Kubinės lygties sprendimas be kompleksinių skaičių
Bendra forma kubinės lygties yra
a
x
3
+
b
x
2
+
c
x
+
d
=
0
(
a
≠
0
)
.
(
1
)
{\displaystyle ax^{3}+bx^{2}+cx+d=0\quad (a\neq 0).\quad (1)}
Padalinus šią lygtį iš
a
≠
0
,
{\displaystyle a\neq 0,}
x
3
+
b
x
2
+
c
x
+
d
=
0.
(
2
)
{\displaystyle x^{3}+bx^{2}+cx+d=0.\quad (2)}
(2) lygtis gali būti perdaryta/suprastina įvedus naują kintamąjį y , kad
x
=
y
+
k
.
{\displaystyle x=y+k.}
k pasirenkamas taip, kad (2) lygtis neturėtų antro laipsnio nario (
b
x
2
{\displaystyle bx^{2}}
x
=
y
+
k
,
{\displaystyle x=y+k,}
x
3
{\displaystyle x^{3}}
b
x
2
{\displaystyle bx^{2}}
y kvadratas (padaugintas iš konstantos).
x
3
+
b
x
2
+
c
x
+
d
=
0.
(
2
)
{\displaystyle x^{3}+bx^{2}+cx+d=0.\quad (2)}
(
y
+
k
)
3
+
b
(
y
+
k
)
2
+
c
(
y
+
k
)
+
d
=
0
,
{\displaystyle (y+k)^{3}+b(y+k)^{2}+c(y+k)+d=0,}
(
y
3
+
3
y
2
k
+
3
y
k
2
+
k
3
)
+
b
(
y
2
+
2
y
k
+
k
2
)
+
c
(
y
+
k
)
+
d
=
0
,
{\displaystyle (y^{3}+3y^{2}k+3yk^{2}+k^{3})+b(y^{2}+2yk+k^{2})+c(y+k)+d=0,}
(
y
3
+
3
y
k
2
+
k
3
)
+
[
3
y
2
k
+
b
y
2
]
+
b
(
2
y
k
+
k
2
)
+
c
(
y
+
k
)
+
d
=
0.
{\displaystyle (y^{3}+3yk^{2}+k^{3})+[3y^{2}k+by^{2}]+b(2yk+k^{2})+c(y+k)+d=0.}
Igriko kvadratas išnyks, kai
[
3
y
2
k
+
b
y
2
]
=
0
{\displaystyle [3y^{2}k+by^{2}]=0}
3
k
+
b
=
0.
{\displaystyle 3k+b=0.}
k
=
−
b
/
3.
{\displaystyle k=-b/3.}
Tai galima gauti ir taip:
x
3
+
b
x
2
=
x
2
(
x
+
b
)
=
(
y
+
k
)
2
(
y
+
k
+
b
)
=
(
y
2
+
2
y
k
+
k
2
)
(
y
+
k
+
b
)
;
{\displaystyle x^{3}+bx^{2}=x^{2}(x+b)=(y+k)^{2}(y+k+b)=(y^{2}+2yk+k^{2})(y+k+b);}
sandaugoje
(
y
2
+
2
y
k
+
k
2
)
(
y
+
k
+
b
)
{\displaystyle (y^{2}+2yk+k^{2})(y+k+b)}
k
y
2
+
b
y
2
+
2
y
2
k
=
3
k
y
2
+
b
y
2
{\displaystyle ky^{2}+by^{2}+2y^{2}k=3ky^{2}+by^{2}}
y
2
.
{\displaystyle y^{2}.}
y
2
{\displaystyle y^{2}}
3
k
+
b
=
0
,
{\displaystyle 3k+b=0,}
k
=
−
b
/
3.
{\displaystyle k=-b/3.}
Gauta lygtis vadinama redukuota (nežinomasis vėl yra x ):
x
3
+
p
x
+
q
=
0.
(
3
)
{\displaystyle x^{3}+px+q=0.\quad (3)}
Kubinė lygtis (3) visada turi bent vieną sprendinį. Funkcija
f
(
x
)
=
x
3
+
p
x
+
q
{\displaystyle f(x)=x^{3}+px+q}
x ir neigiama su dideliais neigiamais x . Todėl f(x) funkcijos grafikas kerta Ox ašį.
