Čia aprašomos paprasčiausios algebrinės lygtys ir jų sprendimai. Aiškinama sunkėjimo tvarka.
Naudosime tokį žymėjimą: x , x 1 , x 2 ir t.t. žymės nežinomuosius, o a , b , c , d ir t.t. – konkrečius duotus skaičius.
Pagrindinė algebros teorema
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n {\displaystyle n} n kompleksinių šaknų (sprendinių).
Tiesinė lygtis
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Bendra forma:
a ⋅ x = b {\displaystyle a\cdot x=b}
Sprendinys:
x = b a {\displaystyle x={\frac {b}{a}}}
Nepilnoji kvadratinė lygtis
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Bendra forma:
a x 2 = b {\displaystyle ax^{2}=b\,}
Sprendimas:
x 2 = b a x 1 , 2 = ± b a {\displaystyle {\begin{aligned}x^{2}&={\frac {b}{a}}\\x_{1,2}&=\pm {\sqrt {\frac {b}{a}}}\end{aligned}}}
Pilnoji kvadratinė lygtis
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Bendra forma:
a x 2 + b x + c = 0 {\displaystyle ax^{2}+bx+c=0\,}
Sprendimas:
randame pagalbini skaičių – diskriminantą D:
D = b 2 − 4 a c {\displaystyle D=b^{2}-4ac\,}
Tada jei D < 0 {\displaystyle D<0}
x 1 , 2 = − b ± D 2 a {\displaystyle x_{1,2}={\frac {-b\pm {\sqrt {D}}}{2a}}}
Pavyzdžiui, reikia surasti kuriuose taškuose kertasi parabolė su Ox ašimi. 3 x 2 + 8 x + 4 = 0 , {\displaystyle 3x^{2}+8x+4=0,} D = b 2 − 4 a c = 8 2 − 4 ⋅ 3 ⋅ 4 = 64 − 48 = 16 , {\displaystyle D=b^{2}-4ac=8^{2}-4\cdot 3\cdot 4=64-48=16,} x 1 , 2 = − b ± D 2 a = − 8 ± 16 2 ⋅ 3 = − 8 ± 4 6 = − 2 3 ; − 2. {\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.} Patikriname:
3 ⋅ ( − 2 3 ) 2 + 8 ⋅ ( − 2 3 ) + 4 = 3 ⋅ 4 9 − 16 3 + 4 = 4 3 − 16 3 + 4 = 4 − 16 3 + 4 = − 12 3 + 4 = − 4 + 4 = 0 ; {\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;} 3 ⋅ ( − 2 ) 2 + 8 ⋅ ( − 2 ) + 4 = 3 ⋅ 4 − 16 + 4 = 12 − 16 + 4 = 0. {\displaystyle 3\cdot (-2)^{2}+8\cdot (-2)+4=3\cdot 4-16+4=12-16+4=0.} Tuo atveju, kai lygties šaknys kompleksiniai skaičiai
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Lygties x 2 + p x + q = 0 {\displaystyle x^{2}+px+q=0}
x 1 = α + i β , {\displaystyle x_{1}=\alpha +i\beta ,} x 2 = α − i β , {\displaystyle x_{2}=\alpha -i\beta ,} kurie yra kompleksiniai skaičia randami taip:
α = − p 2 , {\displaystyle \alpha =-{\frac {p}{2}},} β = q − α 2 = q − p 2 4 ; {\displaystyle \beta ={\sqrt {q-\alpha ^{2}}}={\sqrt {q-{\frac {p^{2}}{4}}}};} β = − D 2 = − ( p 2 − 4 q ) 4 = q − p 2 4 . {\displaystyle \beta ={\frac {\sqrt {-D}}{2}}={\sqrt {\frac {-(p^{2}-4q)}{4}}}={\sqrt {q-{\frac {p^{2}}{4}}}}.}
Lygties a x 2 + b x + c = 0 {\displaystyle ax^{2}+bx+c=0}
x 1 = α + i β , {\displaystyle x_{1}=\alpha +i\beta ,} x 2 = α − i β , {\displaystyle x_{2}=\alpha -i\beta ,} kurie yra kompleksiniai skaičia randami taip:
α = − b 2 a , {\displaystyle \alpha =-{\frac {b}{2a}},} β = − D 2 a = − ( b 2 − 4 a c ) 4 a 2 = c a − b 2 4 a 2 . {\displaystyle \beta ={\frac {\sqrt {-D}}{2a}}={\sqrt {\frac {-(b^{2}-4ac)}{4a^{2}}}}={\sqrt {{\frac {c}{a}}-{\frac {b^{2}}{4a^{2}}}}}.}
Pavyzdis. Rasti sprendinius lygtiesx 2 − 8 x + 25 = 0. {\displaystyle x^{2}-8x+25=0.} Sprendimas. D = b 2 − 4 a c = ( − 8 ) 2 − 4 ⋅ 1 ⋅ 25 = 64 − 100 = − 36 = 36 i 2 , {\displaystyle D=b^{2}-4ac=(-8)^{2}-4\cdot 1\cdot 25=64-100=-36=36i^{2},} x 1 = − b + D 2 = − ( − 8 ) + 36 i 2 2 = 8 + 6 i 2 = 4 + 3 i , {\displaystyle x_{1}={-b+{\sqrt {D}} \over 2}={-(-8)+{\sqrt {36i^{2}}} \over 2}={8+6i \over 2}=4+3i,} x 2 = − b − D 2 = − ( − 8 ) − 36 i 2 2 = 8 − 6 i 2 = 4 − 3 i . {\displaystyle x_{2}={-b-{\sqrt {D}} \over 2}={-(-8)-{\sqrt {36i^{2}}} \over 2}={8-6i \over 2}=4-3i.} Patikriname, kad
( 4 + 3 i ) 2 − 8 ( 4 + 3 i ) + 25 = 0 , {\displaystyle (4+3i)^{2}-8(4+3i)+25=0,} 16 + 2 ⋅ 4 ⋅ 3 i + ( 3 i ) 2 − 32 − 24 i + 25 = 0 , {\displaystyle 16+2\cdot 4\cdot 3i+(3i)^{2}-32-24i+25=0,} 16 + 24 i − 9 − 32 − 24 i + 25 = 0 , {\displaystyle 16+24i-9-32-24i+25=0,} 16 − 41 + 25 = 0 {\displaystyle 16-41+25=0} ir
