Diskriminantas

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

$n$-tojo laipsnio polinomas (taigi, ir lygtis) turi lygiai n kompleksinių šaknų (sprendinių).

Tiesinė lygtis

Bendra forma:

$a\cdot x=b$

Sprendinys:

$x={\frac {b}{a}}$

Nepilnoji kvadratinė lygtis

Bendra forma:

$ax^{2}=b\,$

Sprendimas:

${\begin{aligned}x^{2}&={\frac {b}{a}}\\x_{1,2}&=\pm {\sqrt {\frac {b}{a}}}\end{aligned}}$

Pilnoji kvadratinė lygtis

Bendra forma:

$ax^{2}+bx+c=0\,$

Sprendimas:

randame pagalbini skaičių – diskriminantą D:

$D=b^{2}-4ac\,$

Tada jei $D<0$, tai realiųjų skaičių aibėje sprendinių nėra. Priešingu atveju realiuosius sprendinius rasime taip:

$x_{1,2}={\frac {-b\pm {\sqrt {D}}}{2a}}$

• Pavyzdžiui, reikia surasti kuriuose taškuose kertasi parabolė su Ox ašimi.

$3x^{2}+8x+4=0,$

$D=b^{2}-4ac=8^{2}-4\cdot 3\cdot 4=64-48=16,$

$x_{1,2}={\frac {-b\pm {\sqrt {D}}}{2a}}={\frac {-8\pm {\sqrt {16}}}{2\cdot 3}}={\frac {-8\pm 4}{6}}=-{\frac {2}{3}};\;-2.$

Patikriname:

$3\cdot (-{\frac {2}{3}})^{2}+8\cdot (-{\frac {2}{3}})+4=3\cdot {\frac {4}{9}}-{\frac {16}{3}}+4={\frac {4}{3}}-{\frac {16}{3}}+4={\frac {4-16}{3}}+4={\frac {-12}{3}}+4=-4+4=0;$

$3\cdot (-2)^{2}+8\cdot (-2)+4=3\cdot 4-16+4=12-16+4=0.$

Tuo atveju, kai lygties šaknys kompleksiniai skaičiai

Lygties $x^{2}+px+q=0$ sprendiniai

$x_{1}=\alpha +i\beta ,$

$x_{2}=\alpha -i\beta ,$

kurie yra kompleksiniai skaičia randami taip:

$\alpha =-{\frac {p}{2}},$

$\beta ={\sqrt {q-\alpha ^{2}}}={\sqrt {q-{\frac {p^{2}}{4}}}};$

$\beta ={\frac {\sqrt {-D}}{2}}={\sqrt {\frac {-(p^{2}-4q)}{4}}}={\sqrt {q-{\frac {p^{2}}{4}}}}.$

Lygties $ax^{2}+bx+c=0$ sprendiniai

$x_{1}=\alpha +i\beta ,$

$x_{2}=\alpha -i\beta ,$

kurie yra kompleksiniai skaičia randami taip:

$\alpha =-{\frac {b}{2a}},$

$\beta ={\frac {\sqrt {-D}}{2a}}={\sqrt {\frac {-(b^{2}-4ac)}{4a^{2}}}}={\sqrt {{\frac {c}{a}}-{\frac {b^{2}}{4a^{2}}}}}.$

• Pavyzdis. Rasti sprendinius lygties

$x^{2}-8x+25=0.$

Sprendimas.

$D=b^{2}-4ac=(8)^{2}-4\cdot 1\cdot 25=64-100=-36=36i^{2},$

$x_{1}={-b+{\sqrt {D}} \over 2}={-(-8)+{\sqrt {36i^{2}}} \over 2}={8+6i \over 2}=4+3i,$

$x_{2}={-b-{\sqrt {D}} \over 2}={-(-8)-{\sqrt {36i^{2}}} \over 2}={8-6i \over 2}=4-3i.$

Patikriname, kad

$(4+3i)^{2}-8(4+3i)+25=0,$

$16+2\cdot 4\cdot 3i+(3i)^{2}-32-24i+25=0,$

$16+24i-9-32-24i+25=0,$

$16-41+25=0$

ir

$(4-3i)^{2}-8(4-3i)+25=0,$

$16-2\cdot 4\cdot 3i+(-3i)^{2}-32+24i+25=0,$

$16-24i-9-32+24i+25=0,$

$16-41+25=0.$

Kvadratinė lygtis, kurios $c=0$

Bendra forma:

$ax^{2}+bx=0\,$

Sprendimas:

iškeliame x prieš skliaustus:

$x(ax+b)=0\,$

Tada iš sandaugos savybių išplaukia, kad

${\begin{aligned}x=0\qquad \operatorname {arba} \qquad ax&=-b\\x&=-{\frac {b}{a}}\end{aligned}}$

Kvadratinė lygtis, kurios $a=1$

Duota kvadratinė lygtis:

$x^{2}+bx+c=0,$

kurią perrašome taip:

$\left(x+{\frac {b}{2}}\right)^{2}+\left(c-{\frac {b^{2}}{4}}\right)=0.$

Čia $\left(x+{\frac {b}{2}}\right)^{2}=x^{2}+bx+{\frac {b^{2}}{4}}.$

Todėl:

$\left(x+{\frac {b}{2}}\right)^{2}=-\left(c-{\frac {b^{2}}{4}}\right),$

$\left(x+{\frac {b}{2}}\right)^{2}={\frac {b^{2}}{4}}-c,$

$x+{\frac {b}{2}}=\pm {\sqrt {{\frac {b^{2}}{4}}-c}},$

$x=-{\frac {b}{2}}\pm {\sqrt {{\frac {b^{2}}{4}}-c}},$

$x=-{\frac {b}{2}}\pm {\sqrt {{\frac {1}{4}}\cdot (b^{2}-4c)}},$

$x=-{\frac {b}{2}}\pm {\frac {1}{2}}\cdot {\sqrt {b^{2}-4c}},$

$x={\frac {-b\pm {\sqrt {b^{2}-4c}}}{2}}.$

$x_{1}={\frac {-b+{\sqrt {b^{2}-4c}}}{2}};\quad x_{2}={\frac {-b-{\sqrt {b^{2}-4c}}}{2}}.$

