Čia aprašomos paprasčiausios algebrinės lygtys ir jų sprendimai. Aiškinama sunkėjimo tvarka.
Naudosime tokį žymėjimą: x, x1, x2 ir t.t. žymės nežinomuosius, o a, b, c, d ir t.t. – konkrečius duotus skaičius.
Pagrindinė algebros teorema
[keisti]
-tojo laipsnio polinomas (taigi, ir lygtis) turi lygiai n kompleksinių šaknų (sprendinių).
Bendra forma:
Sprendinys:
Nepilnoji kvadratinė lygtis
[keisti]
Bendra forma:
Sprendimas:
Pilnoji kvadratinė lygtis
[keisti]
Bendra forma:
Sprendimas:
randame pagalbini skaičių – diskriminantą D:
Tada jei
, tai realiųjų skaičių aibėje sprendinių nėra. Priešingu atveju realiuosius sprendinius rasime taip:
- Pavyzdžiui, reikia surasti kuriuose taškuose kertasi parabolė su Ox ašimi.
![{\displaystyle 3x^{2}+8x+4=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a090e76aac42f9d38d3d1f794b42bbcfb91fe5c)
![{\displaystyle D=b^{2}-4ac=8^{2}-4\cdot 3\cdot 4=64-48=16,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2e0ff73a9c619aae4c6f4d284cb07567110cebf)
![{\displaystyle x_{1,2}={\frac {-b\pm {\sqrt {D}}}{2a}}={\frac {-8\pm {\sqrt {16}}}{2\cdot 3}}={\frac {-8\pm 4}{6}}=-{\frac {2}{3}};\;-2.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e71d8ce0808edc6cfa1b598512c35a3d05a27146)
- Patikriname:
![{\displaystyle 3\cdot (-{\frac {2}{3}})^{2}+8\cdot (-{\frac {2}{3}})+4=3\cdot {\frac {4}{9}}-{\frac {16}{3}}+4={\frac {4}{3}}-{\frac {16}{3}}+4={\frac {4-16}{3}}+4={\frac {-12}{3}}+4=-4+4=0;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/714f9d235fb346ca225bdbd4fed595b29e70b6ac)
![{\displaystyle 3\cdot (-2)^{2}+8\cdot (-2)+4=3\cdot 4-16+4=12-16+4=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b829fbc188f88d94d138463417478bd6d0e73c04)
Tuo atveju, kai lygties šaknys kompleksiniai skaičiai
[keisti]
- Lygties
sprendiniai
![{\displaystyle x_{1}=\alpha +i\beta ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce3ae3abdb21268c945bbb98601f7c33482b0ca3)
![{\displaystyle x_{2}=\alpha -i\beta ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa0dc48b55180668c81ffe9cb5652acd70470c9b)
- kurie yra kompleksiniai skaičia randami taip:
![{\displaystyle \alpha =-{\frac {p}{2}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed9773368fd92d50f77f5f432bc0ccceb9f5e0f0)
![{\displaystyle \beta ={\sqrt {q-\alpha ^{2}}}={\sqrt {q-{\frac {p^{2}}{4}}}};}](https://wikimedia.org/api/rest_v1/media/math/render/svg/827578f671ee0d55e3a3652ec745e5ca721d7437)
![{\displaystyle \beta ={\frac {\sqrt {-D}}{2}}={\sqrt {\frac {-(p^{2}-4q)}{4}}}={\sqrt {q-{\frac {p^{2}}{4}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e80dd3850dae1ab02ff631d6ffe1388f83881a3)
- Lygties
sprendiniai
![{\displaystyle x_{1}=\alpha +i\beta ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce3ae3abdb21268c945bbb98601f7c33482b0ca3)
![{\displaystyle x_{2}=\alpha -i\beta ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa0dc48b55180668c81ffe9cb5652acd70470c9b)
- kurie yra kompleksiniai skaičia randami taip:
![{\displaystyle \alpha =-{\frac {b}{2a}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8eb44dc2eb2de4151eabc0985820c98720c963b)
![{\displaystyle \beta ={\frac {\sqrt {-D}}{2a}}={\sqrt {\frac {-(b^{2}-4ac)}{4a^{2}}}}={\sqrt {{\frac {c}{a}}-{\frac {b^{2}}{4a^{2}}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d63a0709f54f58da3f9c49d6cc47f2d05f0650d)
- Pavyzdis. Rasti sprendinius lygties
![{\displaystyle x^{2}-8x+25=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32cf1b9ed85f872e5f4f094d012eadac266b2faa)
- Sprendimas.