( 4 − 3 i ) 2 − 8 ( 4 − 3 i ) + 25 = 0 , {\displaystyle (4-3i)^{2}-8(4-3i)+25=0,} 16 − 2 ⋅ 4 ⋅ 3 i + ( − 3 i ) 2 − 32 + 24 i + 25 = 0 , {\displaystyle 16-2\cdot 4\cdot 3i+(-3i)^{2}-32+24i+25=0,} 16 − 24 i − 9 − 32 + 24 i + 25 = 0 , {\displaystyle 16-24i-9-32+24i+25=0,} 16 − 41 + 25 = 0. {\displaystyle 16-41+25=0.} Kvadratinė lygtis, kurios c = 0 {\displaystyle c=0}
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Bendra forma:
a x 2 + b x = 0 {\displaystyle ax^{2}+bx=0\,}
Sprendimas:
iškeliame x prieš skliaustus:
x ( a x + b ) = 0 {\displaystyle x(ax+b)=0\,}
Tada iš sandaugos savybių išplaukia, kad
x = 0 arba a x = − b x = − b a {\displaystyle {\begin{aligned}x=0\qquad \operatorname {arba} \qquad ax&=-b\\x&=-{\frac {b}{a}}\end{aligned}}}
Kvadratinė lygtis, kurios a = 1 {\displaystyle a=1}
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Duota kvadratinė lygtis:
x 2 + b x + c = 0 , {\displaystyle x^{2}+bx+c=0,} kurią perrašome taip:
( x + b 2 ) 2 + ( c − b 2 4 ) = 0. {\displaystyle \left(x+{\frac {b}{2}}\right)^{2}+\left(c-{\frac {b^{2}}{4}}\right)=0.} Čia ( x + b 2 ) 2 = x 2 + b x + b 2 4 . {\displaystyle \left(x+{\frac {b}{2}}\right)^{2}=x^{2}+bx+{\frac {b^{2}}{4}}.}
Todėl:
( x + b 2 ) 2 = − ( c − b 2 4 ) , {\displaystyle \left(x+{\frac {b}{2}}\right)^{2}=-\left(c-{\frac {b^{2}}{4}}\right),} ( x + b 2 ) 2 = b 2 4 − c , {\displaystyle \left(x+{\frac {b}{2}}\right)^{2}={\frac {b^{2}}{4}}-c,} x + b 2 = ± b 2 4 − c , {\displaystyle x+{\frac {b}{2}}=\pm {\sqrt {{\frac {b^{2}}{4}}-c}},} x = − b 2 ± b 2 4 − c , {\displaystyle x=-{\frac {b}{2}}\pm {\sqrt {{\frac {b^{2}}{4}}-c}},} x = − b 2 ± 1 4 ⋅ ( b 2 − 4 c ) , {\displaystyle x=-{\frac {b}{2}}\pm {\sqrt {{\frac {1}{4}}\cdot (b^{2}-4c)}},} x = − b 2 ± 1 2 ⋅ b 2 − 4 c , {\displaystyle x=-{\frac {b}{2}}\pm {\frac {1}{2}}\cdot {\sqrt {b^{2}-4c}},} x = − b ± b 2 − 4 c 2 . {\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4c}}}{2}}.} x 1 = − b + b 2 − 4 c 2 ; x 2 = − b − b 2 − 4 c 2 . {\displaystyle 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
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Duota kvadratinė lygtis:
a x 2 + b x + c = 0 , {\displaystyle ax^{2}+bx+c=0,} x 2 + b a ⋅ x + c a = 0 , {\displaystyle x^{2}+{\frac {b}{a}}\cdot x+{\frac {c}{a}}=0,} kurią perrašome taip:
( x + b 2 ⋅ a ) 2 + ( c a − b 2 4 ⋅ a 2 ) = 0. {\displaystyle \left(x+{\frac {b}{2\cdot a}}\right)^{2}+\left({\frac {c}{a}}-{\frac {b^{2}}{4\cdot a^{2}}}\right)=0.} Čia ( x + b 2 ⋅ a ) 2 = x 2 + b a ⋅ x + b 2 4 ⋅ a 2 . {\displaystyle \left(x+{\frac {b}{2\cdot a}}\right)^{2}=x^{2}+{\frac {b}{a}}\cdot x+{\frac {b^{2}}{4\cdot a^{2}}}.}
Todėl:
( x + b 2 a ) 2 = − ( c a − b 2 4 a 2 ) , {\displaystyle \left(x+{\frac {b}{2a}}\right)^{2}=-\left({\frac {c}{a}}-{\frac {b^{2}}{4a^{2}}}\right),} ( x + b 2 a ) 2 = b 2 4 a 2 − c a , {\displaystyle \left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}},} x + b 2 a = ± b 2 4 a 2 − c a , {\displaystyle x+{\frac {b}{2a}}=\pm {\sqrt {{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}},} x = − b 2 a ± b 2 4 a 2 − c a , {\displaystyle x=-{\frac {b}{2a}}\pm {\sqrt {{\frac {b^{2}}{4a^{2}}}-{\frac {c}{a}}}},} x = − b 2 a ± 1 4 a 2 ⋅ ( b 2 − 4 a c ) , {\displaystyle x=-{\frac {b}{2a}}\pm {\sqrt {{\frac {1}{4a^{2}}}\cdot (b^{2}-4ac)}},} x = − b 2 a ± 1 2 a ⋅ b 2 − 4 a c , {\displaystyle x=-{\frac {b}{2a}}\pm {\frac {1}{2a}}\cdot {\sqrt {b^{2}-4ac}},} x = − b ± b 2 − 4 a c 2 a . {\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}.} x 1 = − b + b 2 − 4 a c 2 a ; x 2 = − b − b 2 − 4 a c 2 a . {\displaystyle x_{1}={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}};\quad x_{2}={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}.} Bikvadratinė lygtis
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Bendra forma:
a x 4 + b x 2 + c = 0 {\displaystyle ax^{4}+bx^{2}+c=0\,}
Sprendimas:
pažymime x 2 = y {\displaystyle x^{2}=y\,} x 4 = y 2 {\displaystyle x^{4}=y^{2}\,}
a y 2 + b y + c = 0 {\displaystyle ay^{2}+by+c=0\,}
