IR LYGTAIS DAUG KO NEREIKIA DARYTI KAS RAŠOMA ŽEMIAU
Na, o visi realieji sprendiniai yra šie: Tarkime, kai q=1 ir p=1. Tada:
x
0
=
α
+
β
=
1.011780142
−
0.329452338
=
0.682327803.
{\displaystyle x_{0}=\alpha +\beta =1.011780142-0.329452338=0.682327803.}
Įstatome dabar į lygtį
x
3
+
p
x
+
q
=
0
{\displaystyle x^{3}+px+q=0}
x
0
=
0.682327803
{\displaystyle x_{0}=0.682327803}
0.682327803
3
+
1
⋅
0.682327803
+
1
=
0.317672196
+
0.682327803
+
1
=
0.999999998
+
1
=
2.
{\displaystyle 0.682327803^{3}+1\cdot 0.682327803+1=0.317672196+0.682327803+1=0.999999998+1=2.}
Bet reikėjo daryti ne taip. Iš pradžiu reikia parinkti visas a , b , c reikšmes, kurias parenkame, kad a=1, b=1, c=1. Tada įstatome jas čia:
p
=
27
b
−
9
a
2
29
,
q
=
27
c
−
9
a
b
29
{\displaystyle p={\frac {27b-9a^{2}}{29}},\quad q={\frac {27c-9ab}{29}}}
ir gauname:
p
=
27
−
9
29
=
18
29
,
q
=
27
−
9
29
=
18
29
.
{\displaystyle p={\frac {27-9}{29}}={\frac {18}{29}},\quad q={\frac {27-9}{29}}={\frac {18}{29}}.}
Dabar šias reikšmes įstatome į formulę
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}}}})}}=}
=
1
2
⋅
(
−
18
29
+
(
18
29
)
2
+
4
(
18
29
)
3
27
)
3
+
1
2
⋅
(
−
18
29
−
(
18
29
)
2
+
4
(
18
29
)
3
27
)
3
=
{\displaystyle ={\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}+{\sqrt {({\frac {18}{29}})^{2}+{\frac {4({\frac {18}{29}})^{3}}{27}}}})}}+{\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}-{\sqrt {({\frac {18}{29}})^{2}+{\frac {4({\frac {18}{29}})^{3}}{27}}}})}}=}
=
1
2
⋅
(
−
18
29
+
(
18
29
)
2
+
0.956496781
27
)
3
+
1
2
⋅
(
−
18
29
−
(
18
29
)
2
+
0.956496781
27
)
3
=
{\displaystyle ={\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}+{\sqrt {({\frac {18}{29}})^{2}+{\frac {0.956496781}{27}}}})}}+{\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}-{\sqrt {({\frac {18}{29}})^{2}+{\frac {0.956496781}{27}}}})}}=}
=
1
2
⋅
(
−
18
29
+
(
18
29
)
2
+
0.035425806
)
3
+
1
2
⋅
(
−
18
29
−
(
18
29
)
2
+
0.035425806
)
3
=
{\displaystyle ={\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}+{\sqrt {({\frac {18}{29}})^{2}+0.035425806}})}}+{\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}-{\sqrt {({\frac {18}{29}})^{2}+0.035425806}})}}=}
=
1
2
⋅
(
−
18
29
+
0.420681454
)
3
+
1
2
⋅
(
−
18
29
−
0.420681454
)
3
=
1
2
⋅
(
−
18
29
+
0.64859961
)
3
+
1
2
⋅
(
−
18
29
−
0.64859961
)
3
=
{\displaystyle ={\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}+{\sqrt {0.420681454}})}}+{\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}-{\sqrt {0.420681454}})}}={\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}+0.64859961)}}+{\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}-0.64859961)}}=}
=
0.013954977
3
+
−
1.269289265
3
=
0.240755587
−
1.08273008
=
−
0.841974493.