Kvadratinė lygtis, kurios a yra bet koks

Duota kvadratinė lygtis:

$ax^{2}+bx+c=0,$

$x^{2}+{\frac {b}{a}}\cdot x+{\frac {c}{a}}=0,$

kurią perrašome taip:

$\left(x+{\frac {b}{2\cdot a}}\right)^{2}+\left({\frac {c}{a}}-{\frac {b^{2}}{4\cdot a^{2}}}\right)=0.$

Čia $\left(x+{\frac {b}{2\cdot a}}\right)^{2}=x^{2}+{\frac {b}{a}}\cdot x+{\frac {b^{2}}{4\cdot a^{2}}}.$

Todėl:

$\left(x+{\frac {b}{2a}}\right)^{2}=-\left({\frac {c}{a}}-{\frac {b^{2}}{4a^{2}}}\right),$

$\left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}},$

$x+{\frac {b}{2a}}=\pm {\sqrt {{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}},$

$x=-{\frac {b}{2a}}\pm {\sqrt {{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}},$

$x=-{\frac {b}{2a}}\pm {\sqrt {{\frac {1}{4a^{2}}}\cdot (b^{2}-4ac)}},$

$x=-{\frac {b}{2a}}\pm {\frac {1}{2a}}\cdot {\sqrt {b^{2}-4ac}},$

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.$

$x_{1}={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}};\quad x_{2}={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}.$

Bikvadratinė lygtis

Bendra forma:

$ax^{4}+bx^{2}+c=0\,$

Sprendimas:

pažymime $x^{2}=y\,$, tada $x^{4}=y^{2}\,$.

$ay^{2}+by+c=0\,$,

o tai pilnoji kvadratinė lygtis, kuri jau išspręsta anksčiau. Jos sprendiniai yra $y_{1}$ ir $y_{2}$.

Grįžtame prie pažymėjimo:

$y_{1}=x^{2}\qquad \operatorname {ir} \qquad y_{2}=x^{2}$,

o tai kvadratinės lygtys, kurios jau išspręstos anksčiau. Iš jų rasime sprendinius $x_{1},x_{2},x_{3},x_{4}$.

Vijeto teorema

Jei yra lygtis

$a_{n}x^{n}+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+a_{n-3}x^{n-3}+a_{n-4}x^{n-4}+a_{n-5}x^{n-5}+...+a_{1}x+a_{0}=0,$

Tai

$s_{1}=x_{1}+x_{2}+x_{3}+x_{4}+...,$

$s_{2}=x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+...,$

$s_{3}=x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{2}x_{3}x_{4}+...$

ir taip toliau, kur

$s_{i}=(-1)^{i}\cdot {\frac {a_{n-i}}{a_{n}}}.$

Kvadratinė lygtis

Jei yra lygtis

$ax^{2}+bx+c,$

tai lygties sprendiniai:

$x_{1}+x_{2}=-{\frac {b}{a}},$

$x_{1}\cdot x_{2}={\frac {c}{a}}.$

Kubinė lygtis

Jei yra lygtis

$ax^{3}+bx^{2}+cx+d,$

tai lygties sprendiniai:

$x_{1}+x_{2}+x_{3}=-{\frac {b}{a}},$

$x_{1}\cdot x_{2}+x_{1}\cdot x_{3}+x_{2}\cdot x_{3}={\frac {c}{a}},$

$x_{1}\cdot x_{2}\cdot x_{3}=-{\frac {d}{a}}.$

Ketvirto laispnio lygtis

Jei yra lygtis

$ax^{4}+bx^{3}+cx^{2}+dx+e,$

tai lygties sprendiniai:

$x_{1}+x_{2}+x_{3}+x_{4}=-{\frac {b}{a}},$

$x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4}={\frac {c}{a}},$

$x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}=-{\frac {d}{a}},$

$x_{1}\cdot x_{2}\cdot x_{3}\cdot x_{4}={\frac {e}{a}}.$

Pilnoji kubinė lygtis

Bendra forma:

$ax^{3}+bx^{2}+cx+d=0\,$

Sprendimas:

Lygtį padalijame iš a ir keitiniu $x=y-{\frac {b}{3}}$, pertvarkome lygtį į paprastesnį pavidalą

$x^{3}+px+q=0\,$.

Randame pagalbinį skaičių – diskriminantą:

$D=({\frac {q}{2}})^{2}+({\frac {p}{3}})^{3}$

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

$x_{1}={\sqrt[{3}]{-{\frac {p}{2}}+{\sqrt {D}}}}+{\sqrt[{3}]{-{\frac {p}{2}}-{\sqrt {D}}}}$

Kai D > 0, ši šaknis vienintelė

Kai D ≤ 0, tai lygtį $ax^{3}+bx^{2}+cx+d=0\,$ padaliję iš reiškinio $x-x_{1}\,$, gausime kvadratinę lygtį, kurios sprendimas nurodytas aukščiau.

Kubinė lygtis, kurios $d=0$

Bet kokia kūbinė lygtis, kurios $d=0$ yra išsprendžiama be jokių sunkumu.

• Pavyzdis Turime kūbinę lygtį $ax^{3}+bx^{2}+cx=0,$ be skaičiaus d. Tuomet ją sprendžiame taip:

$ax^{3}+bx^{2}+cx=0,$

$x(ax^{2}+bx+c)=0,$

Vadinasi arba x=0 arba $ax^{2}+bx+c.$

Išsprendžiame kvadratinę lygtį ir gauname tris realiasias šaknis arba dvi, arba vieną x=0, kai diskriminantas neigiamas.