o tai pilnoji kvadratinė lygtis, kuri jau išspręsta anksčiau. Jos sprendiniai yra y 1 {\displaystyle y_{1}} y 2 {\displaystyle y_{2}}
Grįžtame prie pažymėjimo:
y 1 = x 2 ir y 2 = x 2 {\displaystyle 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 {\displaystyle x_{1},x_{2},x_{3},x_{4}}
Vijeto teorema
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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 , {\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,} Tai
s 1 = x 1 + x 2 + x 3 + x 4 + . . . , {\displaystyle 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 + . . . , {\displaystyle 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 + . . . {\displaystyle 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 ⋅ a n − i a n . {\displaystyle s_{i}=(-1)^{i}\cdot {\frac {a_{n-i}}{a_{n}}}.} Kvadratinė lygtis
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Jei yra lygtis
a x 2 + b x + c , {\displaystyle ax^{2}+bx+c,} tai lygties sprendiniai:
x 1 + x 2 = − b a , {\displaystyle x_{1}+x_{2}=-{\frac {b}{a}},} x 1 ⋅ x 2 = c a . {\displaystyle x_{1}\cdot x_{2}={\frac {c}{a}}.} Kubinė lygtis
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Jei yra lygtis
a x 3 + b x 2 + c x + d , {\displaystyle ax^{3}+bx^{2}+cx+d,} tai lygties sprendiniai:
x 1 + x 2 + x 3 = − b a , {\displaystyle x_{1}+x_{2}+x_{3}=-{\frac {b}{a}},} x 1 ⋅ x 2 + x 1 ⋅ x 3 + x 2 ⋅ x 3 = c a , {\displaystyle x_{1}\cdot x_{2}+x_{1}\cdot x_{3}+x_{2}\cdot x_{3}={\frac {c}{a}},} x 1 ⋅ x 2 ⋅ x 3 = − d a . {\displaystyle x_{1}\cdot x_{2}\cdot x_{3}=-{\frac {d}{a}}.} Ketvirto laispnio lygtis
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Jei yra lygtis
a x 4 + b x 3 + c x 2 + d x + e , {\displaystyle ax^{4}+bx^{3}+cx^{2}+dx+e,} tai lygties sprendiniai:
x 1 + x 2 + x 3 + x 4 = − b a , {\displaystyle 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 = c a , {\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}},} x 1 x 2 x 3 + x 2 x 3 x 4 + x 1 x 2 x 4 + x 1 x 3 x 4 = − d a , {\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}},} x 1 ⋅ x 2 ⋅ x 3 ⋅ x 4 = e a . {\displaystyle x_{1}\cdot x_{2}\cdot x_{3}\cdot x_{4}={\frac {e}{a}}.} Pilnoji kubinė lygtis
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Bendra forma:
a x 3 + b x 2 + c x + d = 0 {\displaystyle ax^{3}+bx^{2}+cx+d=0\,}
Sprendimas:
Lygtį padalijame iš a ir keitiniu x = y − b 3 {\displaystyle x=y-{\frac {b}{3}}}
x 3 + p x + q = 0 {\displaystyle x^{3}+px+q=0\,}
Randame pagalbinį skaičių – diskriminantą:
D = ( q 2 ) 2 + ( p 3 ) 3 {\displaystyle 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 = − p 2 + D 3 + − p 2 − D 3 {\displaystyle 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 a x 3 + b x 2 + c x + d = 0 {\displaystyle ax^{3}+bx^{2}+cx+d=0\,} x − x 1 {\displaystyle x-x_{1}\,}
Kubinė lygtis, kurios d = 0 {\displaystyle d=0}
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Kubinės lygties sprendimas Kordano metodu
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Yra kubinė lygtis:
y 3 + a y 2 + b y + c = 0. {\displaystyle y^{3}+ay^{2}+by+c=0.} Pakeičiame y = x − a 3 , {\displaystyle y=x-{\frac {a}{3}},}
( x − a 3 ) 3 + a ( x − a 3 ) 2 + b ( x − a 3 ) + c = 0 , {\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,} ( x 3 − 3 ⋅ x 2 ⋅ a 3 + 3 ⋅ x ⋅ ( a 3 ) 2 − ( a 3 ) 3 ) + a ( x 2 − 2 ⋅ x ⋅ a 3 + ( a 3 ) 2 ) + b ( x − a 3 ) + c = 0 , {\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,} ( x 3 − a x 2 + 3 ⋅ x ⋅ a 2 9 − a 3 27 ) + a ( x 2 − 2 a x 3 + a 2 9 ) + b x − a b 3 + c = 0 , {\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,} ( x 3 − a x 2 + a 2 x 3 − a 3 27 ) + a x 2 − 2 a 2 x 3 + a 3 9 + b x − a b 3 + c = 0 , {\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,} x 3 − a 3 27 − a 2 x 3 + a 3 9 + b x − a b 3 + c = 0 , {\displaystyle x^{3}-{\frac {a^{3}}{27}}-{\frac {a^{2}x}{3}}+{\frac {a^{3}}{9}}+bx-{\frac {ab}{3}}+c=0,} x 3 − a 2 x 3 + b x + a 3 9 − a 3 27 − a b 3 + c = 0 , {\displaystyle x^{3}-{\frac {a^{2}x}{3}}+bx+{\frac {a^{3}}{9}}-{\frac {a^{3}}{27}}-{\frac {ab}{3}}+c=0,} x 3 + ( b − a 2 3 ) x + 3 a 3 27 − a 3 27 − a b 3 + c = 0 , {\displaystyle x^{3}+(b-{\frac {a^{2}}{3}})x+{\frac {3a^{3}}{27}}-{\frac {a^{3}}{27}}-{\frac {ab}{3}}+c=0,} x 3 + ( b − a 2 3 ) x + 2 a 3 27 − a b 3 + c = 0. {\displaystyle x^{3}+(b-{\frac {a^{2}}{3}})x+{\frac {2a^{3}}{27}}-{\frac {ab}{3}}+c=0.} Pažymime p = b − a 2 3 , q = 2 a 3 27 − a b 3 + c {\displaystyle p=b-{\frac {a^{2}}{3}},\quad q={\frac {2a^{3}}{27}}-{\frac {ab}{3}}+c}