![{\displaystyle D=b^{2}-4ac=(-8)^{2}-4\cdot 1\cdot 25=64-100=-36=36i^{2},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/38ee9aeb367a9d3f6948caf99adce91d8d6073c6)
![{\displaystyle x_{1}={-b+{\sqrt {D}} \over 2}={-(-8)+{\sqrt {36i^{2}}} \over 2}={8+6i \over 2}=4+3i,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4515a91917a251d3989e53ae05a84889d528a629)
![{\displaystyle x_{2}={-b-{\sqrt {D}} \over 2}={-(-8)-{\sqrt {36i^{2}}} \over 2}={8-6i \over 2}=4-3i.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2a95d1feb53ebc146efa0c3b6f1701ae675913e)
- Patikriname, kad
![{\displaystyle (4+3i)^{2}-8(4+3i)+25=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ba5fef92983d22be11c882d0480b44110ce6d02)
![{\displaystyle 16+2\cdot 4\cdot 3i+(3i)^{2}-32-24i+25=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3d4e6aa516e4b8ed83eb343dc2c5654e97d6a9f)
![{\displaystyle 16+24i-9-32-24i+25=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d9d22336ad34ca792edd143012e79c011ffe544)
![{\displaystyle 16-41+25=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42f69b073f3de08f5721303d47761c90964bc56f)
- ir
![{\displaystyle (4-3i)^{2}-8(4-3i)+25=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a21f67959e284e8c5d8d97e4e26e7f977f6c512)
![{\displaystyle 16-2\cdot 4\cdot 3i+(-3i)^{2}-32+24i+25=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4768a39ce553ee8e059a43891c9375d76822cb0)
![{\displaystyle 16-24i-9-32+24i+25=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eed7ca445cf56e9652b54f20ed307c8aea583d8b)
![{\displaystyle 16-41+25=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df0e7e921b3ad9d628d8a9e86edfc8eab73d877e)
Kvadratinė lygtis, kurios ![{\displaystyle c=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9ee918699d0cb4b8c633cc1f520a8a7a174f44a)
[keisti]
Bendra forma:
Sprendimas:
iškeliame x prieš skliaustus:
Tada iš sandaugos savybių išplaukia, kad
Kvadratinė lygtis, kurios ![{\displaystyle a=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6104442ed30596ef4d7795d3186273f68d796ea4)
[keisti]
Duota kvadratinė lygtis:
![{\displaystyle x^{2}+bx+c=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/553992b06d1e563354e3a1092dd968dcd9231343)
kurią perrašome taip:
![{\displaystyle \left(x+{\frac {b}{2}}\right)^{2}+\left(c-{\frac {b^{2}}{4}}\right)=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7162b03186f25a18d9768ea1e147ba680d9a9a7c)
- Čia
![{\displaystyle \left(x+{\frac {b}{2}}\right)^{2}=x^{2}+bx+{\frac {b^{2}}{4}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a78e214e6c10580f1d57d52d12679e68eb8f0ecf)
- Todėl:
![{\displaystyle \left(x+{\frac {b}{2}}\right)^{2}=-\left(c-{\frac {b^{2}}{4}}\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/82b379660a4bbec6b25d34a471f8dfc2bcd27362)
![{\displaystyle \left(x+{\frac {b}{2}}\right)^{2}={\frac {b^{2}}{4}}-c,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b1a7c95a34cf1e2e6983c8f034a0dcd52524d50d)
![{\displaystyle x+{\frac {b}{2}}=\pm {\sqrt {{\frac {b^{2}}{4}}-c}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ddad67a71b435f950cba87580893f11514fbc6f7)
![{\displaystyle x=-{\frac {b}{2}}\pm {\sqrt {{\frac {b^{2}}{4}}-c}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff15c09960226e5b1c194275095b4e4648575dff)
![{\displaystyle x=-{\frac {b}{2}}\pm {\sqrt {{\frac {1}{4}}\cdot (b^{2}-4c)}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c1906458eaadb49ea78f7dfa71855b944b52841)
![{\displaystyle x=-{\frac {b}{2}}\pm {\frac {1}{2}}\cdot {\sqrt {b^{2}-4c}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6527dbc329d4be80f3b66ed1cb539b148d9fd9c1)
![{\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4c}}}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2837c33dea9444ba18a0e6513db3ce96fc8b2042)
![{\displaystyle x_{1}={\frac {-b+{\sqrt {b^{2}-4c}}}{2}};\quad x_{2}={\frac {-b-{\sqrt {b^{2}-4c}}}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f08b7f415269b8c02ebb4d4d7594936431dea867)
Kvadratinė lygtis, kurios a yra bet koks
[keisti]
Duota kvadratinė lygtis:
![{\displaystyle ax^{2}+bx+c=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24c2ce44ca552049d96088988f5d83f6763c059a)
![{\displaystyle x^{2}+{\frac {b}{a}}\cdot x+{\frac {c}{a}}=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7746b669421aab538227b75bdc4ec86d3b52e2f)
kurią perrašome taip:
![{\displaystyle \left(x+{\frac {b}{2\cdot a}}\right)^{2}+\left({\frac {c}{a}}-{\frac {b^{2}}{4\cdot a^{2}}}\right)=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f03f3a23be60a0cc850457789b466218beddc01)
- Čia
![{\displaystyle \left(x+{\frac {b}{2\cdot a}}\right)^{2}=x^{2}+{\frac {b}{a}}\cdot x+{\frac {b^{2}}{4\cdot a^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0da2a807d263f8125cdcb9dcfa9f9cc087ee7b1a)
- Todėl:
![{\displaystyle \left(x+{\frac {b}{2a}}\right)^{2}=-\left({\frac {c}{a}}-{\frac {b^{2}}{4a^{2}}}\right),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e8d0f467851b67db9bb9ade090250100a4e154c)
![{\displaystyle \left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52b47d246c1057ea3a6d779fc85ef135a5e99edf)
![{\displaystyle x+{\frac {b}{2a}}=\pm {\sqrt {{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35badb180eebe002d8f82dc55670fd099131c6db)
![{\displaystyle x=-{\frac {b}{2a}}\pm {\sqrt {{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/903cedd22650f1787188fa5826b03cabe3111877)
![{\displaystyle x=-{\frac {b}{2a}}\pm {\sqrt {{\frac {1}{4a^{2}}}\cdot (b^{2}-4ac)}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/73c5e5d25d42ca20a2acbc7d768841bd814e460a)
![{\displaystyle x=-{\frac {b}{2a}}\pm {\frac {1}{2a}}\cdot {\sqrt {b^{2}-4ac}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/818bfa6ff6b547b0d992fe1d63eca46d97a09a13)
![{\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a9804ca8ce019507e3199ca8fced800fb5b7d7c)
![{\displaystyle x_{1}={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}};\quad x_{2}={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e794e438735523911e435089ef65c873808e9654)
Bendra forma:
Sprendimas:
pažymime
, tada
.
,
o tai pilnoji kvadratinė lygtis, kuri jau išspręsta anksčiau. Jos sprendiniai yra
ir
.
Grįžtame prie pažymėjimo:
,
o tai kvadratinės lygtys, kurios jau išspręstos anksčiau. Iš jų rasime sprendinius
.