Kubinės lygties sprendimas Kordano metodu

Yra kubinė lygtis:

$y^{3}+ay^{2}+by+c=0.$

Pakeičiame $y=x-{\frac {a}{3}},$ gauname:

$\left(x-{\frac {a}{3}}\right)^{3}+a\left(x-{\frac {a}{3}}\right)^{2}+b\left(x-{\frac {a}{3}}\right)+c=0,$

$\left(x^{3}-3\cdot x^{2}\cdot {\frac {a}{3}}+3\cdot x\cdot \left({\frac {a}{3}}\right)^{2}-\left({\frac {a}{3}}\right)^{3}\right)+a\left(x^{2}-2\cdot x\cdot {\frac {a}{3}}+\left({\frac {a}{3}}\right)^{2}\right)+b\left(x-{\frac {a}{3}}\right)+c=0,$

$\left(x^{3}-ax^{2}+3\cdot x\cdot {\frac {a^{2}}{9}}-{\frac {a^{3}}{27}}\right)+a\left(x^{2}-{\frac {2ax}{3}}+{\frac {a^{2}}{9}}\right)+bx-{\frac {ab}{3}}+c=0,$

$\left(x^{3}-ax^{2}+{\frac {a^{2}x}{3}}-{\frac {a^{3}}{27}}\right)+ax^{2}-{\frac {2a^{2}x}{3}}+{\frac {a^{3}}{9}}+bx-{\frac {ab}{3}}+c=0,$

$x^{3}-{\frac {a^{3}}{27}}-{\frac {a^{2}x}{3}}+{\frac {a^{3}}{9}}+bx-{\frac {ab}{3}}+c=0,$

$x^{3}-{\frac {a^{2}x}{3}}+bx+{\frac {a^{3}}{9}}-{\frac {a^{3}}{27}}-{\frac {ab}{3}}+c=0,$

$x^{3}+(b-{\frac {a^{2}}{3}})x+{\frac {3a^{3}}{27}}-{\frac {a^{3}}{27}}-{\frac {ab}{3}}+c=0,$

$x^{3}+(b-{\frac {a^{2}}{3}})x+{\frac {2a^{3}}{27}}-{\frac {ab}{3}}+c=0.$

Pažymime $p=b-{\frac {a^{2}}{3}},\quad q={\frac {2a^{3}}{27}}-{\frac {ab}{3}}+c$ ir pakeitę gauname:

$x^{3}+px+q=0.$

Tegu $x_{0}$ yra sprendinis lygties $x^{3}+px+q=0$ (pagal teorema lygtis $x^{3}+px+q=0$ turi 3 kompleksines šaknis). Įvedame pagalbinį u ir tikimes, kad polinomas

$f(u)=u^{2}-x_{0}u-{\frac {p}{3}}$

padės surasti $x_{0}$ [jei lygtis $u^{2}-x_{0}u-{\frac {p}{3}}=0$ bus teisingai išspresta].

Polinomo koeficientai - kompleksiniai skaičiai, ir todėl jis turi dvi kompleksines šaknis $\alpha$ ir $\beta$, be to, pagal Vijeto formulę,

$\alpha +\beta =x_{0},$

$\alpha \cdot \beta =-{\frac {p}{3}}.$

Įstatę $\alpha +\beta =x_{0}$ į lygtį $x^{3}+px+q=0$ gauname:

$(\alpha +\beta )^{3}+p(\alpha +\beta )+q=0,$

$(\alpha ^{3}+3\alpha ^{2}\beta +3\alpha \beta ^{2}+\beta ^{3})+p(\alpha +\beta )+q=0,$

$\alpha ^{3}+\beta ^{3}+3\alpha \beta (\alpha +\beta )+p(\alpha +\beta )+q=0,$

$\alpha ^{3}+\beta ^{3}+(3\alpha \beta +p)(\alpha +\beta )+q=0,$

Iš lygties $\alpha \cdot \beta =-{\frac {p}{3}}$ turime, kad:

$\alpha \cdot \beta +{\frac {p}{3}}=0,$

$3\cdot \alpha \cdot \beta +p=0.$

Todėl gauname:

$\alpha ^{3}+\beta ^{3}+0\cdot (\alpha +\beta )+q=0,$

$\alpha ^{3}+\beta ^{3}+q=0,$

$\alpha ^{3}+\beta ^{3}=-q.$

Dabar turime nauja gabaliuką iš Vijeto teoremos, tai yra lygtis $\alpha ^{3}+\beta ^{3}=-q.$ Mes žinome, kad koeficientas q priklauso lygčiai $x^{3}+px+q=0$. 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 $\alpha \cdot \beta =-{\frac {p}{3}}$ narius $\alpha$, $\beta$ ir kitą pusę. Ir gauname:

$\alpha ^{3}\cdot \beta ^{3}=-{\frac {p^{3}}{27}}.$

Šios dalys $g=q,$ $s=-{\frac {p^{3}}{27}}$ yra g ir s koeficientai kvadratinės lygties $z^{2}+gz+s=0,$ kuri turi sprendinius $z_{1}=\alpha ^{3}$ ir $z_{2}=\beta ^{3}$ (iš Vijeto teoremos). Taigi, užtenka paaiškinimų, o dabar įstatome koeficientus į kvadratinę lygtį ir gauname:

$z^{2}+qz-{\frac {p^{3}}{27}}=0.$

Randame diskriminantą:

$D=g^{2}-4s=q^{2}-4\cdot (-{\frac {p^{3}}{27}})=q^{2}+{\frac {4p^{3}}{27}}.$

Randame sprendinius:

$z_{1}={\frac {-g+{\sqrt {D}}}{2}}={\frac {-q+{\sqrt {q^{2}+{\frac {4p^{3}}{27}}}}}{2}}={\frac {1}{2}}\cdot (-q+{\sqrt {q^{2}+{\frac {4p^{3}}{27}}}}),\quad \alpha ={\sqrt[{3}]{z_{1}}}={\sqrt[{3}]{{\frac {1}{2}}\cdot (-q+{\sqrt {q^{2}+{\frac {4p^{3}}{27}}}})}};$

$z_{2}={\frac {-g-{\sqrt {D}}}{2}}={\frac {-q-{\sqrt {q^{2}+{\frac {4p^{3}}{27}}}}}{2}}={\frac {1}{2}}\cdot (-q-{\sqrt {q^{2}+{\frac {4p^{3}}{27}}}}),\quad \beta ={\sqrt[{3}]{z_{2}}}={\sqrt[{3}]{{\frac {1}{2}}\cdot (-q-{\sqrt {q^{2}+{\frac {4p^{3}}{27}}}})}}.$