x 3 + p x + q = 0. {\displaystyle x^{3}+px+q=0.} Tegu x 0 {\displaystyle x_{0}} x 3 + p x + q = 0 {\displaystyle x^{3}+px+q=0} x 3 + p x + q = 0 {\displaystyle x^{3}+px+q=0} u ir tikimes, kad polinomas
f ( u ) = u 2 − x 0 u − p 3 {\displaystyle f(u)=u^{2}-x_{0}u-{\frac {p}{3}}} padės surasti x 0 {\displaystyle x_{0}} u 2 − x 0 u − p 3 = 0 {\displaystyle u^{2}-x_{0}u-{\frac {p}{3}}=0}
Polinomo koeficientai - kompleksiniai skaičiai, ir todėl jis turi dvi kompleksines šaknis α {\displaystyle \alpha } β {\displaystyle \beta }
α + β = x 0 , {\displaystyle \alpha +\beta =x_{0},} α ⋅ β = − p 3 . {\displaystyle \alpha \cdot \beta =-{\frac {p}{3}}.} Įstatę α + β = x 0 {\displaystyle \alpha +\beta =x_{0}} x 3 + p x + q = 0 {\displaystyle x^{3}+px+q=0}
( α + β ) 3 + p ( α + β ) + q = 0 , {\displaystyle (\alpha +\beta )^{3}+p(\alpha +\beta )+q=0,} ( α 3 + 3 α 2 β + 3 α β 2 + β 3 ) + p ( α + β ) + q = 0 , {\displaystyle (\alpha ^{3}+3\alpha ^{2}\beta +3\alpha \beta ^{2}+\beta ^{3})+p(\alpha +\beta )+q=0,} α 3 + β 3 + 3 α β ( α + β ) + p ( α + β ) + q = 0 , {\displaystyle \alpha ^{3}+\beta ^{3}+3\alpha \beta (\alpha +\beta )+p(\alpha +\beta )+q=0,} α 3 + β 3 + ( 3 α β + p ) ( α + β ) + q = 0 , {\displaystyle \alpha ^{3}+\beta ^{3}+(3\alpha \beta +p)(\alpha +\beta )+q=0,} Iš lygties α ⋅ β = − p 3 {\displaystyle \alpha \cdot \beta =-{\frac {p}{3}}}
α ⋅ β + p 3 = 0 , {\displaystyle \alpha \cdot \beta +{\frac {p}{3}}=0,} 3 ⋅ α ⋅ β + p = 0. {\displaystyle 3\cdot \alpha \cdot \beta +p=0.} Todėl gauname:
α 3 + β 3 + 0 ⋅ ( α + β ) + q = 0 , {\displaystyle \alpha ^{3}+\beta ^{3}+0\cdot (\alpha +\beta )+q=0,} α 3 + β 3 + q = 0 , {\displaystyle \alpha ^{3}+\beta ^{3}+q=0,} α 3 + β 3 = − q . {\displaystyle \alpha ^{3}+\beta ^{3}=-q.} Dabar turime nauja gabaliuką iš Vijeto teoremos, tai yra lygtis α 3 + β 3 = − q . {\displaystyle \alpha ^{3}+\beta ^{3}=-q.} q priklauso lygčiai x 3 + p x + q = 0 {\displaystyle x^{3}+px+q=0} b ir c Vijeto teoremoje žymimi kvadratinėje lygtyje), taigi pakeliame kubu lygtyje α ⋅ β = − p 3 {\displaystyle \alpha \cdot \beta =-{\frac {p}{3}}} α {\displaystyle \alpha } β {\displaystyle \beta }
α 3 ⋅ β 3 = − p 3 27 . {\displaystyle \alpha ^{3}\cdot \beta ^{3}=-{\frac {p^{3}}{27}}.} Šios dalys g = q , {\displaystyle g=q,} s = − p 3 27 {\displaystyle s=-{\frac {p^{3}}{27}}} g ir s koeficientai kvadratinės lygties z 2 + g z + s = 0 , {\displaystyle z^{2}+gz+s=0,} z 1 = α 3 {\displaystyle z_{1}=\alpha ^{3}} z 2 = β 3 {\displaystyle z_{2}=\beta ^{3}}
z 2 + q z − p 3 27 = 0. {\displaystyle z^{2}+qz-{\frac {p^{3}}{27}}=0.} Randame diskriminantą:
D = g 2 − 4 s = q 2 − 4 ⋅ ( − p 3 27 ) = q 2 + 4 p 3 27 . {\displaystyle D=g^{2}-4s=q^{2}-4\cdot (-{\frac {p^{3}}{27}})=q^{2}+{\frac {4p^{3}}{27}}.} Randame sprendinius:
z 1 = − g + D 2 = − q + q 2 + 4 p 3 27 2 = 1 2 ⋅ ( − q + q 2 + 4 p 3 27 ) , α = z 1 3 = 1 2 ⋅ ( − q + q 2 + 4 p 3 27 ) 3 ; {\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}}}})}};} z 2 = − g − D 2 = − q − q 2 + 4 p 3 27 2 = 1 2 ⋅ ( − q − q 2 + 4 p 3 27 ) , β = z 2 3 = 1 2 ⋅ ( − q − q 2 + 4 p 3 27 ) 3 . {\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}}}})}}.} Toliau α {\displaystyle \alpha } β {\displaystyle \beta } α + β = x 0 , {\displaystyle \alpha +\beta =x_{0},} x 3 + p x + q = 0 {\displaystyle x^{3}+px+q=0} x 0 {\displaystyle x_{0}}
