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.
keisti
n
{\displaystyle n}
n kompleksinių šaknų (sprendinių).
Bendra forma:
a
⋅
x
=
b
{\displaystyle a\cdot x=b}
Sprendinys:
x
=
b
a
{\displaystyle x={\frac {b}{a}}}
keisti
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}}}
keisti
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.}
keisti
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 lygties
x
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.}
c
=
0
{\displaystyle c=0}
keisti
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}}}
a
=
1
{\displaystyle a=1}
keisti
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}}.}
a yra bet koks
keisti
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}}.}
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}}
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}}}.}
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}}.}
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
keisti
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}}.}
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}\,}
d
=
0
{\displaystyle d=0}
keisti
keisti
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 lygtį
x
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 lygtį
x
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
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.}
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)