Toliau $\alpha$ ir $\beta$ įsistatome į lygtį $\alpha +\beta =x_{0},$ kad surasti lygties $x^{3}+px+q=0$ sprendinį (šaknį) $x_{0}$. Taigi, gauname:

$x_{0}=\alpha +\beta ={\sqrt[{3}]{{\frac {1}{2}}\cdot (-q+{\sqrt {q^{2}+{\frac {4p^{3}}{27}}}})}}+{\sqrt[{3}]{{\frac {1}{2}}\cdot (-q-{\sqrt {q^{2}+{\frac {4p^{3}}{27}}}})}}.$

$x_{0}=\alpha +\beta ={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}.$

Kalbant apie kompleksinius sprendinius, negalima imti tokių sprendinių, kurie netenkina salygos $\alpha \cdot \beta =-{\frac {p}{3}}.$

$\alpha _{2}\beta _{3}=\alpha _{1}\epsilon \cdot \beta _{1}\epsilon ^{2}=\alpha _{1}\beta _{1}\epsilon ^{3}=\alpha _{1}\beta _{1}={\frac {-p}{3}}.$

Na, o visi sprendiniai yra šie:

$x_{1}=\alpha _{1}+\beta _{1},$

$x_{2}=\alpha _{2}+\beta _{3}=\alpha _{1}\epsilon +\beta _{1}\epsilon ^{2}=\alpha _{1}\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)+\beta _{1}\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right)=-{\frac {\alpha _{1}+\beta _{1}}{2}}+i{\sqrt {3}}{\frac {\alpha _{1}-\beta _{1}}{2}},$

$x_{3}=\alpha _{3}+\beta _{2}=\alpha _{1}\epsilon ^{2}+\beta _{1}\epsilon =\alpha _{1}\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right)+\beta _{1}\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)=-{\frac {\alpha _{1}+\beta _{1}}{2}}-i{\sqrt {3}}{\frac {\alpha _{1}-\beta _{1}}{2}},$

Jei sudėti ant apskritimo, kurio spindulys r=1, taškus $\epsilon$ ir $\epsilon ^{2}$, tai $\epsilon =120$ laipsnių, o $\epsilon ^{2}=240$ laipsnių. Na o $\epsilon ^{3}=1+i\cdot 0=1$, todėl $\epsilon ^{3}=360$ laipsnių (arba 0 laipsnių).

$\epsilon ^{2}=\left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)\cdot \left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)={\frac {1}{4}}-i{\frac {\sqrt {3}}{4}}-i{\frac {\sqrt {3}}{4}}+i\cdot i\cdot {\frac {3}{4}}={\frac {1}{4}}-i{\frac {2{\sqrt {3}}}{4}}-{\frac {3}{4}}=-{\frac {2}{4}}-i{\frac {\sqrt {3}}{2}}=-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}.$

$\epsilon ^{3}=\epsilon ^{2}\cdot \epsilon =\left(-{\frac {1}{2}}-i{\frac {\sqrt {3}}{2}}\right)\cdot \left(-{\frac {1}{2}}+i{\frac {\sqrt {3}}{2}}\right)={\frac {1}{4}}-i{\frac {\sqrt {3}}{4}}+i{\frac {\sqrt {3}}{4}}-i\cdot i\cdot {\frac {3}{4}}={\frac {1}{4}}-(-1)\cdot {\frac {3}{4}}={\frac {1}{4}}+{\frac {3}{4}}=1.$

• Pavyzdis. Išspresti lygtį $y^{3}+3y^{2}-3y-14=0.$

Keitinys $y=x-{\frac {a}{3}}=x-{\frac {3}{3}}=x-1$ (čia a yra koeficientas esantis prie $y^{2}$) suprastina šitą lygtį į tokią lygtį:

$(x-1)^{3}+3(x-1)^{2}-3(x-1)-14=0,$

$x^{3}-3x^{2}+3x-1^{2}+3(x^{2}-2x+1^{2})-3x+1-14=0,$

$x^{3}-3x^{2}+3x-1+3x^{2}-6x+3-3x+1-14=0,$

$x^{3}-6x+3-14=0,$

$x^{3}-6x-9=0.$

Čia $p=-6$, $q=-9$, todėl

${\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}={\frac {(-9)^{2}}{4}}+{\frac {(-6)^{3}}{27}}={\frac {81}{4}}+{\frac {-216}{27}}={\frac {81}{4}}-8={\frac {81-32}{4}}={\frac {49}{4}}>0,$

t. y. lygtis $x^{3}-6x-9=0$ turi vieną tikrąjį ir du kompleksinius sprendinius.

Pagal formulę:

$\alpha ={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}={\sqrt[{3}]{-{\frac {-9}{2}}+{\sqrt {{\frac {(-9)^{2}}{4}}+{\frac {(-6)^{3}}{27}}}}}}={\sqrt[{3}]{{\frac {9}{2}}+{\frac {7}{2}}}}={\sqrt[{3}]{\frac {16}{2}}}={\sqrt[{3}]{8}}=2,$

$\beta ={\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}={\sqrt[{3}]{-{\frac {-9}{2}}-{\sqrt {{\frac {(-9)^{2}}{4}}+{\frac {(-6)^{3}}{27}}}}}}={\sqrt[{3}]{{\frac {9}{2}}-{\frac {7}{2}}}}={\sqrt[{3}]{\frac {2}{2}}}={\sqrt[{3}]{1}}=1.$

Todėl $\alpha _{1}=2,$ $\beta _{1}=1,$ t. y. $x_{1}=\alpha _{1}+\beta _{1}=2+1=3$. Du kitus sprendinius rasime pagal formules:

$x_{2}=-{\frac {\alpha _{1}+\beta _{1}}{2}}+i{\sqrt {3}}\cdot {\frac {\alpha _{1}-\beta _{1}}{2}}=-{\frac {2+1}{2}}+i{\sqrt {3}}\cdot {\frac {2-1}{2}}=-{\frac {3}{2}}+i{\frac {\sqrt {3}}{2}},$