x 0 = α + β = 1 2 ⋅ ( − q + q 2 + 4 p 3 27 ) 3 + 1 2 ⋅ ( − q − q 2 + 4 p 3 27 ) 3 . {\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}}}})}}.} x 0 = α + β = − q 2 + q 2 4 + p 3 27 3 + − q 2 − q 2 4 + p 3 27 3 . {\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}}}}}}.} Kalbant apie kompleksinius sprendinius, negalima imti tokių sprendinių, kurie netenkina salygos α ⋅ β = − p 3 . {\displaystyle \alpha \cdot \beta =-{\frac {p}{3}}.}
α 2 β 3 = α 1 ϵ ⋅ β 1 ϵ 2 = α 1 β 1 ϵ 3 = α 1 β 1 = − p 3 . {\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}}.} Na, o visi sprendiniai yra šie:
x 1 = α 1 + β 1 , {\displaystyle x_{1}=\alpha _{1}+\beta _{1},} x 2 = α 2 + β 3 = α 1 ϵ + β 1 ϵ 2 = α 1 ( − 1 2 + i 3 2 ) + β 1 ( − 1 2 − i 3 2 ) = − α 1 + β 1 2 + i 3 α 1 − β 1 2 , {\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}},} x 3 = α 3 + β 2 = α 1 ϵ 2 + β 1 ϵ = α 1 ( − 1 2 − i 3 2 ) + β 1 ( − 1 2 + i 3 2 ) = − α 1 + β 1 2 − i 3 α 1 − β 1 2 , {\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}},} Jei sudėti ant apskritimo, kurio spindulys r=1, taškus ϵ {\displaystyle \epsilon } ϵ 2 {\displaystyle \epsilon ^{2}} ϵ = 120 {\displaystyle \epsilon =120} ϵ 2 = 240 {\displaystyle \epsilon ^{2}=240} ϵ 3 = 1 + i ⋅ 0 = 1 {\displaystyle \epsilon ^{3}=1+i\cdot 0=1} ϵ 3 = 360 {\displaystyle \epsilon ^{3}=360}
ϵ 2 = ( − 1 2 + i 3 2 ) ⋅ ( − 1 2 + i 3 2 ) = 1 4 − i 3 4 − i 3 4 + i ⋅ i ⋅ 3 4 = 1 4 − i 2 3 4 − 3 4 = − 2 4 − i 3 2 = − 1 2 − i 3 2 . {\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}}.} ϵ 3 = ϵ 2 ⋅ ϵ = ( − 1 2 − i 3 2 ) ⋅ ( − 1 2 + i 3 2 ) = 1 4 − i 3 4 + i 3 4 − i ⋅ i ⋅ 3 4 = 1 4 − ( − 1 ) ⋅ 3 4 = 1 4 + 3 4 = 1. {\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.}
Pavyzdis . Išspresti lygtį y 3 + 3 y 2 − 3 y − 14 = 0. {\displaystyle y^{3}+3y^{2}-3y-14=0.} Keitinys y = x − a 3 = x − 3 3 = x − 1 {\displaystyle y=x-{\frac {a}{3}}=x-{\frac {3}{3}}=x-1} a yra koeficientas esantis prie y 2 {\displaystyle y^{2}}
( x − 1 ) 3 + 3 ( x − 1 ) 2 − 3 ( x − 1 ) − 14 = 0 , {\displaystyle (x-1)^{3}+3(x-1)^{2}-3(x-1)-14=0,} x 3 − 3 x 2 + 3 x − 1 2 + 3 ( x 2 − 2 x + 1 2 ) − 3 x + 1 − 14 = 0 , {\displaystyle x^{3}-3x^{2}+3x-1^{2}+3(x^{2}-2x+1^{2})-3x+1-14=0,} x 3 − 3 x 2 + 3 x − 1 + 3 x 2 − 6 x + 3 − 3 x + 1 − 14 = 0 , {\displaystyle x^{3}-3x^{2}+3x-1+3x^{2}-6x+3-3x+1-14=0,} x 3 − 6 x + 3 − 14 = 0 , {\displaystyle x^{3}-6x+3-14=0,} x 3 − 6 x − 9 = 0. {\displaystyle x^{3}-6x-9=0.} Čia p = − 6 {\displaystyle p=-6} q = − 9 {\displaystyle q=-9}
q 2 4 + p 3 27 = ( − 9 ) 2 4 + ( − 6 ) 3 27 = 81 4 + − 216 27 = 81 4 − 8 = 81 − 32 4 = 49 4 > 0 , {\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,} t. y. lygtis x 3 − 6 x − 9 = 0 {\displaystyle x^{3}-6x-9=0}
Pagal formulę:
α = − q 2 + q 2 4 + p 3 27 3 = − − 9 2 + ( − 9 ) 2 4 + ( − 6 ) 3 27 3 = 9 2 + 7 2 3 = 16 2 3 = 8 3 = 2 , {\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,} β = − q 2 − q 2 4 + p 3 27 3 = − − 9 2 − ( − 9 ) 2 4 + ( − 6 ) 3 27 3 = 9 2 − 7 2 3 = 2 2 3 = 1 3 = 1. {\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.} Todėl α 1 = 2 , {\displaystyle \alpha _{1}=2,} β 1 = 1 , {\displaystyle \beta _{1}=1,} x 1 = α 1 + β 1 = 2 + 1 = 3 {\displaystyle x_{1}=\alpha _{1}+\beta _{1}=2+1=3}
x 2 = − α 1 + β 1 2 + i 3 ⋅ α 1 − β 1 2 = − 2 + 1 2 + i 3 ⋅ 2 − 1 2 = − 3 2 + i 3 2 , {\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}},} x 3 = − α 1 + β 1 2 − i 3 ⋅ α 1 − β 1 2 = − 2 + 1 2 − i 3 ⋅ 2 − 1 2 = − 3 2 − i 3 2 . {\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}}.} Iš čia gauname, kad sprendiniai užduotos lygties yra skaičiai:
y 1 = x − a 3 = x 0 − 1 = 3 − 1 = 2 , {\displaystyle y_{1}=x-{\frac {a}{3}}=x_{0}-1=3-1=2,} y 2 = − 5 2 + i 3 2 , {\displaystyle y_{2}=-{\frac {5}{2}}+i{\frac {\sqrt {3}}{2}},} y 3 = − 5 2 − i 3 2 . {\displaystyle y_{3}=-{\frac {5}{2}}-i{\frac {\sqrt {3}}{2}}.} Patikriname, kai y 1 = 2 {\displaystyle y_{1}=2} y 3 + 3 y 2 − 3 y − 14 = 2 3 + 3 ⋅ 2 2 − 3 ⋅ 2 − 14 = 8 + 3 ⋅ 4 − 6 − 14 = 8 + 12 − 6 − 14 = 0. {\displaystyle 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 = α + β = 2 + 1 = 3 {\displaystyle x_{0}=\alpha +\beta =2+1=3}