Kubinė lygtis gali turėti vieną, du ar trys sprendinius – pavyzdžiui, lygtys
(
x
−
1
)
3
=
0
,
(
x
−
1
)
2
(
x
−
2
)
=
0
,
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
=
0
{\displaystyle (x-1)^{3}=0,\quad (x-1)^{2}(x-2)=0,\quad (x-1)(x-2)(x-3)=0}
turi atitinkamai sprendinius
{
1
}
,
{
1
,
2
}
,
{
1
,
2
,
3
}
.
{\displaystyle \{1\},\;\;\{1,\;2\},\;\;\{1,\;2,\;3\}.}
Daugiau nei 3 sprendinius kubinė lygtis negali turėti. Tarkim, kad lygtis (3) turi sprendinį
α
.
{\displaystyle \alpha .}
α
3
+
p
α
+
q
=
0.
{\displaystyle \alpha ^{3}+p\alpha +q=0.}
x
3
+
p
x
+
q
=
x
3
+
p
x
+
q
−
α
3
−
p
α
−
q
=
x
3
−
α
3
+
p
x
−
p
α
=
(
x
−
α
)
(
x
2
+
x
α
+
α
2
+
p
)
.
{\displaystyle x^{3}+px+q=x^{3}+px+q-\alpha ^{3}-p\alpha -q=x^{3}-\alpha ^{3}+px-p\alpha =(x-\alpha )(x^{2}+x\alpha +\alpha ^{2}+p).}
Taigi kitus sprendinius lygties (3) rasime išsprenę kvadratinę lygtį (nuo nežinomojo x)
x
2
+
x
α
+
α
2
+
p
=
0.
(
4
)
{\displaystyle x^{2}+x\alpha +\alpha ^{2}+p=0.\quad (4)}
Lygtis (4) gali turėti 0 arba 1, arba 2 sprendinius, kurie nėra
α
.
{\displaystyle \alpha .}
[
x
3
+
p
x
+
q
=
0.
(
3
)
{\displaystyle x^{3}+px+q=0.\quad (3)}
Matome, kad žinant bent vieną (3) lygties spreninį, galima rasti visus kitus (3) lygties sprendinius iš (4) lygties. Štai kodėl labai lengva spręsti kubinę lygtį, kuri turi racionalų sprendinį. Beje, galima lengvai surasti bet kokio laipsnio lygties sprendinius, kai jos koeficientai yra racionalieji skaičiai. Pavyzdžiui, jeigu lygties koeficientai yra sveikieji skaičiai ir pirmas koeficientas yra 1 (tokią formą galima suteikti bet kokiai lygčiai su racionaliais koeficientais), tada racionalieji sprendiniai gali būti tik sveikieji skaičiai – teigiami ir neigiami dalikliai laisvojo nario (kubinėje lygtyje (3) laisvasis narys yra q ; žr. čia ). Todėl apsimoka sprendžiant bet kokią lygtį su racionaliaisiais koeficientais, pradėti lygties sprendimą nuo racionaliųjų sprendinių ieškojimo.
Kubinė lygtis (3) gali būti toliau suprastinta, padarius koeficiento
p
≠
0
{\displaystyle p\neq 0}
x
3
=
−
q
,
{\displaystyle x^{3}=-q,}
x
=
−
q
3
{\displaystyle x=-{\sqrt[{3}]{q}}}
Vėl mes panaudojame paprastą, tiesinį keitinį
x
=
k
y
{\displaystyle x=ky}
k . Mūsų (3) lygtis konvertuojama į
k
3
y
3
+
p
k
y
+
q
=
0
,
y
3
+
p
y
/
k
2
+
q
/
k
3
=
0.
{\displaystyle k^{3}y^{3}+pky+q=0,\quad y^{3}+py/k^{2}+q/k^{3}=0.}
Atveju
p
>
0
,
{\displaystyle p>0,}
k taip:
p
/
k
2
=
3
,
p
/
3
=
k
2
,
k
=
p
/
3
.
{\displaystyle p/k^{2}=3,\;\;p/3=k^{2},\;\;k={\sqrt {p/3}}.}
Todėl su tokiu k , koeficientas prie y tampa lygus 3.
x
=
k
y
=
p
/
3
y
,
{\displaystyle x=ky={\sqrt {p/3}}\;y,}
q
/
k
3
=
q
/
(
p
/
3
)
3
=
q
p
3
/
(
3
⋅
9
)
=
3
q
p
p
/
3
=
3
3
q
p
p
.