$x_{3}=-{\frac {\alpha _{1}+\beta _{1}}{2}}-i{\sqrt {3}}\cdot {\frac {\alpha _{1}-\beta _{1}}{2}}=-{\frac {2+1}{2}}-i{\sqrt {3}}\cdot {\frac {2-1}{2}}=-{\frac {3}{2}}-i{\frac {\sqrt {3}}{2}}.$

Iš čia gauname, kad sprendiniai užduotos lygties yra skaičiai:

$y_{1}=x-{\frac {a}{3}}=x_{0}-1=3-1=2,$

$y_{2}=-{\frac {5}{2}}+i{\frac {\sqrt {3}}{2}},$

$y_{3}=-{\frac {5}{2}}-i{\frac {\sqrt {3}}{2}}.$

Patikriname, kai $y_{1}=2$, tai:

$y^{3}+3y^{2}-3y-14=2^{3}+3\cdot 2^{2}-3\cdot 2-14=8+3\cdot 4-6-14=8+12-6-14=0.$

Patikriname, kai $x_{0}=\alpha +\beta =2+1=3$, tai:

$x^{3}-6x-9=3^{3}-6\cdot 3-9=27-18-9=0.$

• Pavyzdis. Išspręsti lygtį

$x^{3}-12x+16=0.$

Čia $p=-12$, $q=16$, todėl

${\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}={\frac {16^{2}}{4}}+{\frac {(-12)^{3}}{27}}={\frac {256}{4}}+{\frac {-1728}{27}}=64-64=0.$

$\alpha ={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}={\sqrt[{3}]{-{\frac {16}{2}}+{\sqrt {0}}}}={\sqrt[{3}]{-8+0}}={\sqrt[{3}]{-8}}=-2,$

$\beta ={\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}={\sqrt[{3}]{-{\frac {16}{2}}-{\sqrt {0}}}}={\sqrt[{3}]{-8}}=-2.$

Iš čia seka: $\alpha ={\sqrt[{3}]{-8}},$ t. y. $\alpha _{1}=\beta _{1}=-2.$ Todėl

$x_{1}=\alpha _{1}+\beta _{1}=-2-2=-4,$

$x_{2}=-{\frac {\alpha _{1}+\beta _{1}}{2}}+i{\sqrt {3}}\cdot {\frac {\alpha _{1}-\beta _{1}}{2}}=-{\frac {-2-2}{2}}+i{\sqrt {3}}\cdot {\frac {-2-(-2)}{2}}=-{\frac {-4}{2}}+i\cdot 0\cdot {\frac {\sqrt {3}}{2}}=2,$

$x_{3}=-{\frac {\alpha _{1}+\beta _{1}}{2}}-i{\sqrt {3}}\cdot {\frac {\alpha _{1}-\beta _{1}}{2}}=-{\frac {-2-2}{2}}-i{\sqrt {3}}\cdot {\frac {2-(-2)}{2}}=-{\frac {-4}{2}}=2.$

Patikriname įstatę $x_{1}=-4$ ir gauname:

$x^{3}-12x+16=(-4)^{3}-12\cdot (-4)+16=-64+48+16=0.$

Patikriname įstatę $x_{2}=2$ ir gauname:

$x^{3}-12x+16=2^{3}-12\cdot 2+16=8-24+16=0.$

Pasinaudodami šiuo pavyzdžiu patvirtinsime šias formules:

$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=-2p;$

$x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=-3q;$

$x_{1}^{4}+x_{2}^{4}+x_{3}^{4}=2p^{2};$

$x_{1}^{5}+x_{2}^{5}+x_{3}^{5}=5pq;$

čia $p=-12$, $q=16$, $x_{1}=-4$, $x_{2}=2$, $x_{3}=2$. Atitinkamai turime:

$(-4)^{2}+2^{2}+2^{2}=16+4+4=24=-2\cdot 12;$

$(-4)^{3}+2^{3}+2^{3}=-64+8+8=-48=-3\cdot 16;$

$(-4)^{4}+2^{4}+2^{4}=256+16+16=288=2\cdot (-12)^{2}=2\cdot 144;$

$(-4)^{5}+2^{5}+2^{5}=-1024+32+32=-960=5\cdot (-12)\cdot 16.$

Kai $x_{2}=x_{3}=-{\frac {x_{1}}{2}},$ tai

$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=x_{1}^{2}+\left(-{\frac {x_{1}}{2}}\right)^{2}+\left(-{\frac {x_{1}}{2}}\right)^{2}=x_{1}^{2}+{\frac {x_{1}^{2}}{2}}=-2p;$

$x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3}=x_{1}\cdot \left(-{\frac {x_{1}}{2}}\right)+x_{1}\cdot \left(-{\frac {x_{1}}{2}}\right)+\left(-{\frac {x_{1}}{2}}\right)\cdot \left(-{\frac {x_{1}}{2}}\right)=-x_{1}^{2}+{\frac {x_{1}^{2}}{4}}=p,$

$-2(-x_{1}^{2}+{\frac {x_{1}^{2}}{4}})=2x_{1}^{2}-{\frac {x_{1}^{2}}{2}}=-2p.$

Taip pat ir su q, kai $x_{2}=x_{3}=-{\frac {x_{1}}{2}},$ tai

$x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=x_{1}^{3}+\left(-{\frac {x_{1}}{2}}\right)^{3}+\left(-{\frac {x_{1}}{2}}\right)^{3}=x_{1}^{3}-2\cdot {\frac {x_{1}^{3}}{8}}=x_{1}^{3}-{\frac {x_{1}^{3}}{4}}={\frac {4x_{1}^{3}-x_{1}^{3}}{4}}={\frac {3x_{1}^{3}}{4}}=-3q;$

$x_{1}x_{2}x_{3}=x_{1}\cdot \left(-{\frac {x_{1}}{2}}\right)\cdot \left(-{\frac {x_{1}}{2}}\right)={\frac {x_{1}^{3}}{4}}=-q,$

$3\cdot {\frac {x_{1}^{3}}{4}}=3\cdot (-q).$

• Pavyzdis. Išspręsti lygtį

$x^{3}-19x+30=0.$

Čia $p=-19$, $q=30$, todėl

${\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}={\frac {30^{2}}{4}}+{\frac {(-19)^{3}}{27}}={\frac {900}{4}}+{\frac {-6859}{27}}=225-254.037037=-29.037037037<0.$

Tokiu atveju, jeigu pasilikti srityje realiųjų skaičių, Kardano formulė šiai lygybei netinka, nors šios lygties sprendiniai ir yra 2, 3 ir $-5$.