Jei yra lygtis
![{\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+a_{n-3}x^{n-3}+a_{n-4}x^{n-4}+a_{n-5}x^{n-5}+...+a_{1}x+a_{0}=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4e18f04fce62c84b0d50a34174addc8b0741ba0)
- Tai
![{\displaystyle s_{1}=x_{1}+x_{2}+x_{3}+x_{4}+...,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10ba66531334b6f062ef93d148589df1461119d7)
![{\displaystyle s_{2}=x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+...,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc9d2db30204b7a3cf623fecc6b8ca6c15f5f28c)
![{\displaystyle s_{3}=x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{2}x_{3}x_{4}+...}](https://wikimedia.org/api/rest_v1/media/math/render/svg/11dd10e8813027f2407a12cf5d1c37659d9f940a)
- ir taip toliau, kur
![{\displaystyle s_{i}=(-1)^{i}\cdot {\frac {a_{n-i}}{a_{n}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7124cca69c654d53c561eec4e9e31d75fa46d09)
Jei yra lygtis
![{\displaystyle ax^{2}+bx+c,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8fb8b910705f051903130697415ef9679d69f6a5)
- tai lygties sprendiniai:
![{\displaystyle x_{1}+x_{2}=-{\frac {b}{a}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a939a7e2f5e05085c2c52941b2d4ea21720991d4)
![{\displaystyle x_{1}\cdot x_{2}={\frac {c}{a}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0977e9d608f31849eca1a1d86ec071c7dcb404e)
Jei yra lygtis
![{\displaystyle ax^{3}+bx^{2}+cx+d,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c98496e4cc23d7e631f28f096dc0f45d1401e9b0)
- tai lygties sprendiniai:
![{\displaystyle x_{1}+x_{2}+x_{3}=-{\frac {b}{a}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/459508c549e53d9f42fb04ebe9091c8a4d716478)
![{\displaystyle x_{1}\cdot x_{2}+x_{1}\cdot x_{3}+x_{2}\cdot x_{3}={\frac {c}{a}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a551ac235228e1c9c689389db3b8073c834590a2)
![{\displaystyle x_{1}\cdot x_{2}\cdot x_{3}=-{\frac {d}{a}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4539c765ae93504d8ee0e1e2c16b32085ecfbe94)
Ketvirto laispnio lygtis
[keisti]
Jei yra lygtis
![{\displaystyle ax^{4}+bx^{3}+cx^{2}+dx+e,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00de4d9e61cfa964b75dbbf5ba593578e0a98feb)
- tai lygties sprendiniai:
![{\displaystyle x_{1}+x_{2}+x_{3}+x_{4}=-{\frac {b}{a}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/159584dcfec69cc952a5998fc176de67812f0afb)
![{\displaystyle x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4}={\frac {c}{a}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ab89ac99fe925b461ba48d276c032b4baa6ccb6)
![{\displaystyle x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}=-{\frac {d}{a}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/11a5108c1cd76d4e88835ad68fdbb89955b54adf)
![{\displaystyle x_{1}\cdot x_{2}\cdot x_{3}\cdot x_{4}={\frac {e}{a}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/36c74567b9a5111b9bbd2f3bed1cf9b52fd93df8)
Bendra forma:
Sprendimas:
Lygtį padalijame iš a ir keitiniu
,
pertvarkome lygtį į paprastesnį pavidalą
.
Randame pagalbinį skaičių – diskriminantą:
Kubinės lygties su realiaisiais koeficientais diskriminantas apibrėžia, kokias šaknis turi lygtis:
1. Jei D > 0, viena šaknis yra realioji ir dvi kompleksinės.
2. Jei D = 0, visos šaknys yra realiosios ir bent dvi iš jų yra vienodos.
3. Jei D < 0, visos trys šaknys yra realiosios ir skirtingos.
Pagal Kardano formulę, viena lygties šaknis
Kai D > 0, ši šaknis vienintelė
Kai D ≤ 0, tai lygtį
padaliję iš reiškinio
, gausime kvadratinę lygtį, kurios sprendimas nurodytas aukščiau.
Kubinė lygtis, kurios ![{\displaystyle d=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c87f7389ad2498c0f93551ec4fc92a882548484f)
[keisti]
Bet kokia kūbinė lygtis, kurios
yra išsprendžiama be jokių sunkumu.
- Pavyzdis Turime kūbinę lygtį
be skaičiaus d. Tuomet ją sprendžiame taip:
![{\displaystyle ax^{3}+bx^{2}+cx=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e475daae4a9c6c79e72fddd32824294d55bda6d0)
![{\displaystyle x(ax^{2}+bx+c)=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03c90402aefac371262196ef9bedb32a2c20867d)
- Vadinasi arba x=0 arba
![{\displaystyle ax^{2}+bx+c.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bebe03b03f574e0dd75bb8ddd5d4899bf4c6d6a8)
Išsprendžiame kvadratinę lygtį ir gauname tris realiasias šaknis arba dvi, arba vieną x=0, kai diskriminantas neigiamas.
Kubinės lygties sprendimas Kordano metodu
[keisti]
Yra kubinė lygtis:
![{\displaystyle y^{3}+ay^{2}+by+c=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/43b04882a0116fd56fe5d100de5a18068fe35a19)
- Pakeičiame
gauname:
![{\displaystyle \left(x-{\frac {a}{3}}\right)^{3}+a\left(x-{\frac {a}{3}}\right)^{2}+b\left(x-{\frac {a}{3}}\right)+c=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92ec5c36b0be81de5c323f2a3b9e4165f72adf06)
![{\displaystyle \left(x^{3}-3\cdot x^{2}\cdot {\frac {a}{3}}+3\cdot x\cdot \left({\frac {a}{3}}\right)^{2}-\left({\frac {a}{3}}\right)^{3}\right)+a\left(x^{2}-2\cdot x\cdot {\frac {a}{3}}+\left({\frac {a}{3}}\right)^{2}\right)+b\left(x-{\frac {a}{3}}\right)+c=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8747b284caaf937a0497e382d5d903db91464e4c)
![{\displaystyle \left(x^{3}-ax^{2}+3\cdot x\cdot {\frac {a^{2}}{9}}-{\frac {a^{3}}{27}}\right)+a\left(x^{2}-{\frac {2ax}{3}}+{\frac {a^{2}}{9}}\right)+bx-{\frac {ab}{3}}+c=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e9a1b485f036d7ab9d82fd3ce8dc3ab31be1b7fc)
![{\displaystyle \left(x^{3}-ax^{2}+{\frac {a^{2}x}{3}}-{\frac {a^{3}}{27}}\right)+ax^{2}-{\frac {2a^{2}x}{3}}+{\frac {a^{3}}{9}}+bx-{\frac {ab}{3}}+c=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/703e947b27830bc19c52107eea06c0abc32fbd0d)
![{\displaystyle x^{3}-{\frac {a^{3}}{27}}-{\frac {a^{2}x}{3}}+{\frac {a^{3}}{9}}+bx-{\frac {ab}{3}}+c=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae2197eee1b0848491fc8ec2ed0062c9ae996eed)
![{\displaystyle x^{3}-{\frac {a^{2}x}{3}}+bx+{\frac {a^{3}}{9}}-{\frac {a^{3}}{27}}-{\frac {ab}{3}}+c=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b8350c47629ddf8675c24d9e35900b0cd3d18e3e)
![{\displaystyle x^{3}+(b-{\frac {a^{2}}{3}})x+{\frac {3a^{3}}{27}}-{\frac {a^{3}}{27}}-{\frac {ab}{3}}+c=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/30afeba34bdf5c9e7936c062ad92eaca090f645d)
![{\displaystyle x^{3}+(b-{\frac {a^{2}}{3}})x+{\frac {2a^{3}}{27}}-{\frac {ab}{3}}+c=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/45c18af7104c691a515bf5b25d9b3b02f5b17e3f)
- Pažymime
ir pakeitę gauname:
![{\displaystyle x^{3}+px+q=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6dca36e86799ff88aa104efe8a0d66498ce3eaa7)
- Tegu
yra sprendinis lygties
(pagal teorema lygtis
turi 3 kompleksines šaknis). Įvedame pagalbinį u ir tikimes, kad polinomas
![{\displaystyle f(u)=u^{2}-x_{0}u-{\frac {p}{3}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5b945912dd673a4c3573a41613cdf3262b7b941)
- padės surasti
[jei lygtis
bus teisingai išspresta].