{\displaystyle q/k^{3}=q/({\sqrt {p/3}})^{3}={\frac {q}{\sqrt {p^{3}/(3\cdot 9)}}}={\frac {3q}{p{\sqrt {p/3}}}}={\frac {3{\sqrt {3}}q}{p{\sqrt {p}}}}.}
Pažymėkime
m
=
−
3
3
q
2
p
p
.
{\displaystyle m=-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}.}
Tada mes turime lygtį (gautą iš lygties
y
3
+
p
y
/
k
2
+
q
/
k
3
=
0
{\displaystyle y^{3}+py/k^{2}+q/k^{3}=0}
y
3
+
p
y
/
k
2
+
q
/
k
3
=
0
,
{\displaystyle y^{3}+py/k^{2}+q/k^{3}=0,}
y
3
+
3
y
−
2
m
=
0
,
(
5
)
{\displaystyle y^{3}+3y-2m=0,\quad (5)}
kurioje
p
/
k
2
=
3
,
3
3
q
p
p
=
−
2
m
.
{\displaystyle p/k^{2}=3,\;\;{\frac {3{\sqrt {3}}q}{p{\sqrt {p}}}}=-2m.}
Dabar mes bandysime rasti sprendinius, parinkę keitinį
y
=
z
−
1
/
z
:
{\displaystyle y=z-1/z:}
(
z
−
1
/
z
)
3
+
3
(
z
−
1
/
z
)
−
2
m
=
0
,
{\displaystyle (z-1/z)^{3}+3(z-1/z)-2m=0,}
(
z
3
−
3
z
2
⋅
1
/
z
+
3
z
⋅
1
/
z
2
−
1
/
z
3
)
+
3
(
z
−
1
/
z
)
−
2
m
=
0
,
{\displaystyle (z^{3}-3z^{2}\cdot 1/z+3z\cdot 1/z^{2}-1/z^{3})+3(z-1/z)-2m=0,}
(
z
3
−
3
z
+
3
/
z
−
1
/
z
3
)
+
3
z
−
3
/
z
−
2
m
=
0
,
{\displaystyle (z^{3}-3z+3/z-1/z^{3})+3z-3/z-2m=0,}
(
z
3
−
1
/
z
3
)
−
2
m
=
0
,
{\displaystyle (z^{3}-1/z^{3})-2m=0,}
z
3
−
1
/
z
3
−
2
m
=
0
,
|
⋅
z
3
{\displaystyle z^{3}-1/z^{3}-2m=0,\quad |\cdot z^{3}}
z
6
−
1
−
2
m
z
3
=
0
,
{\displaystyle z^{6}-1-2mz^{3}=0,}
z
6
−
2
m
z
3
=
1
,
{\displaystyle z^{6}-2mz^{3}=1,}
z
6
−
2
m
z
3
+
m
2
=
1
+
m
2
,
{\displaystyle z^{6}-2mz^{3}+m^{2}=1+m^{2},}
(
z
3
−
m
)
2
=
1
+
m
2
,
{\displaystyle (z^{3}-m)^{2}=1+m^{2},}
z
3
−
m
=
±
1
+
m
2
,
{\displaystyle z^{3}-m=\pm {\sqrt {1+m^{2}}},}
z
3
=
m
±
1
+
m
2
,
{\displaystyle z^{3}=m\pm {\sqrt {1+m^{2}}},}
z
=
m
±
1
+
m
2
3
.
{\displaystyle z={\sqrt[{3}]{m\pm {\sqrt {1+m^{2}}}}}.}
Abi šios z reikšmės duoda tą patį sprendinį (5) lygties.