x 3 − 6 x − 9 = 3 3 − 6 ⋅ 3 − 9 = 27 − 18 − 9 = 0. {\displaystyle x^{3}-6x-9=3^{3}-6\cdot 3-9=27-18-9=0.}
Pavyzdis . Išspręsti lygtįx 3 − 12 x + 16 = 0. {\displaystyle x^{3}-12x+16=0.}
Čia p = − 12 {\displaystyle p=-12} q = 16 {\displaystyle q=16}
q 2 4 + p 3 27 = 16 2 4 + ( − 12 ) 3 27 = 256 4 + − 1728 27 = 64 − 64 = 0. {\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.} α = − q 2 + q 2 4 + p 3 27 3 = − 16 2 + 0 3 = − 8 + 0 3 = − 8 3 = − 2 , {\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,} β = − q 2 − q 2 4 + p 3 27 3 = − 16 2 − 0 3 = − 8 3 = − 2. {\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.} Iš čia seka: α = − 8 3 , {\displaystyle \alpha ={\sqrt[{3}]{-8}},} α 1 = β 1 = − 2. {\displaystyle \alpha _{1}=\beta _{1}=-2.}
x 1 = α 1 + β 1 = − 2 − 2 = − 4 , {\displaystyle x_{1}=\alpha _{1}+\beta _{1}=-2-2=-4,} x 2 = − α 1 + β 1 2 + i 3 ⋅ α 1 − β 1 2 = − − 2 − 2 2 + i 3 ⋅ − 2 − ( − 2 ) 2 = − − 4 2 + i ⋅ 0 ⋅ 3 2 = 2 , {\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,} x 3 = − α 1 + β 1 2 − i 3 ⋅ α 1 − β 1 2 = − − 2 − 2 2 − i 3 ⋅ 2 − ( − 2 ) 2 = − − 4 2 = 2. {\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.} Patikriname įstatę x 1 = − 4 {\displaystyle x_{1}=-4}
x 3 − 12 x + 16 = ( − 4 ) 3 − 12 ⋅ ( − 4 ) + 16 = − 64 + 48 + 16 = 0. {\displaystyle x^{3}-12x+16=(-4)^{3}-12\cdot (-4)+16=-64+48+16=0.} Patikriname įstatę x 2 = 2 {\displaystyle x_{2}=2}
x 3 − 12 x + 16 = 2 3 − 12 ⋅ 2 + 16 = 8 − 24 + 16 = 0. {\displaystyle 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 = − 2 p ; {\displaystyle x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=-2p;} x 1 3 + x 2 3 + x 3 3 = − 3 q ; {\displaystyle x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=-3q;} x 1 4 + x 2 4 + x 3 4 = 2 p 2 ; {\displaystyle x_{1}^{4}+x_{2}^{4}+x_{3}^{4}=2p^{2};} x 1 5 + x 2 5 + x 3 5 = 5 p q ; {\displaystyle x_{1}^{5}+x_{2}^{5}+x_{3}^{5}=5pq;} čia p = − 12 {\displaystyle p=-12} q = 16 {\displaystyle q=16} x 1 = − 4 {\displaystyle x_{1}=-4} x 2 = 2 {\displaystyle x_{2}=2} x 3 = 2 {\displaystyle x_{3}=2}
( − 4 ) 2 + 2 2 + 2 2 = 16 + 4 + 4 = 24 = − 2 ⋅ 12 ; {\displaystyle (-4)^{2}+2^{2}+2^{2}=16+4+4=24=-2\cdot 12;} ( − 4 ) 3 + 2 3 + 2 3 = − 64 + 8 + 8 = − 48 = − 3 ⋅ 16 ; {\displaystyle (-4)^{3}+2^{3}+2^{3}=-64+8+8=-48=-3\cdot 16;} ( − 4 ) 4 + 2 4 + 2 4 = 256 + 16 + 16 = 288 = 2 ⋅ ( − 12 ) 2 = 2 ⋅ 144 ; {\displaystyle (-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 ⋅ ( − 12 ) ⋅ 16. {\displaystyle (-4)^{5}+2^{5}+2^{5}=-1024+32+32=-960=5\cdot (-12)\cdot 16.} Kai x 2 = x 3 = − x 1 2 , {\displaystyle x_{2}=x_{3}=-{\frac {x_{1}}{2}},}
x 1 2 + x 2 2 + x 3 2 = x 1 2 + ( − x 1 2 ) 2 + ( − x 1 2 ) 2 = x 1 2 + x 1 2 2 = − 2 p ; {\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;} x 1 x 2 + x 1 x 3 + x 2 x 3 = x 1 ⋅ ( − x 1 2 ) + x 1 ⋅ ( − x 1 2 ) + ( − x 1 2 ) ⋅ ( − x 1 2 ) = − x 1 2 + x 1 2 4 = p , {\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,} − 2 ( − x 1 2 + x 1 2 4 ) = 2 x 1 2 − x 1 2 2 = − 2 p . {\displaystyle -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 = − x 1 2 , {\displaystyle x_{2}=x_{3}=-{\frac {x_{1}}{2}},}
x 1 3 + x 2 3 + x 3 3 = x 1 3 + ( − x 1 2 ) 3 + ( − x 1 2 ) 3 = x 1 3 − 2 ⋅ x 1 3 8 = x 1 3 − x 1 3 4 = 4 x 1 3 − x 1 3 4 = 3 x 1 3 4 = − 3 q ; {\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;} x 1 x 2 x 3 = x 1 ⋅ ( − x 1 2 ) ⋅ ( − x 1 2 ) = x 1 3 4 = − q , {\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,} 3 ⋅ x 1 3 4 = 3 ⋅ ( − q ) . {\displaystyle 3\cdot {\frac {x_{1}^{3}}{4}}=3\cdot (-q).}
Pavyzdis . Išspręsti lygtįx 3 − 19 x + 30 = 0. {\displaystyle x^{3}-19x+30=0.}
Čia p = − 19 {\displaystyle p=-19} q = 30 {\displaystyle q=30}
q 2 4 + p 3 27 = 30 2 4 + ( − 19 ) 3 27 = 900 4 + − 6859 27 = 225 − 254.037037 = − 29.037037037 < 0. {\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.} Tokiu atveju, jeigu pasilikti srityje realiųjų skaičių, Kardano formulė šiai lygybei netinka, nors šios lygties sprendiniai ir yra 2, 3 ir − 5 {\displaystyle -5}