- Polinomo koeficientai - kompleksiniai skaičiai, ir todėl jis turi dvi kompleksines šaknis
ir
, be to, pagal Vijeto formulę,
![{\displaystyle \alpha +\beta =x_{0},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/20120de3b282fcd8d5d9435e4cbfeb1b8fec33fd)
![{\displaystyle \alpha \cdot \beta =-{\frac {p}{3}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b60bdcdad87ac0db7559574bd6d800fc3db0aaa0)
- Įstatę
į lygtį
gauname:
![{\displaystyle (\alpha +\beta )^{3}+p(\alpha +\beta )+q=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab303c01ced22f50e425a1bc4e49b53d54eb25f5)
![{\displaystyle (\alpha ^{3}+3\alpha ^{2}\beta +3\alpha \beta ^{2}+\beta ^{3})+p(\alpha +\beta )+q=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b159eb46afbc7df73026bdd172125966a8629606)
![{\displaystyle \alpha ^{3}+\beta ^{3}+3\alpha \beta (\alpha +\beta )+p(\alpha +\beta )+q=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b8f69a8118ddf1f5940ffee768821abbd5e32b00)
![{\displaystyle \alpha ^{3}+\beta ^{3}+(3\alpha \beta +p)(\alpha +\beta )+q=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/197005c86086259e8dfddbc4c4a76bcdc37558eb)
- Iš lygties
turime, kad:
![{\displaystyle \alpha \cdot \beta +{\frac {p}{3}}=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6bfc4edd17d64b232f747ec184e75ef01a41d2d)
![{\displaystyle 3\cdot \alpha \cdot \beta +p=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67d2b56ed017c54eff8660cf618252d5847febcc)
- Todėl gauname:
![{\displaystyle \alpha ^{3}+\beta ^{3}+0\cdot (\alpha +\beta )+q=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68a7ee9cbfb11f57177aa5f2ed8bdb1d951079d6)
![{\displaystyle \alpha ^{3}+\beta ^{3}+q=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f891fde83080242f12fe77a2a1b91a2be61fa85)
![{\displaystyle \alpha ^{3}+\beta ^{3}=-q.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65a34e7052653cc5005438ac2ac0346abac8a230)
- Dabar turime nauja gabaliuką iš Vijeto teoremos, tai yra lygtis
Mes žinome, kad koeficientas q priklauso lygčiai
. Todėl taip pat turime padaryti ir su kitu gabaliuku, kad sudėtos ir sudaugintos dalys duotų koeficientus (b ir c Vijeto teoremoje žymimi kvadratinėje lygtyje), taigi pakeliame kubu lygtyje
narius
,
ir kitą pusę. Ir gauname:
![{\displaystyle \alpha ^{3}\cdot \beta ^{3}=-{\frac {p^{3}}{27}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3c8e63baae858ee30baa99fe831e731d0811988)
- Šios dalys
yra g ir s koeficientai kvadratinės lygties
kuri turi sprendinius
ir
(iš Vijeto teoremos). Taigi, užtenka paaiškinimų, o dabar įstatome koeficientus į kvadratinę lygtį ir gauname:
![{\displaystyle z^{2}+qz-{\frac {p^{3}}{27}}=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db13af40a487a3b84cf2646888f836a76efa1eb8)
- Randame diskriminantą:
![{\displaystyle D=g^{2}-4s=q^{2}-4\cdot (-{\frac {p^{3}}{27}})=q^{2}+{\frac {4p^{3}}{27}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f749a25c6649606d0a06f8f12d54e0611666ed0)
- Randame sprendinius:
![{\displaystyle z_{1}={\frac {-g+{\sqrt {D}}}{2}}={\frac {-q+{\sqrt {q^{2}+{\frac {4p^{3}}{27}}}}}{2}}={\frac {1}{2}}\cdot (-q+{\sqrt {q^{2}+{\frac {4p^{3}}{27}}}}),\quad \alpha ={\sqrt[{3}]{z_{1}}}={\sqrt[{3}]{{\frac {1}{2}}\cdot (-q+{\sqrt {q^{2}+{\frac {4p^{3}}{27}}}})}};}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93b116a7e202b419da6bd7501bbd40be9b90b531)
![{\displaystyle z_{2}={\frac {-g-{\sqrt {D}}}{2}}={\frac {-q-{\sqrt {q^{2}+{\frac {4p^{3}}{27}}}}}{2}}={\frac {1}{2}}\cdot (-q-{\sqrt {q^{2}+{\frac {4p^{3}}{27}}}}),\quad \beta ={\sqrt[{3}]{z_{2}}}={\sqrt[{3}]{{\frac {1}{2}}\cdot (-q-{\sqrt {q^{2}+{\frac {4p^{3}}{27}}}})}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7986a6136b34758bb61f77a0db293dd99a868adc)
- Toliau
ir
įsistatome į lygtį
kad surasti lygties
sprendinį (šaknį)
. Taigi, gauname:
![{\displaystyle x_{0}=\alpha +\beta ={\sqrt[{3}]{{\frac {1}{2}}\cdot (-q+{\sqrt {q^{2}+{\frac {4p^{3}}{27}}}})}}+{\sqrt[{3}]{{\frac {1}{2}}\cdot (-q-{\sqrt {q^{2}+{\frac {4p^{3}}{27}}}})}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/da7304dbbaefa6301fcd991f8acd1a5142a39f6d)
![{\displaystyle x_{0}=\alpha +\beta ={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5af86908d9d0f935dc6837ee811aa6d4e1fe82b9)
- Kalbant apie kompleksinius sprendinius, negalima imti tokių sprendinių, kurie netenkina salygos
![{\displaystyle \alpha \cdot \beta =-{\frac {p}{3}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b60bdcdad87ac0db7559574bd6d800fc3db0aaa0)