{\displaystyle ={\sqrt[{3}]{0.013954977}}+{\sqrt[{3}]{-1.269289265}}=0.240755587-1.08273008=-0.841974493.}
Įstatome šią
β
=
−
1.08273008
{\displaystyle \beta =-1.08273008}
y
=
x
−
a
3
{\displaystyle y=x-{\frac {a}{3}}}
a nepasikeite ir todel a=1):
y
=
x
−
a
3
=
x
0
−
a
3
=
−
1.08273008
−
1
3
=
−
1.416063413.
{\displaystyle y=x-{\frac {a}{3}}=x_{0}-{\frac {a}{3}}=-1.08273008-{\frac {1}{3}}=-1.416063413.}
Toliau belieka y įstatyti į pradinę lygtį:
y
3
+
a
y
2
+
b
y
+
c
=
(
−
1.416063413
)
3
+
(
−
1.416063413
)
2
−
1.416063413
+
1
=
−
1.250368578.
{\displaystyle y^{3}+ay^{2}+by+c=(-1.416063413)^{3}+(-1.416063413)^{2}-1.416063413+1=-1.250368578.}
Gali būti, kad lygtis su tokiais koeficientais, kai visi koeficientai vienetai, turi tik kompleksinių skaičių sprendinius.
Dabar įstatome
β
=
−
1.08273008
{\displaystyle \beta =-1.08273008}
x
3
+
p
x
+
q
=
0
{\displaystyle x^{3}+px+q=0}
(
−
1.08273008
)
3
+
1
(
−
1.08273008
)
+
1
=
−
1.352019345.
{\displaystyle (-1.08273008)^{3}+1(-1.08273008)+1=-1.352019345.}
Be abejonės, kubinės lygties sprendiniai atspėti, o tik pavaidinta, kad išspresta, nebent pagal Vijeto teoremą, jei
x
3
=
0
{\displaystyle x_{3}=0}
x
1
x
2
+
x
1
x
3
+
x
2
x
3
=
c
a
{\displaystyle x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3}={\frac {c}{a}}}
p
(
X
)
=
a
X
3
+
b
X
2
+
c
X
+
d
{\displaystyle p(X)=aX^{3}+bX^{2}+cX+d\,}
x
1
x
2
=
c
a
.
{\displaystyle x_{1}x_{2}={\frac {c}{a}}.}
"Vijeto formulės kvadratiniam polinomui
p
(
X
)
=
a
X
2
+
b
X
+
c
{\displaystyle p(X)=aX^{2}+bX+c\,}
x
1
,
x
2
{\displaystyle x_{1},x_{2}\,}
kvadratinėje lygtyje
p
(
X
)
=
0
{\displaystyle p(X)=0\,}
x
1
+
x
2
=
−
b
a
,
x
1
⋅
x
2
=
c
a
{\displaystyle x_{1}+x_{2}=-{\frac {b}{a}},\quad x_{1}\cdot x_{2}={\frac {c}{a}}}
Pavyzdžiui, jei turime kvadratinę lygtį
x
2
−
x
−
6
=
0
,
{\displaystyle x^{2}-x-6=0,\,}
ją išspręsti galime pasinaudoję Vijeto teorema ir sudarę lygčių sistemą
{
x
1
+
x
2
=
−
−
1
1
x
1
⋅
x
2
=
−
6
1
{\displaystyle {\begin{cases}x_{1}+x_{2}=-{\frac {-1}{1}}\\x_{1}\cdot x_{2}={\frac {-6}{1}}\end{cases}}}
Jei šią sistemą bandytume spręsti formaliai (pvz., išsireikšdami vieną iš kintamųjų), vėl gautume tą pačią lygtį. Praktikoje, naudojant Vijeto teoremą lygčių sprendimui, sprendinius x1 ir x2 bandoma „atspėti“ - sugalvoti tokius x1 ir x2 , kad jie tenkintų lygčių sistemą. Šiuo atveju sprendiniai yra -2 ir 3.