Kaip galima išspręsti šitą lygtį žiūrėti čia https://lt.wikibooks.org/wiki/Kompleksiniai_skaičiai#Šaknies_traukimo_operacijos_trigonometrinėje_formoje Kubinė lygtis
keisti
Kanoninė forma:
a x 3 + b x 2 + c x + d = 0. {\displaystyle ax^{3}+bx^{2}+cx+d=0.} Padaliname iš a ir įvedame vietoje x naują kintamjį y 3 + 3 p y + 2 q = 0 , {\displaystyle y^{3}+3py+2q=0,}
kur 2 q = 2 b 3 27 a 3 − b c 3 a 2 + d a {\displaystyle 2q={\frac {2b^{3}}{27a^{3}}}-{\frac {bc}{3a^{2}}}+{\frac {d}{a}}} 3 p = 3 a c − b 2 3 a 2 . {\displaystyle 3p={\frac {3ac-b^{2}}{3a^{2}}}.}
Kardano sprendiniai y 1 = u + v , y 2 = ϵ 1 u + ϵ 2 v , y 3 = ϵ 2 u + ϵ 1 v , {\displaystyle y_{1}=u+v,\quad y_{2}=\epsilon _{1}u+\epsilon _{2}v,\quad y_{3}=\epsilon _{2}u+\epsilon _{1}v,} kur u = − q + q 2 + p 3 3 , v = − q − q 2 + p 3 3 , {\displaystyle u={\sqrt[{3}]{-q+{\sqrt {q^{2}+p^{3}}}}},\quad v={\sqrt[{3}]{-q-{\sqrt {q^{2}+p^{3}}}}},}
o ϵ 1 {\displaystyle \epsilon _{1}} ϵ 2 {\displaystyle \epsilon _{2}} x 2 + x + 1 = 0 , {\displaystyle x^{2}+x+1=0,} ϵ 1 , ϵ 2 = − 1 2 ± i 3 2 . {\displaystyle \epsilon _{1},\epsilon _{2}=-{\frac {1}{2}}\pm i{\frac {\sqrt {3}}{2}}.}
Tuo atveju, kai D = q 2 + p 3 < 0 {\displaystyle D=q^{2}+p^{3}<0} Pavyzdis . y 3 + 6 y + 2 = 0. {\displaystyle y^{3}+6y+2=0.} q 2 + p 3 = 1 2 + 2 3 = 1 + 8 = 9 ; {\displaystyle q^{2}+p^{3}=1^{2}+2^{3}=1+8=9;} u = − q + q 2 + p 3 3 = − 1 + 9 3 = − 1 + 3 3 = 2 3 = 1.25992105 , {\displaystyle u={\sqrt[{3}]{-q+{\sqrt {q^{2}+p^{3}}}}}={\sqrt[{3}]{-1+{\sqrt {9}}}}={\sqrt[{3}]{-1+3}}={\sqrt[{3}]{2}}=1.25992105,} v = − q − q 2 + p 3 3 = − 1 − 9 3 = − 1 − 3 3 = − 4 3 = − 1.587401052 , {\displaystyle 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 = 2 3 + − 4 3 = 1.25992105 − 1.587401052 = − 0.327480002 ; {\displaystyle y_{1}=u+v={\sqrt[{3}]{2}}+{\sqrt[{3}]{-4}}=1.25992105-1.587401052=-0.327480002;}
Kompleksiniai sprendiniai: y 2 , 3 = − 1 2 ( u + v ) ± i ⋅ 3 2 ⋅ ( u − v ) = 0.163740001 ± i ⋅ 3 2 ⋅ 2.847322102 = 0.163740001 ± i ⋅ 2.465853273. {\displaystyle 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 + 6 y + 2 = ( − 0.327480002 ) 3 + 6 ( − 0.327480002 ) + 2 = − 0.035119987 − 1.964880012 + 2 = − 2 + 2 = 0. {\displaystyle 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. {\displaystyle y^{3}+y+1=0.} q 2 + p 3 = ( 1 2 ) 2 + ( 1 3 ) 3 = 1 4 + 1 27 = 0.287037037 ; {\displaystyle q^{2}+p^{3}=({\frac {1}{2}})^{2}+({\frac {1}{3}})^{3}={\frac {1}{4}}+{\frac {1}{27}}=0.287037037;} u = − q + q 2 + p 3 3 = − 1 2 + 0.287037037 3 = − 1 2 + 0.535758375 3 = 0.035758375 3 = 0.329452338 , {\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,} v = − q − q 2 + p 3 3 = − 1 2 − 0.287037037 3 = − 1 2 − 0.535758375 3 = − 1.035758375 3 = − 1.011780142 , {\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,} Tikrasis sprendinis yra y 1 = u + v = 0.329452338 − 1.011780142 = − 0.682327803 ; {\displaystyle 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. {\displaystyle y^{3}+y+1=(-0.682327803)^{3}-0.682327803+1=0.317672196-0.682327803+1=-0.999999999+1=0.}
Pavyzdis . y 3 + 7 y + 18 = 0. {\displaystyle y^{3}+7y+18=0.} q 2 + p 3 = 9 2 + ( 2.333333333 ) 3 = 81 + 12.7037037 = 93.7037037 ; {\displaystyle q^{2}+p^{3}=9^{2}+(2.333333333)^{3}=81+12.7037037=93.7037037;} u = − q + q 2 + p 3 3 = − 9 + 93.7037037 3 = − 9 + 9.68006734 3 = 0.680067339 3 = 0.879394961 , {\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,} v = − q − q 2 + p 3 3 = − 9 − 93.7037037 3 = − 9 − 9.68006734 3 = − 18.68006734 3 = − 2.65333944 , {\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,} Tikrasis sprendinis yra y 1 = u + v = 0.879394961 − 2.65333944 = − 1.773944479 ; {\displaystyle y_{1}=u+v=0.879394961-2.65333944=-1.773944479;}
Patikriname:
y 3 + 7 y + 18 = ( − 1.773944479 ) 3 + 7 ( − 1.773944479 ) + 18 = − 5.582388651 − 12.41761135 + 18 = − 18 + 18 = 0. {\displaystyle y^{3}+7y+18=(-1.773944479)^{3}+7(-1.773944479)+18=-5.582388651-12.41761135+18=-18+18=0.} Kūbinės lygties sprendiniai