![{\displaystyle \alpha _{2}\beta _{3}=\alpha _{1}\epsilon \cdot \beta _{1}\epsilon ^{2}=\alpha _{1}\beta _{1}\epsilon ^{3}=\alpha _{1}\beta _{1}={\frac {-p}{3}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1408622b0c8edb9a0cba974d5a88f30ad7586c44)
- Na, o visi sprendiniai yra šie:
![{\displaystyle x_{1}=\alpha _{1}+\beta _{1},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d82f8e9b4f18af086bad7072400312cb5b353dec)
![{\displaystyle x_{2}=\alpha _{2}+\beta _{3}=\alpha _{1}\epsilon +\beta _{1}\epsilon ^{2}=\alpha _{1}\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)+\beta _{1}\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right)=-{\frac {\alpha _{1}+\beta _{1}}{2}}+i{\sqrt {3}}{\frac {\alpha _{1}-\beta _{1}}{2}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e1c4930b00d109a48fcab6e1e18bf4c8a3cc7f0)
![{\displaystyle x_{3}=\alpha _{3}+\beta _{2}=\alpha _{1}\epsilon ^{2}+\beta _{1}\epsilon =\alpha _{1}\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right)+\beta _{1}\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)=-{\frac {\alpha _{1}+\beta _{1}}{2}}-i{\sqrt {3}}{\frac {\alpha _{1}-\beta _{1}}{2}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4dee23eba71f310c0b6ec7fa6d73761e7f139e7a)
- Jei sudėti ant apskritimo, kurio spindulys r=1, taškus
ir
, tai
laipsnių, o
laipsnių. Na o
, todėl
laipsnių (arba 0 laipsnių).
![{\displaystyle \epsilon ^{2}=\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)\cdot \left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)={\frac {1}{4}}-i{\frac {\sqrt {3}}{4}}-i{\frac {\sqrt {3}}{4}}+i\cdot i\cdot {\frac {3}{4}}={\frac {1}{4}}-i{\frac {2{\sqrt {3}}}{4}}-{\frac {3}{4}}=-{\frac {2}{4}}-i{\frac {\sqrt {3}}{2}}=-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb91a41db4d58aaabfafd1b464469249407f5218)
![{\displaystyle \epsilon ^{3}=\epsilon ^{2}\cdot \epsilon =\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right)\cdot \left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)={\frac {1}{4}}-i{\frac {\sqrt {3}}{4}}+i{\frac {\sqrt {3}}{4}}-i\cdot i\cdot {\frac {3}{4}}={\frac {1}{4}}-(-1)\cdot {\frac {3}{4}}={\frac {1}{4}}+{\frac {3}{4}}=1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be20d72ffbe21bd436d0ecd8725b5ff333ce97a8)
- Pavyzdis. Išspresti lygtį
![{\displaystyle y^{3}+3y^{2}-3y-14=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7fd2a4a7226d29e27769a8ec4118f0e50ded3d17)
- Keitinys
(čia a yra koeficientas esantis prie
) suprastina šitą lygtį į tokią lygtį:
![{\displaystyle (x-1)^{3}+3(x-1)^{2}-3(x-1)-14=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f5ae725d36bf41fe20d7c4ee5684388aaf21ef28)
![{\displaystyle x^{3}-3x^{2}+3x-1^{2}+3(x^{2}-2x+1^{2})-3x+1-14=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fdc33d169b84f5ea7fd20b5e4f1d2ee3209f9ef7)
![{\displaystyle x^{3}-3x^{2}+3x-1+3x^{2}-6x+3-3x+1-14=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a169f97261ccdd936b1a36bf2611030f056e9176)
![{\displaystyle x^{3}-6x+3-14=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb6b9962677fb1b1fb5c94b5312b6f3c616280b1)
![{\displaystyle x^{3}-6x-9=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f147b98f96e421d81ec14d92ceba78c24445e95)
- Čia
,
, todėl
![{\displaystyle {\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}={\frac {(-9)^{2}}{4}}+{\frac {(-6)^{3}}{27}}={\frac {81}{4}}+{\frac {-216}{27}}={\frac {81}{4}}-8={\frac {81-32}{4}}={\frac {49}{4}}>0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d38878f173d37cc235cce076e4962ad06f1a2e6b)
- t. y. lygtis
turi vieną tikrąjį ir du kompleksinius sprendinius.
- Pagal formulę:
![{\displaystyle \alpha ={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}={\sqrt[{3}]{-{\frac {-9}{2}}+{\sqrt {{\frac {(-9)^{2}}{4}}+{\frac {(-6)^{3}}{27}}}}}}={\sqrt[{3}]{{\frac {9}{2}}+{\frac {7}{2}}}}={\sqrt[{3}]{\frac {16}{2}}}={\sqrt[{3}]{8}}=2,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02632476cac911a9cac17c410dcf43b51d9b750a)
![{\displaystyle \beta ={\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}={\sqrt[{3}]{-{\frac {-9}{2}}-{\sqrt {{\frac {(-9)^{2}}{4}}+{\frac {(-6)^{3}}{27}}}}}}={\sqrt[{3}]{{\frac {9}{2}}-{\frac {7}{2}}}}={\sqrt[{3}]{\frac {2}{2}}}={\sqrt[{3}]{1}}=1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7167b33f98d49f2ee2f28832f10cba1cd2e887dd)
- Todėl
t. y.