Vijeto formulės kubiniam polinomui
p
(
X
)
=
a
X
3
+
b
X
2
+
c
X
+
d
{\displaystyle p(X)=aX^{3}+bX^{2}+cX+d\,}
x
1
,
x
2
,
x
3
{\displaystyle x_{1},x_{2},x_{3}\,}
p
(
X
)
=
0
{\displaystyle p(X)=0\,}
x
1
+
x
2
+
x
3
=
−
b
a
,
x
1
x
2
+
x
1
x
3
+
x
2
x
3
=
c
a
,
x
1
x
2
x
3
=
−
d
a
{\displaystyle x_{1}+x_{2}+x_{3}=-{\frac {b}{a}},\quad x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3}={\frac {c}{a}},\quad x_{1}x_{2}x_{3}=-{\frac {d}{a}}}
![{\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/5425b02f9e2d1c4a6a78a6629a71996a587ab94c)
![{\displaystyle ={\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}+{\sqrt {({\frac {18}{29}})^{2}+{\frac {4({\frac {18}{29}})^{3}}{27}}}})}}+{\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}-{\sqrt {({\frac {18}{29}})^{2}+{\frac {4({\frac {18}{29}})^{3}}{27}}}})}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84ac0e4ebeb1cd331829509431cd13f618252eaa)
![{\displaystyle ={\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}+{\sqrt {({\frac {18}{29}})^{2}+{\frac {0.956496781}{27}}}})}}+{\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}-{\sqrt {({\frac {18}{29}})^{2}+{\frac {0.956496781}{27}}}})}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/79034eebbf885770514ece367ffd3637e852d614)
![{\displaystyle ={\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}+{\sqrt {({\frac {18}{29}})^{2}+0.035425806}})}}+{\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}-{\sqrt {({\frac {18}{29}})^{2}+0.035425806}})}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3bd9e6aa64249ce904d8258044948306b0e8bd4)
![{\displaystyle ={\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}+{\sqrt {0.420681454}})}}+{\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}-{\sqrt {0.420681454}})}}={\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}+0.64859961)}}+{\sqrt[{3}]{{\frac {1}{2}}\cdot (-{\frac {18}{29}}-0.64859961)}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c17df5dbbd3813471d71106fedb1bc68b397c2c)
![{\displaystyle ={\sqrt[{3}]{0.013954977}}+{\sqrt[{3}]{-1.269289265}}=0.240755587-1.08273008=-0.841974493.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f4e93453b8ff3aac896de618c00d1a9931afa99)
![{\displaystyle {\begin{aligned}x_{1}=&-{\frac {b}{3a}}\\&-{\frac {1}{3a}}{\sqrt[{3}]{{\tfrac {1}{2}}\left[2b^{3}-9abc+27a^{2}d+{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}\right]}}\\&-{\frac {1}{3a}}{\sqrt[{3}]{{\tfrac {1}{2}}\left[2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}\right]}}\\x_{2}=&-{\frac {b}{3a}}\\&+{\frac {1+i{\sqrt {3}}}{6a}}{\sqrt[{3}]{{\tfrac {1}{2}}\left[2b^{3}-9abc+27a^{2}d+{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}\right]}}\\&+{\frac {1-i{\sqrt {3}}}{6a}}{\sqrt[{3}]{{\tfrac {1}{2}}\left[2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}\right]}}\\x_{3}=&-{\frac {b}{3a}}\\&+{\frac {1-i{\sqrt {3}}}{6a}}{\sqrt[{3}]{{\tfrac {1}{2}}\left[2b^{3}-9abc+27a^{2}d+{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}\right]}}\\&+{\frac {1+i{\sqrt {3}}}{6a}}{\sqrt[{3}]{{\tfrac {1}{2}}\left[2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}\right]}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49ca47638af5e9d4c062420c03a61b60273c1abe)
![{\displaystyle x_{1}=-{\frac {b}{3a}}-{\frac {1}{3a}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(2b^{3}-9abc+27a^{2}d+{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}\right)}}-}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7a917ad068bfc86a051a227f4bf0887bb7eb5ff)
![{\displaystyle -{\frac {1}{3a}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}\right)}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4271b47f6f9beb206c52f9419f329c9bd403551)
![{\displaystyle =-{\frac {0}{3}}-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(2\cdot 0^{3}-0+27\cdot 9+{\sqrt {\left(2\cdot 0^{3}-0+27\cdot 9\right)^{2}-4\left(0^{2}-3\cdot (-6)\right)^{3}}}\right)}}-}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e90925a9a4e3f05872df667b2a1937edf699ef6c)
