikole: kufanele ukwazi ukuphinda ifomula yesixazululo yezibalo zequadratic ngisho noma uvuswa phakathi nobusuku. Emadigri esithathu nesine, lo mdlalo wokucabanga uba nesidingo kakhulu. Kukhona futhi amafomula avaliwe alokhu kusetshenziswa i-arithmetic eyisisekelo nezimpande, kodwa made kakhulu. Lapha akuma-coefficient angempela \(a,b,c,d,e\) , ngaphandle kokuguquguquka okuyizisekelo kanye nezimo ezidingekayo.
I-coefficient ehamba phambili ihlala yanelisa \(a\ne0\) ; kungenjalo i-degree iyehla. Sifuna bonke oziro abayinkimbinkimbi okuhlanganisa nokuphindaphinda. Endabeni yomugqa \(n=1\) ama-coefficients \(a\) kanye \(b\) anele: isibalo singaxazululwa ngokuqondile kokuthi \(x\) futhi inoziro oyedwa olula. Lokhu kusebenza futhi kuma-coefficients ayinkimbinkimbi. Ku- \(a=0\) nokho, asikho isixazululo sokuthi \(b\ne0\) ; are \(a=b=0\) , zonke eziyinkimbinkimbi \(x\) ziyisixazululo. Ngokwezibalo zomugqa zangempela umuntu azitholayo:
$$ax+b=0,\qquad a\ne0.$$
$$\boxed{x_k=-\frac{b}{a},\qquad k\in\{1\}.}$$
Ku- \(n=2\) isibalo se-quadratic siqukethe ama-coefficient \(a,b,c\) . Nge \(k=1,2\) \((-1)^k\) ishintsha uphawu futhi ibuyisele kokubili oziro kuhlanganise nokuphindaphinda. Kuma-coefficients wangempela kunezixazululo ezimbili zangempela ezihlukile ze- \(b^2-4ac>0\) , ze- \(b^2-4ac=0\) okuphindwe kabili kanye nenani elingalungile izinhlanganisela ezimbili ze-conjugate: Bese ungakhetha izimpande ukuze \(\sqrt{b^2-4ac}=i\sqrt{4ac-b^2}\) , lapho iyunithi ye-imanary itholakala khona \(i\) \(i^2=-1\) ngu. Awukho umehluko wecala lama-coefficients ayinkimbinkimbi; ifomula isebenza kuzo zombili izimpande eziyisikwele:
$$ax^2+bx+c=0,\qquad a\ne0.$$
$$\boxed{x_k=\frac{-b+(-1)^k\sqrt{b^2-4ac}}{2a},\qquad k\in\{1,2\}.}$$
Ku- \(n=3\) Ifomula ye-Cardano ibhalwe ngokuqondile kokuthi \(a,b,c,d\) . Uhlamvu \(k\) luyinkomba yesixazululo: amanani \(1,2,3\) akhetha oziro abathathu, alikho elinye inani elisizayo. Amandla aqukethe impande yesithathu ebhaliwe yobunye . Uphawu lwempande ngayinye lumelela noma iyiphi impande eyinkimbinkimbi; isisho esiphindaphindiwe kumelwe sisho inani elifanayo yonke indawo. Kuphela ngokuthi \(b^2-3ac=0\) lapho impande eyisikwele engaphakathi kufanele ibe nenani \(2b^3-9abc+27a^2d\) , ngaphandle kwalokho i-radicand yempande ye-cube iba nguziro.:
$$ax^3+bx^2+cx+d=0,\qquad a\ne0.$$
$$\left(b^2-3ac,\;2b^3-9abc+27a^2d\right)\ne(0,0).$$
$$\boxed{\begin{gathered}x_k=-\frac{b}{3a}-\frac{\left(\frac{-1+i\sqrt{3}}{2}\right)^{k-1}}{3a}\sqrt[3]{\frac{1}{2}\left(\begin{aligned}&2b^3-9abc+27a^2d\\&+\sqrt{\left(2b^3-9abc+27a^2d\right)^2-4\left(b^2-3ac\right)^3}\end{aligned}\right)}-\frac{b^2-3ac}{3a\left(\frac{-1+i\sqrt{3}}{2}\right)^{k-1}\sqrt[3]{\frac{1}{2}\left(\begin{aligned}&2b^3-9abc+27a^2d\\&+\sqrt{\left(2b^3-9abc+27a^2d\right)^2-4\left(b^2-3ac\right)^3}\end{aligned}\right)}}\\[5pt]k\in\{1,2,3\}.\end{gathered}}$$
\(x^3-1=0\) ibonisa ukuthi ukucaciswa kwempande eyisikwele engaphakathi kusebenza kanjani: Lapha \(b^2-3ac=0\) kanye \(2b^3-9abc+27a^2d=-27\) . Ngevelu \(27\) esikhundleni sokuthi \(-27\) impande yekhyubhu izoba \(\frac{-27+27}{2}=0\) futhi ifomula ngeke ilinganiseke. Isimo esingaphansi kwesibalo sokuqala asifaki ncamashi isimo lapho zombili izisho zinguziro futhi impande yekhyubhu ihlale inyamalala. Isibalo sibe sinoziro kathathu futhi asihlanganiswa nefomula enikezwe lapha.