. Du kitus sprendinius rasime pagal formules:
![{\displaystyle x_{2}=-{\frac {\alpha _{1}+\beta _{1}}{2}}+i{\sqrt {3}}\cdot {\frac {\alpha _{1}-\beta _{1}}{2}}=-{\frac {2+1}{2}}+i{\sqrt {3}}\cdot {\frac {2-1}{2}}=-{\frac {3}{2}}+i{\frac {\sqrt {3}}{2}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f5ec3b5aacd3561006da3fd192d97496eb16558)
![{\displaystyle x_{3}=-{\frac {\alpha _{1}+\beta _{1}}{2}}-i{\sqrt {3}}\cdot {\frac {\alpha _{1}-\beta _{1}}{2}}=-{\frac {2+1}{2}}-i{\sqrt {3}}\cdot {\frac {2-1}{2}}=-{\frac {3}{2}}-i{\frac {\sqrt {3}}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/486ba9dd8f84b74d06195458b33307f8324bad71)
- Iš čia gauname, kad sprendiniai užduotos lygties yra skaičiai:
![{\displaystyle y_{1}=x-{\frac {a}{3}}=x_{0}-1=3-1=2,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d1ed9c2befbd5aabe3c8411d70cafbf7f6628a4)
![{\displaystyle y_{2}=-{\frac {5}{2}}+i{\frac {\sqrt {3}}{2}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e6b77c74269be426c5de905b983304972fd2ccab)
![{\displaystyle y_{3}=-{\frac {5}{2}}-i{\frac {\sqrt {3}}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8e3ab97a4f9963fedd5373e6af07dbde5238e17)
- Patikriname, kai
, tai:
- Patikriname, kai
, tai:
![{\displaystyle x^{3}-6x-9=3^{3}-6\cdot 3-9=27-18-9=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c092838dc9ffd31c2ac8b3fad91e11367b50313)
- Pavyzdis. Išspręsti lygtį
- Čia
,
, todėl
![{\displaystyle {\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}={\frac {16^{2}}{4}}+{\frac {(-12)^{3}}{27}}={\frac {256}{4}}+{\frac {-1728}{27}}=64-64=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aab2d89755068071622192e6bd4c5fddf5d35cf0)
![{\displaystyle \alpha ={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}={\sqrt[{3}]{-{\frac {16}{2}}+{\sqrt {0}}}}={\sqrt[{3}]{-8+0}}={\sqrt[{3}]{-8}}=-2,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ce4cac29467da66c6dbbc513910382b090d5bc29)
![{\displaystyle \beta ={\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}={\sqrt[{3}]{-{\frac {16}{2}}-{\sqrt {0}}}}={\sqrt[{3}]{-8}}=-2.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9af97c4a74706e7779564949c24a28444402da5f)
- Iš čia seka:
t. y.
Todėl
![{\displaystyle x_{1}=\alpha _{1}+\beta _{1}=-2-2=-4,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9297c930190c0bc6eed3cddfa1811d30bf49d417)
![{\displaystyle x_{2}=-{\frac {\alpha _{1}+\beta _{1}}{2}}+i{\sqrt {3}}\cdot {\frac {\alpha _{1}-\beta _{1}}{2}}=-{\frac {-2-2}{2}}+i{\sqrt {3}}\cdot {\frac {-2-(-2)}{2}}=-{\frac {-4}{2}}+i\cdot 0\cdot {\frac {\sqrt {3}}{2}}=2,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/16b4df41043d2a7623c374820b4da6a10bce0e32)
![{\displaystyle x_{3}=-{\frac {\alpha _{1}+\beta _{1}}{2}}-i{\sqrt {3}}\cdot {\frac {\alpha _{1}-\beta _{1}}{2}}=-{\frac {-2-2}{2}}-i{\sqrt {3}}\cdot {\frac {2-(-2)}{2}}=-{\frac {-4}{2}}=2.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/525f84e97cfc7dd6938fbb2b4d976c6bf22baf2a)
- Patikriname įstatę
ir gauname:
![{\displaystyle x^{3}-12x+16=(-4)^{3}-12\cdot (-4)+16=-64+48+16=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/968da92e704cebe38639a565848470019e6b2626)
- Patikriname įstatę
ir gauname:
![{\displaystyle x^{3}-12x+16=2^{3}-12\cdot 2+16=8-24+16=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/edccdebe2ec6e871b44412de4bb60ecfe8bcc2c3)
- Pasinaudodami šiuo pavyzdžiu patvirtinsime šias formules:
![{\displaystyle x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=-2p;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/602a7b6018057f64f4d090869260f9e5cc35da10)
![{\displaystyle x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=-3q;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db1962f71dd14a5d36326037d2fda0b938bcb0ff)
![{\displaystyle x_{1}^{4}+x_{2}^{4}+x_{3}^{4}=2p^{2};}](https://wikimedia.org/api/rest_v1/media/math/render/svg/72911428356c4ced2e742b81ecb884ad09049dbd)
![{\displaystyle x_{1}^{5}+x_{2}^{5}+x_{3}^{5}=5pq;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/487477477ef9fad4e754471b9f05cda2d0088829)
- čia
,
,
,
,
. Atitinkamai turime:
![{\displaystyle (-4)^{2}+2^{2}+2^{2}=16+4+4=24=-2\cdot 12;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c438ffd483a3f585d93d9ac92cb27689ad3b43d)
![{\displaystyle (-4)^{3}+2^{3}+2^{3}=-64+8+8=-48=-3\cdot 16;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf198c512074ad09b10074523685a821737be900)
![{\displaystyle (-4)^{4}+2^{4}+2^{4}=256+16+16=288=2\cdot (-12)^{2}=2\cdot 144;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f1c6b8e1baf559c4a3d3c86c18531d766f5ebfd)
![{\displaystyle (-4)^{5}+2^{5}+2^{5}=-1024+32+32=-960=5\cdot (-12)\cdot 16.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d653424e7a072982273bdc6dcd2eeae0d6778ce3)
- Kai
tai
![{\displaystyle x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=x_{1}^{2}+\left(-{\frac {x_{1}}{2}}\right)^{2}+\left(-{\frac {x_{1}}{2}}\right)^{2}=x_{1}^{2}+{\frac {x_{1}^{2}}{2}}=-2p;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67dc6c0b545ed5a6c23e3f56bddd298a605f03bc)