Kaip galima išspręsti šitą lygtį žiūrėti čia https://lt.wikibooks.org/wiki/Kompleksiniai_skaičiai#Šaknies_traukimo_operacijos_trigonometrinėje_formoje

Kubinė lygtis

Kanoninė forma:

$ax^{3}+bx^{2}+cx+d=0.$

Padaliname iš a ir įvedame vietoje x naują kintamjį $y^{3}+3py+2q=0,$

kur $2q={\frac {2b^{3}}{27a^{3}}}-{\frac {bc}{3a^{2}}}+{\frac {d}{a}}$ ir $3p={\frac {3ac-b^{2}}{3a^{2}}}.$

Kardano sprendiniai $y_{1}=u+v,\quad y_{2}=\epsilon _{1}u+\epsilon _{2}v,\quad y_{3}=\epsilon _{2}u+\epsilon _{1}v,$

kur $u={\sqrt[{3}]{-q+{\sqrt {q^{2}+p^{3}}}}},\quad v={\sqrt[{3}]{-q-{\sqrt {q^{2}+p^{3}}}}},$

o $\epsilon _{1}$ ir $\epsilon _{2}$ yra sprendiniai lygties $x^{2}+x+1=0,$ t. y. $\epsilon _{1},\epsilon _{2}=-{\frac {1}{2}}\pm i{\frac {\sqrt {3}}{2}}.$

Tuo atveju, kai $D=q^{2}+p^{3}<0$ tris tikrieji sprendiniai išreiškiami kompleksiniais dydžiais, ir protinga naudotis lentelės skaičiavimo budu.

• Pavyzdis. $y^{3}+6y+2=0.$ Čia p=2, q=1; $q^{2}+p^{3}=1^{2}+2^{3}=1+8=9;$

$u={\sqrt[{3}]{-q+{\sqrt {q^{2}+p^{3}}}}}={\sqrt[{3}]{-1+{\sqrt {9}}}}={\sqrt[{3}]{-1+3}}={\sqrt[{3}]{2}}=1.25992105,$

$v={\sqrt[{3}]{-q-{\sqrt {q^{2}+p^{3}}}}}={\sqrt[{3}]{-1-{\sqrt {9}}}}={\sqrt[{3}]{-1-3}}={\sqrt[{3}]{-4}}=-1.587401052,$

Tikrasis sprendinis yra $y_{1}=u+v={\sqrt[{3}]{2}}+{\sqrt[{3}]{-4}}=1.25992105-1.587401052=-0.327480002;$

Kompleksiniai sprendiniai:

$y_{2,3}=-{\frac {1}{2}}(u+v)\pm i\cdot {\frac {\sqrt {3}}{2}}\cdot (u-v)=0.163740001\pm i\cdot {\frac {\sqrt {3}}{2}}\cdot 2.847322102=0.163740001\pm i\cdot 2.465853273.$

Patikriname:

$y^{3}+6y+2=(-0.327480002)^{3}+6(-0.327480002)+2=-0.035119987-1.964880012+2=-2+2=0.$

• Pavyzdis. $y^{3}+y+1=0.$ Čia p=1/3, q=1/2; $q^{2}+p^{3}=({\frac {1}{2}})^{2}+({\frac {1}{3}})^{3}={\frac {1}{4}}+{\frac {1}{27}}=0.287037037;$

$u={\sqrt[{3}]{-q+{\sqrt {q^{2}+p^{3}}}}}={\sqrt[{3}]{-{\frac {1}{2}}+{\sqrt {0.287037037}}}}={\sqrt[{3}]{-{\frac {1}{2}}+0.535758375}}={\sqrt[{3}]{0.035758375}}=0.329452338,$

$v={\sqrt[{3}]{-q-{\sqrt {q^{2}+p^{3}}}}}={\sqrt[{3}]{-{\frac {1}{2}}-{\sqrt {0.287037037}}}}={\sqrt[{3}]{-{\frac {1}{2}}-0.535758375}}={\sqrt[{3}]{-1.035758375}}=-1.011780142,$

Tikrasis sprendinis yra $y_{1}=u+v=0.329452338-1.011780142=-0.682327803;$

Patikriname:

$y^{3}+y+1=(-0.682327803)^{3}-0.682327803+1=0.317672196-0.682327803+1=-0.999999999+1=0.$

• Pavyzdis. $y^{3}+7y+18=0.$ Čia p=7/3=2.3(3), q=18/2=9; $q^{2}+p^{3}=9^{2}+(2.333333333)^{3}=81+12.7037037=93.7037037;$

$u={\sqrt[{3}]{-q+{\sqrt {q^{2}+p^{3}}}}}={\sqrt[{3}]{-9+{\sqrt {93.7037037}}}}={\sqrt[{3}]{-9+9.68006734}}={\sqrt[{3}]{0.680067339}}=0.879394961,$

$v={\sqrt[{3}]{-q-{\sqrt {q^{2}+p^{3}}}}}={\sqrt[{3}]{-9-{\sqrt {93.7037037}}}}={\sqrt[{3}]{-9-9.68006734}}={\sqrt[{3}]{-18.68006734}}=-2.65333944,$

Tikrasis sprendinis yra $y_{1}=u+v=0.879394961-2.65333944=-1.773944479;$

Patikriname:

$y^{3}+7y+18=(-1.773944479)^{3}+7(-1.773944479)+18=-5.582388651-12.41761135+18=-18+18=0.$