keisti
![{\displaystyle x_{1}={\sqrt[{3}]{-{\frac {p}{2}}+{\sqrt {D}}}}+{\sqrt[{3}]{-{\frac {p}{2}}-{\sqrt {D}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b901c0922dcf47d0f30a714dde4acdf03cbbb243)
![{\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)
![{\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)
![{\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)
![{\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)
![{\displaystyle \alpha ={\sqrt[{3}]{-8}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c31dae31059bae5c442359413a5a0af28681cddd)
![{\displaystyle u={\sqrt[{3}]{-q+{\sqrt {q^{2}+p^{3}}}}},\quad v={\sqrt[{3}]{-q-{\sqrt {q^{2}+p^{3}}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f7518479812f626e321961c8cb7d38d058d118c7)
![{\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)
![{\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)
![{\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)
![{\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)
![{\displaystyle 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}}}}};}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ac3f432aa74ce8ee38ea6e54f4ea93f59559bf0)
![{\displaystyle 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)+}](https://wikimedia.org/api/rest_v1/media/math/render/svg/52a030deb8f5b4248860cc25573b7b87d9b6729c)
![{\displaystyle +{\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);}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e9e7a1c2dc851fe8dbf7e1529e7ba8e915b70882)
![{\displaystyle 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)-}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6fa5364c87fd92481051fd57aab497ea6394c145)
![{\displaystyle -{\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).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/04755363e1adeebeb7a3ccd0a869607da97face5)
![{\displaystyle 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}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5fb5c67d73c08a4b2123e80c457127549f78ff99)
![{\displaystyle =-{\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}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b0035185cace9b6ea9bfaebbff1b8d7f7d503ca)
![{\displaystyle =-{\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}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/072648e05dd0da520e30e06aec467867b5ed8505)
![{\displaystyle =-{\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}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/972373d8f9f9b31856ebea9dd4dfa00fb143d3ac)
![{\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)
![{\displaystyle =-{\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}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/60d607b1eeba2562709da157c754ee1979669519)
![{\displaystyle =-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}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f3770459e383c59a1f7612ab68d457ba2634327e)
![{\displaystyle =-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}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d8bfbac874fbbcb1a9cc1938a648dda4f067c52)
![{\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)
![{\displaystyle =-{\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}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6254a77bb43e28685982ea8ecb8efb1c3b2c0af8)
![{\displaystyle =-{\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}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4b3252a337f88db5f6b7f5ad32477a3d7526f79a)
![{\displaystyle =-{\frac {2}{3}}+{\sqrt[{3}]{{\frac {308}{54}}+{\sqrt {-3.013717421+32.53223594}}}}+{\sqrt[{3}]{{\frac {308}{54}}+{\sqrt {-3.013717421+32.53223594}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1317868734cdf2363c90c9a82ac498f907018129)
![{\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)