![{\displaystyle x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3}=x_{1}\cdot \left(-{\frac {x_{1}}{2}}\right)+x_{1}\cdot \left(-{\frac {x_{1}}{2}}\right)+\left(-{\frac {x_{1}}{2}}\right)\cdot \left(-{\frac {x_{1}}{2}}\right)=-x_{1}^{2}+{\frac {x_{1}^{2}}{4}}=p,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7e9c95002e108eb14b9bda768fd30be36df8772)
![{\displaystyle -2(-x_{1}^{2}+{\frac {x_{1}^{2}}{4}})=2x_{1}^{2}-{\frac {x_{1}^{2}}{2}}=-2p.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4d67b12397bbefb7c6e71170f47ded40e9e336a)
- Taip pat ir su q, kai
tai
![{\displaystyle x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=x_{1}^{3}+\left(-{\frac {x_{1}}{2}}\right)^{3}+\left(-{\frac {x_{1}}{2}}\right)^{3}=x_{1}^{3}-2\cdot {\frac {x_{1}^{3}}{8}}=x_{1}^{3}-{\frac {x_{1}^{3}}{4}}={\frac {4x_{1}^{3}-x_{1}^{3}}{4}}={\frac {3x_{1}^{3}}{4}}=-3q;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42ff62f0a0dbb9a3807faed198b092b67f2f0879)
![{\displaystyle x_{1}x_{2}x_{3}=x_{1}\cdot \left(-{\frac {x_{1}}{2}}\right)\cdot \left(-{\frac {x_{1}}{2}}\right)={\frac {x_{1}^{3}}{4}}=-q,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e030c649d322f827d632bfec004672830e56273a)
![{\displaystyle 3\cdot {\frac {x_{1}^{3}}{4}}=3\cdot (-q).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50d657af2321147444f1e0ac25240e5eb5f4daa1)
- Pavyzdis. Išspręsti lygtį
- Čia
,
, todėl
![{\displaystyle {\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}={\frac {30^{2}}{4}}+{\frac {(-19)^{3}}{27}}={\frac {900}{4}}+{\frac {-6859}{27}}=225-254.037037=-29.037037037<0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3167373fb78dbb6a974c72bb026c1727c7da3598)
- Tokiu atveju, jeigu pasilikti srityje realiųjų skaičių, Kardano formulė šiai lygybei netinka, nors šios lygties sprendiniai ir yra 2, 3 ir
.
- Kaip galima išspręsti šitą lygtį žiūrėti čia https://lt.wikibooks.org/wiki/Kompleksiniai_skaičiai#Šaknies_traukimo_operacijos_trigonometrinėje_formoje
Kanoninė forma:
![{\displaystyle ax^{3}+bx^{2}+cx+d=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7002ed87fcbeaf757671f3535ef1706d108e963)
- Padaliname iš a ir įvedame vietoje x naują kintamjį
![{\displaystyle y^{3}+3py+2q=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/760d36085eed58fc1c07eb698c600699cdd2f5b3)
- kur
ir ![{\displaystyle 3p={\frac {3ac-b^{2}}{3a^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3d0a401ee43c6141426af53dcc1baa2a09d8663)
- Kardano sprendiniai
![{\displaystyle y_{1}=u+v,\quad y_{2}=\epsilon _{1}u+\epsilon _{2}v,\quad y_{3}=\epsilon _{2}u+\epsilon _{1}v,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81151100168341646ee68250525d104c5a6a050d)
kur
- o
ir
yra sprendiniai lygties
t. y. ![{\displaystyle \epsilon _{1},\epsilon _{2}=-{\frac {1}{2}}\pm i{\frac {\sqrt {3}}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0485a4bcbad8e4270be76a83e170463dec5e0b50)
- Tuo atveju, kai
tris tikrieji sprendiniai išreiškiami kompleksiniais dydžiais, ir protinga naudotis lentelės skaičiavimo budu.
- Pavyzdis.
Čia p=2, q=1; ![{\displaystyle q^{2}+p^{3}=1^{2}+2^{3}=1+8=9;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/670284f152ab7ce836fa24a31eee32d7f48e4bd6)
![{\displaystyle u={\sqrt[{3}]{-q+{\sqrt {q^{2}+p^{3}}}}}={\sqrt[{3}]{-1+{\sqrt {9}}}}={\sqrt[{3}]{-1+3}}={\sqrt[{3}]{2}}=1.25992105,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3deb984685c0b34825a46a695ad2f50b81c0b47)
![{\displaystyle v={\sqrt[{3}]{-q-{\sqrt {q^{2}+p^{3}}}}}={\sqrt[{3}]{-1-{\sqrt {9}}}}={\sqrt[{3}]{-1-3}}={\sqrt[{3}]{-4}}=-1.587401052,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fa1575a4b542483dbc82ff0fde0576bb838f1e03)
- Tikrasis sprendinis yra
![{\displaystyle y_{1}=u+v={\sqrt[{3}]{2}}+{\sqrt[{3}]{-4}}=1.25992105-1.587401052=-0.327480002;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b15f56e9fd32b44eaab90e3579ddf838a2a7a2bc)
- Kompleksiniai sprendiniai:
- Patikriname:
- Pavyzdis.
Čia p=1/3, q=1/2; ![{\displaystyle q^{2}+p^{3}=({\frac {1}{2}})^{2}+({\frac {1}{3}})^{3}={\frac {1}{4}}+{\frac {1}{27}}=0.287037037;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb634a350698dcabaaf9f58d95221d5f8b8451e5)
![{\displaystyle u={\sqrt[{3}]{-q+{\sqrt {q^{2}+p^{3}}}}}={\sqrt[{3}]{-{\frac {1}{2}}+{\sqrt {0.287037037}}}}={\sqrt[{3}]{-{\frac {1}{2}}+0.535758375}}={\sqrt[{3}]{0.035758375}}=0.329452338,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/327a5d565ce2e6dd58ec5f741240eb1360816db3)
![{\displaystyle v={\sqrt[{3}]{-q-{\sqrt {q^{2}+p^{3}}}}}={\sqrt[{3}]{-{\frac {1}{2}}-{\sqrt {0.287037037}}}}={\sqrt[{3}]{-{\frac {1}{2}}-0.535758375}}={\sqrt[{3}]{-1.035758375}}=-1.011780142,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99c21ebe6d62050c93bfb74ec18a52445463154b)
- Tikrasis sprendinis yra
![{\displaystyle y_{1}=u+v=0.329452338-1.011780142=-0.682327803;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06a5abc8d33e122eb77104ad77674c7b80bc9cdb)
- Patikriname:
![{\displaystyle y^{3}+y+1=(-0.682327803)^{3}-0.682327803+1=0.317672196-0.682327803+1=-0.999999999+1=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf832135d4187502a6a26973f9eea66b81b3baff)
- Pavyzdis.