Kuma-cubic coefficients wangempela, uphawu lokuthi \(b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd\) lunquma oziro: Inani eliphozithivu lisho izixazululo zangempela ezintathu ezihlukene, inani elinegethivu lisho okukodwa okungokoqobo nezimbili okungezona ezangempela, kokuthi uziro zonke zingokoqobo ngokuphindaphinda. Esimweni sokuqala, le fomula ye-Cardano iphoqa amanani aphakathi ayinkimbinkimbi. Amafomula e-Radical ane-algebra enembile, kodwa awahlali ezinzile ngokwezinombolo: amanani acishe afane amaphakathi angabiza amadijithi ekubalweni kwamaphoyinti antantayo; uma kunesidingo, kusetshenziswa ukunemba okuphezulu.
Ku- \(n=4\) ifomula ye-Ferrari nayo ibhalwe ngokuphelele kokuthi \(a,b,c,d,e\) . Inkomba \(k\) iqala \(1\) iye ku \(4\) ; amandla okuthi \(-1\) athwebule zonke izinhlanganisela zezimpawu ezine futhi abhangqe izimpande ngendlela efanele. Izisho ezifanayo zempande zisho inani elifanayo, ngisho nakwehlukile \(k\) . Ngokuba \(c^2-3bd+12ae=0\) \(2c^3-9bcd+27b^2e+27ad^2-72ace\) inani lempande eyisikwele engaphakathi. Izimo ezingaphansi kwesibalo sokuqala zivimbela ukuhlukaniswa ngoziro; wonke uziro urekhodwa ngobuningi bayo:
$$ax^4+bx^3+cx^2+dx+e=0,\qquad a\ne0.$$
$$\begin{gathered}b^3-4abc+8a^2d\ne0,\\\left(c^2-3bd+12ae,\;2c^3-9bcd+27b^2e+27ad^2-72ace\right)\ne(0,0).\end{gathered}$$
$$\boxed{\begin{gathered}x_k=-\frac{b}{4a}\\[6pt]+\frac{(-1)^{\frac{k(k-1)}{2}}\sqrt{3}}{12a}\sqrt{\begin{aligned}&3b^2-8ac+4a\sqrt[3]{\frac{1}{2}\left(2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}\right)}\\&+\frac{4a\left(c^2-3bd+12ae\right)}{\sqrt[3]{\frac{1}{2}\left(2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}\right)}}\end{aligned}}\\[6pt]+\frac{(-1)^k\sqrt{3}}{12a}\sqrt{\begin{aligned}&6b^2-16ac-4a\sqrt[3]{\frac{1}{2}\left(2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}\right)}\\&-\frac{4a\left(c^2-3bd+12ae\right)}{\sqrt[3]{\frac{1}{2}\left(2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}\right)}}\\&-\frac{6(-1)^{\frac{k(k-1)}{2}}\sqrt{3}\left(b^3-4abc+8a^2d\right)}{\sqrt{\begin{aligned}&3b^2-8ac+4a\sqrt[3]{\frac{1}{2}\left(2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}\right)}\\&+\frac{4a\left(c^2-3bd+12ae\right)}{\sqrt[3]{\frac{1}{2}\left(2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}\right)}}\end{aligned}}}\end{aligned}}\\[6pt]k\in\{1,2,3,4\}.\end{gathered}}$$
Umbandela wokuqala awubandakanyi ama-polynomial ano-symmetric mayelana nendawo \(x=-\frac{b}{4a}\) , okuhlanganisa izibalo ze-biquadratic ne- \(b=d=0\) njengokuthi \(x^4-5x^2+4=0\) . Lapho impande yokuqala enkulu ingaba nguziro bese iba sedinomineyitha. Isimo sesibili asifaki okungenani uziro ophindwe kathathu, isibonelo ngokuthi \(x^4-6x^2+8x-3=(x-1)^3(x+3)=0\) : Lapha impande yekhyubhu iyanyamalala ku-denominator. Amacala akhethekile angabaliwe awahlanganiswa nefomula esebhokisini; amafomula esixazululo engeziwe akafakiwe lapha.
Kusukela ku \(n=5\) ngisho namakhono angcono kakhulu okubamba ngekhanda awasizi: ayikho ifomula evamile eyenziwe ngenani elilinganiselwe lemisebenzi ye-arithmetic eyisisekelo nezimpande. Yilokhu okushiwo yi -theorem ka-Abel–Ruffini ; ubufakazi bungatholakala ku- Milne's Fields kanye ne-Galois Theory . Izibalo ngazinye zisengaxazululwa kusetshenziswa ama-radicals. Futhi amanye amanani alinganiselwe nemisebenzi ekhethekile isala, njenge-Bring's radical at the five degree. Ngakho-ke ama-zero awanyamalali, kuphela ithemba lefomula yempande yendawo yonke.