![{\displaystyle -{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(2\cdot 0^{3}-0+27\cdot 9-{\sqrt {\left(2\cdot 0^{3}-0+27\cdot 9\right)^{2}-4(0^{2}-3\cdot (-6))^{3}}}\right)}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f5b8394dc33f1eb15beb889f96aaa5dc5a501c56)
![{\displaystyle =-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(243+{\sqrt {243^{2}-4\cdot 18^{3}}}\right)}}-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(243-{\sqrt {243^{2}-4\cdot 18^{3}}}\right)}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00cbdf4f132820c7f92a1fba11d37d6b1ee7119e)
![{\displaystyle =-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(243+{\sqrt {59049-23328}}\right)}}-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(243-{\sqrt {59049-23328}}\right)}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4f97005b7e6ea3c1cc3b5c506d600f3fac4da7f)
![{\displaystyle =-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(243+{\sqrt {35721}}\right)}}-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(243-{\sqrt {35721}}\right)}}=-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(243+189\right)}}-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(243-189\right)}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee2420326919fb67f3973692ea9e32de9dcbfa08)
![{\displaystyle =-{\frac {1}{3}}{\sqrt[{3}]{216}}-{\frac {1}{3}}{\sqrt[{3}]{54}}=-{\frac {1}{3}}\cdot 6-{\frac {1}{3}}\cdot 3{\sqrt[{3}]{2}}=-2-{\sqrt[{3}]{2}}=-3.25992105.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e97a39a3bf13ee06a71aa92693be73d7eda0d2d)
![{\displaystyle x_{2}=-{\frac {1}{3a}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}\right)}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24c94579a293dab3f26e25e987aa829594b768e1)
![{\displaystyle =-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(2\cdot 3^{3}-9\cdot 3\cdot (-3)+27\cdot 1^{2}\cdot (-14)-{\sqrt {\left(2\cdot 3^{3}-9\cdot 3\cdot (-3)+27\cdot 1^{2}\cdot (-14)\right)^{2}-4\left(3^{2}-3\cdot (-3)\right)^{3}}}\right)}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f8be5a4acc9c03dec881c6ab0dd05e364c551f2)
![{\displaystyle =-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(2\cdot 9-9\cdot (-9)+27\cdot (-14)-{\sqrt {\left(2\cdot 27-9\cdot (-9)+27\cdot (-14)\right)^{2}-4\left(9+9\right)^{3}}}\right)}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/210824fb27fc129305d0d8cee993823f6916dfdf)
![{\displaystyle =-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(18+81-378-{\sqrt {(54+81-378)^{2}-4\cdot 18^{3}}}\right)}}=-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(99-378-{\sqrt {(135-378)^{2}-4\cdot 5832}}\right)}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99a22938450e75695b7b495a6579d24c3050d1fc)
![{\displaystyle =-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(-279-{\sqrt {(-243)^{2}-23328}}\right)}}=-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(-279-{\sqrt {59049-23328}}\right)}}=-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(-279-{\sqrt {35721}}\right)}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08b45145c8f868280de11921a6e1eec10dcecfa2)
![{\displaystyle =-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot \left(-279-189\right)}}=-{\frac {1}{3}}{\sqrt[{3}]{{\frac {1}{2}}\cdot (-468)}}=-{\frac {1}{3}}{\sqrt[{3}]{-234}}={\frac {6.162240148}{3}}=2.054080049.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06dd7dcfe8ee098a2c680d91e08d8674e6e967e3)
![{\displaystyle \left({\frac {\sqrt[{3}]{-234}}{3}}\right)^{3}+3\left({\frac {\sqrt[{3}]{-234}}{3}}\right)^{2}-3\left({\frac {\sqrt[{3}]{-234}}{3}}\right)-14=1.162161065.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42249381d18bfd3d8b6751c2df30844014dbb031)