Čia p=7/3=2.3(3), q=18/2=9; ![{\displaystyle q^{2}+p^{3}=9^{2}+(2.333333333)^{3}=81+12.7037037=93.7037037;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff92cfc23831204b2342b573893684196def95dc)
![{\displaystyle u={\sqrt[{3}]{-q+{\sqrt {q^{2}+p^{3}}}}}={\sqrt[{3}]{-9+{\sqrt {93.7037037}}}}={\sqrt[{3}]{-9+9.68006734}}={\sqrt[{3}]{0.680067339}}=0.879394961,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d324be3fe7e5424a3a66a3e8745d5b4d7ce57229)
![{\displaystyle v={\sqrt[{3}]{-q-{\sqrt {q^{2}+p^{3}}}}}={\sqrt[{3}]{-9-{\sqrt {93.7037037}}}}={\sqrt[{3}]{-9-9.68006734}}={\sqrt[{3}]{-18.68006734}}=-2.65333944,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d623d6363d53856aa0e1add31d58c1c7d06501a)
- Tikrasis sprendinis yra
![{\displaystyle y_{1}=u+v=0.879394961-2.65333944=-1.773944479;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/363f8d813b8de09ee829d5d47852477b96430948)
- Patikriname:
![{\displaystyle y^{3}+7y+18=(-1.773944479)^{3}+7(-1.773944479)+18=-5.582388651-12.41761135+18=-18+18=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e88b816907f4ab57357615ad5c55051a694db3a)
Kūbinės lygties sprendiniai
[keisti]
Jei duota lygtis
![{\displaystyle z^{3}+a_{2}z^{2}+a_{1}z+a_{0}=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b41d430cabdfba3012690089dc856073575cdade)
- tai jos 3 sprendiniai yra šie:
- Pavyzdis. Rasime lygties
realųjį sprendinį. Gauname:
![{\displaystyle =-{\frac {5}{3}}+{\sqrt[{3}]{{\frac {-223}{54}}+{\sqrt {{\frac {-4096}{729}}+17.05384088}}}}+{\sqrt[{3}]{{\frac {-223}{54}}-{\sqrt {{\frac {-4096}{729}}+17.05384088}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2cf4e07335382ce5151bdd9db585db7bd77ca2e3)
![{\displaystyle =-{\frac {5}{3}}+{\sqrt[{3}]{{\frac {-223}{54}}+{\sqrt {11.43518519}}}}+{\sqrt[{3}]{{\frac {-223}{54}}-{\sqrt {11.43518519}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea7410f4867c1d7a3305c1964b259618789ecf94)
![{\displaystyle =-{\frac {5}{3}}+{\sqrt[{3}]{{\frac {-223}{54}}+3.381595065}}+{\sqrt[{3}]{{\frac {-223}{54}}-3.381595065}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31e14919f699ad8f91f78918a1cc9f9f1f9b405d)
![{\displaystyle =-{\frac {5}{3}}+{\sqrt[{3}]{-0.748034565}}+{\sqrt[{3}]{-7.511224695}}=-{\frac {5}{3}}-0.907765950-1.958409849=-{\frac {5}{3}}-2.866175799=-4.532842466.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3075315cec70d48e3dd108ec7da627585929a2e)
- Patikriname:
![{\displaystyle z^{3}+5z^{2}+3z+4=(-4.532842466)^{3}+5(-4.532842466)^{2}+3(-4.532842466)+4=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1facfcc6195ae9796491f0eb9bd1b4e50c9a25b0)
![{\displaystyle =-93.13477672+102.73330411-13.598527398+4=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7fb3128645b163b126ea4af27ca199f86182c75)
- Pavyzdis. Rasime lygties
realųjį sprendinį. Gauname:
![{\displaystyle =-1+{\sqrt[{3}]{{\frac {9}{2}}+{\sqrt {-8+20.25}}}}+{\sqrt[{3}]{{\frac {9}{2}}-{\sqrt {-8+20.25}}}}=-1+{\sqrt[{3}]{{\frac {9}{2}}+{\sqrt {12.25}}}}+{\sqrt[{3}]{{\frac {9}{2}}-{\sqrt {12.25}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8fdc76cd37e262291ad562b9378d063a39c8e368)
![{\displaystyle =-1+{\sqrt[{3}]{4.5+3.5}}+{\sqrt[{3}]{4.5-3.5}}=-1+{\sqrt[{3}]{8}}+{\sqrt[{3}]{1}}=-1+2+1=2.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/541de74179b8f578aefdf61bd9a3c63d4030538c)
- Patikriname:
![{\displaystyle z^{3}+3z^{2}-3z-14=2^{3}+3\cdot 2^{2}-3\cdot 2-14=8+12-6-14=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18a3b892e2ac9a60601211d770f3ed2ebc4b68ff)
- Pavyzdis. Rasime lygties
realųjį sprendinį. Gauname:
![{\displaystyle =-{\frac {2}{3}}+{\sqrt[{3}]{{\frac {308}{54}}+{\sqrt {29.51851852}}}}+{\sqrt[{3}]{{\frac {308}{54}}-{\sqrt {29.51851852}}}}=-{\frac {2}{3}}+{\sqrt[{3}]{11.13679845}}+{\sqrt[{3}]{0.270608957}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5980539f2701fe4bf9f9c909ab7fc6e5bc7ee49d)
![{\displaystyle =-{\frac {2}{3}}+2.233161439+0.646815953=-{\frac {2}{3}}+2.879977392=2.213310725.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4299a27ec2277b0b4011167f335a3d6de62d136)
- Patikriname:
![{\displaystyle 2.213310725^{3}+2\cdot 2.213310725^{2}-3\cdot 2.213310725-14=10.84244344+9.797488731-6.639932175-14=20.63993217-20.63993218=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e197c6db490aa873e2734432bf55fafcc4526a35)