Kūbinės lygties sprendiniai

Jei duota lygtis

$z^{3}+a_{2}z^{2}+a_{1}z+a_{0}=0,$

tai jos 3 sprendiniai yra šie:

$z_{1}=-{\frac {a_{2}}{3}}+{\sqrt[{3}]{{\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}+{\sqrt {\left({\frac {3a_{1}-a_{2}^{2}}{9}}\right)^{3}+\left({\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}\right)^{2}}}}}+{\sqrt[{3}]{{\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}-{\sqrt {\left({\frac {3a_{1}-a_{2}^{2}}{9}}\right)^{3}+\left({\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}\right)^{2}}}}};$

$z_{2}=-{\frac {a_{2}}{3}}-{\frac {1}{2}}\cdot \left({\sqrt[{3}]{{\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}+{\sqrt {\left({\frac {3a_{1}-a_{2}^{2}}{9}}\right)^{3}+\left({\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}\right)^{2}}}}}+{\sqrt[{3}]{{\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}-{\sqrt {\left({\frac {3a_{1}-a_{2}^{2}}{9}}\right)^{3}+\left({\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}\right)^{2}}}}}\right)+$ $+{\frac {i{\sqrt {3}}}{2}}\cdot \left({\sqrt[{3}]{{\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}+{\sqrt {\left({\frac {3a_{1}-a_{2}^{2}}{9}}\right)^{3}+\left({\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}\right)^{2}}}}}-{\sqrt[{3}]{{\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}-{\sqrt {\left({\frac {3a_{1}-a_{2}^{2}}{9}}\right)^{3}+\left({\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}\right)^{2}}}}}\right);$

$z_{3}=-{\frac {a_{2}}{3}}-{\frac {1}{2}}\cdot \left({\sqrt[{3}]{{\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}+{\sqrt {\left({\frac {3a_{1}-a_{2}^{2}}{9}}\right)^{3}+\left({\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}\right)^{2}}}}}+{\sqrt[{3}]{{\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}-{\sqrt {\left({\frac {3a_{1}-a_{2}^{2}}{9}}\right)^{3}+\left({\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}\right)^{2}}}}}\right)-$ $-{\frac {i{\sqrt {3}}}{2}}\cdot \left({\sqrt[{3}]{{\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}+{\sqrt {\left({\frac {3a_{1}-a_{2}^{2}}{9}}\right)^{3}+\left({\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}\right)^{2}}}}}-{\sqrt[{3}]{{\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}-{\sqrt {\left({\frac {3a_{1}-a_{2}^{2}}{9}}\right)^{3}+\left({\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}\right)^{2}}}}}\right).$

• Pavyzdis. Rasime lygties $z^{3}+5z^{2}+3z+4=0$ realųjį sprendinį. Gauname:

$z_{1}=-{\frac {a_{2}}{3}}+{\sqrt[{3}]{{\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}+{\sqrt {\left({\frac {3a_{1}-a_{2}^{2}}{9}}\right)^{3}+\left({\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}\right)^{2}}}}}+{\sqrt[{3}]{{\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}-{\sqrt {\left({\frac {3a_{1}-a_{2}^{2}}{9}}\right)^{3}+\left({\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}\right)^{2}}}}}=$ $=-{\frac {5}{3}}+{\sqrt[{3}]{{\frac {9\cdot 5\cdot 3-27\cdot 4-2\cdot 5^{3}}{54}}+{\sqrt {\left({\frac {3\cdot 3-5^{2}}{9}}\right)^{3}+\left({\frac {9\cdot 5\cdot 3-27\cdot 4-2\cdot 5^{3}}{54}}\right)^{2}}}}}+{\sqrt[{3}]{{\frac {9\cdot 5\cdot 3-27\cdot 4-2\cdot 5^{3}}{54}}-{\sqrt {\left({\frac {3\cdot 3-5^{2}}{9}}\right)^{3}+\left({\frac {9\cdot 5\cdot 3-27\cdot 4-2\cdot 5^{3}}{54}}\right)^{2}}}}}=$ $=-{\frac {5}{3}}+{\sqrt[{3}]{{\frac {135-108-250}{54}}+{\sqrt {\left({\frac {9-25}{9}}\right)^{3}+\left({\frac {135-108-250}{54}}\right)^{2}}}}}+{\sqrt[{3}]{{\frac {135-108-250}{54}}-{\sqrt {\left({\frac {9-25}{9}}\right)^{3}+\left({\frac {135-108-250}{54}}\right)^{2}}}}}=$ $=-{\frac {5}{3}}+{\sqrt[{3}]{{\frac {-223}{54}}+{\sqrt {\left({\frac {-16}{9}}\right)^{3}+\left({\frac {-223}{54}}\right)^{2}}}}}+{\sqrt[{3}]{{\frac {-223}{54}}-{\sqrt {\left({\frac {-16}{9}}\right)^{3}+\left({\frac {-223}{54}}\right)^{2}}}}}=$

$=-{\frac {5}{3}}+{\sqrt[{3}]{{\frac {-223}{54}}+{\sqrt {{\frac {-4096}{729}}+17.05384088}}}}+{\sqrt[{3}]{{\frac {-223}{54}}-{\sqrt {{\frac {-4096}{729}}+17.05384088}}}}=$

$=-{\frac {5}{3}}+{\sqrt[{3}]{{\frac {-223}{54}}+{\sqrt {11.43518519}}}}+{\sqrt[{3}]{{\frac {-223}{54}}-{\sqrt {11.43518519}}}}=$

$=-{\frac {5}{3}}+{\sqrt[{3}]{{\frac {-223}{54}}+3.381595065}}+{\sqrt[{3}]{{\frac {-223}{54}}-3.381595065}}=$

$=-{\frac {5}{3}}+{\sqrt[{3}]{-0.748034565}}+{\sqrt[{3}]{-7.511224695}}=-{\frac {5}{3}}-0.907765950-1.958409849=-{\frac {5}{3}}-2.866175799=-4.532842466.$

Patikriname:

$z^{3}+5z^{2}+3z+4=(-4.532842466)^{3}+5(-4.532842466)^{2}+3(-4.532842466)+4=$

$=-93.13477672+102.73330411-13.598527398+4=0.$

• Pavyzdis. Rasime lygties $z^{3}+a_{2}z^{2}+a_{1}z+a_{0}=z^{3}+3z^{2}-3z-14=0$ realųjį sprendinį. Gauname:

$z_{1}=-{\frac {a_{2}}{3}}+{\sqrt[{3}]{{\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}+{\sqrt {\left({\frac {3a_{1}-a_{2}^{2}}{9}}\right)^{3}+\left({\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}\right)^{2}}}}}+{\sqrt[{3}]{{\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}-{\sqrt {\left({\frac {3a_{1}-a_{2}^{2}}{9}}\right)^{3}+\left({\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}\right)^{2}}}}}=$ $=-{\frac {3}{3}}+{\sqrt[{3}]{{\frac {9\cdot 3\cdot (-3)-27\cdot (-14)-2\cdot 3^{3}}{54}}+{\sqrt {\left({\frac {3\cdot (-3)-3^{2}}{9}}\right)^{3}+\left({\frac {-81+378-54}{54}}\right)^{2}}}}}+{\sqrt[{3}]{{\frac {-81+378-54}{54}}-{\sqrt {\left({\frac {-9-9}{9}}\right)^{3}+\left({\frac {-81+378-54}{54}}\right)^{2}}}}}=$ $=-1+{\sqrt[{3}]{{\frac {243}{54}}+{\sqrt {\left({\frac {-18}{9}}\right)^{3}+\left({\frac {243}{54}}\right)^{2}}}}}+{\sqrt[{3}]{{\frac {243}{54}}-{\sqrt {\left({\frac {-18}{9}}\right)^{3}+\left({\frac {243}{54}}\right)^{2}}}}}=$ $=-1+{\sqrt[{3}]{{\frac {9}{2}}+{\sqrt {\left(-2\right)^{3}+\left({\frac {9}{2}}\right)^{2}}}}}+{\sqrt[{3}]{{\frac {9}{2}}-{\sqrt {(-2)^{3}+\left({\frac {9}{2}}\right)^{2}}}}}=$

$=-1+{\sqrt[{3}]{{\frac {9}{2}}+{\sqrt {-8+20.25}}}}+{\sqrt[{3}]{{\frac {9}{2}}-{\sqrt {-8+20.25}}}}=-1+{\sqrt[{3}]{{\frac {9}{2}}+{\sqrt {12.25}}}}+{\sqrt[{3}]{{\frac {9}{2}}-{\sqrt {12.25}}}}=$

$=-1+{\sqrt[{3}]{4.5+3.5}}+{\sqrt[{3}]{4.5-3.5}}=-1+{\sqrt[{3}]{8}}+{\sqrt[{3}]{1}}=-1+2+1=2.$

Patikriname:

$z^{3}+3z^{2}-3z-14=2^{3}+3\cdot 2^{2}-3\cdot 2-14=8+12-6-14=0.$

• Pavyzdis. Rasime lygties $z^{3}+2z^{2}-3z-14=0$ realųjį sprendinį. Gauname:

$z_{1}=-{\frac {a_{2}}{3}}+{\sqrt[{3}]{{\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}+{\sqrt {\left({\frac {3a_{1}-a_{2}^{2}}{9}}\right)^{3}+\left({\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}\right)^{2}}}}}+{\sqrt[{3}]{{\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}-{\sqrt {\left({\frac {3a_{1}-a_{2}^{2}}{9}}\right)^{3}+\left({\frac {9a_{2}a_{1}-27a_{0}-2a_{2}^{3}}{54}}\right)^{2}}}}}=$ $=-{\frac {2}{3}}+{\sqrt[{3}]{{\frac {9\cdot 2\cdot (-3)-27\cdot (-14)-2\cdot 2^{3}}{54}}+{\sqrt {\left({\frac {3\cdot (-3)-2^{2}}{9}}\right)^{3}+\left({\frac {-54+378-16}{54}}\right)^{2}}}}}+{\sqrt[{3}]{{\frac {-54+378-16}{54}}-{\sqrt {\left({\frac {-9-4}{9}}\right)^{3}+\left({\frac {-54+378-16}{54}}\right)^{2}}}}}=$ $=-{\frac {2}{3}}+{\sqrt[{3}]{{\frac {308}{54}}+{\sqrt {\left({\frac {-13}{9}}\right)^{3}+\left({\frac {308}{54}}\right)^{2}}}}}+{\sqrt[{3}]{{\frac {308}{54}}-{\sqrt {\left({\frac {-13}{9}}\right)^{3}+\left({\frac {308}{54}}\right)^{2}}}}}=$ $=-{\frac {2}{3}}+{\sqrt[{3}]{{\frac {308}{54}}+{\sqrt {-3.013717421+32.53223594}}}}+{\sqrt[{3}]{{\frac {308}{54}}+{\sqrt {-3.013717421+32.53223594}}}}=$

$=-{\frac {2}{3}}+{\sqrt[{3}]{{\frac {308}{54}}+{\sqrt {29.51851852}}}}+{\sqrt[{3}]{{\frac {308}{54}}-{\sqrt {29.51851852}}}}=-{\frac {2}{3}}+{\sqrt[{3}]{11.13679845}}+{\sqrt[{3}]{0.270608957}}=$

$=-{\frac {2}{3}}+2.233161439+0.646815953=-{\frac {2}{3}}+2.879977392=2.213310725.$

Patikriname:

$2.213310725^{3}+2\cdot 2.213310725^{2}-3\cdot 2.213310725-14=10.84244344+9.797488731-6.639932175-14=20.63993217-20.63993218=0.$

Nuorodos

• Trečio ir ketvirto laipsnio lygties sprendinių radimas
• Kūbinės lygties sprendimas
• Ketvirto laipsnio lygties sprendinių radimas
• Ketvirto laipsnio lygties sprendinių radimas ir pavyzdžiai
• Po-Shen Loh’o naujas kvadratinės lygties sprendimo metodas https://www.poshenloh.com/quadraticdetail/ https://arxiv.org/pdf/1910.06709v2.pdf