y
=
z
−
1
/
z
=
m
+
1
+
m
2
3
−
1
m
+
1
+
m
2
3
=
{\displaystyle y=z-1/z={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}-{\frac {1}{\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}}=}
=
m
+
1
+
m
2
3
−
m
−
1
+
m
2
3
m
+
1
+
m
2
3
⋅
m
−
1
+
m
2
3
=
m
+
1
+
m
2
3
−
m
−
1
+
m
2
3
m
2
−
(
1
+
m
2
)
2
3
=
{\displaystyle ={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}-{\frac {\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}{{\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}\cdot {\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}}}={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}-{\frac {\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}{\sqrt[{3}]{m^{2}-({\sqrt {1+m^{2}}})^{2}}}}=}
=
m
+
1
+
m
2
3
−
m
−
1
+
m
2
3
m
2
−
(
1
+
m
2
)
3
=
m
+
1
+
m
2
3
−
m
−
1
+
m
2
3
−
1
=
{\displaystyle ={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}-{\frac {\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}{\sqrt[{3}]{m^{2}-(1+m^{2})}}}={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}-{\frac {\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}{-1}}=}
=
m
+
1
+
m
2
3
+
m
−
1
+
m
2
3
;
{\displaystyle ={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}+{\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}};}
y
=
z
−
1
/
z
=
m
−
1
+
m
2
3
−
1
m
−
1
+
m
2
3
=
{\displaystyle y=z-1/z={\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}-{\frac {1}{\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}}=}
=
m
−
1
+
m
2
3
−
m
+
1
+
m
2
3
m
+
1
+
m
2
3
⋅
m
−
1
+
m
2
3
=
m
−
1
+
m
2
3
−
m
+
1
+
m
2
3
m
2
−
(
1
+
m
2
)
2
3
=
{\displaystyle ={\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}-{\frac {\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}{{\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}\cdot {\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}}}={\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}-{\frac {\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}{\sqrt[{3}]{m^{2}-({\sqrt {1+m^{2}}})^{2}}}}=}
=
m
−
1
+
m
2
3
−
m
+
1
+
m
2
3
m
2
−
(
1
+
m
2
)
3
=
m
−
1
+
m
2
3
−
m
+
1
+
m
2
3
−
1
=
{\displaystyle ={\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}-{\frac {\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}{\sqrt[{3}]{m^{2}-(1+m^{2})}}}={\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}-{\frac {\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}{-1}}=}
=
m
−
1
+
m
2
3
+
m
+
1
+
m
2
3
.
{\displaystyle ={\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}+{\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}.}
Taigi, abiais atvejais (su dviem skirtingom z reikšmėm) gavome tą patį sprendinį
y
=
m
+
1
+
m
2
3
+
m
−
1
+
m
2
3
.
{\displaystyle y={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}+{\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}.}
Tai skiriasi nuo sprendinio iš PDF failo, kuris yra toks:
y
=
m
2
+
1
+
m
3
+
m
2
+
1
−
m
3
.
{\displaystyle y={\sqrt[{3}]{{\sqrt {m^{2}+1}}+m}}+{\sqrt[{3}]{{\sqrt {m^{2}+1}}-m}}.}
Įsitikinkime, kad (5) lygtis neturiu daugiau sprendinių. Pagal (4) lygtį (iš (5) lygties), dabar turime:
[
x
2
+
x
α
+
α
2
+
p
=
0.
(
4
)
{\displaystyle x^{2}+x\alpha +\alpha ^{2}+p=0.\quad (4)}
[
y
3
+
3
y
−
2
m
=
0
,
(
5
)
{\displaystyle y^{3}+3y-2m=0,\quad (5)}
y
2
+
y
α
+
α
2
+
3
=
0
,
(
5.1
)
{\displaystyle y^{2}+y\alpha +\alpha ^{2}+3=0,\quad (5.1)}
kur
α
=
m
+
1
+
m
2
3
+
m
−
1
+
m
2
3
.
{\displaystyle \alpha ={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}+{\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}.}
Diskriminantas (5.1) lygties lygus
D
=
α
2
−
4
(
α
2
+
3
)
=
−
3
α
2
−
12
<
0.
{\displaystyle D=\alpha ^{2}-4(\alpha ^{2}+3)=-3\alpha ^{2}-12<0.}
Kadangi diskriminantas neigiamas, (5) lygtis daugiau sprendinių neturi. Todėl (3) lygtis, atveju p>0, turi tik vieną sprendinį. Mes rasime jį perėję nuo y prie
x
=
y
p
/
3
.
{\displaystyle x=y{\sqrt {p/3}}.}
Rasime kam lygus x (kuris yra sprendinys (3) lygties).
m
=
−
3
3
q
2
p
p
.
{\displaystyle m=-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}.}
y
=
m
+
1
+
m
2
3
+
m
−
1
+
m
2
3
.
{\displaystyle y={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}+{\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}.}
x
=
k
y
=
p
/
3
y
=
p
/
3
(
m
+
1
+
m
2
3
+
m
−
1
+
m
2
3
)
=
{\displaystyle x=ky={\sqrt {p/3}}\;y={\sqrt {p/3}}\left({\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}+{\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}\right)=}
=
p
/
3
(
−
3
3
q
2
p
p
+
1
+
(
−
3
3
q
2
p
p
)
2
3
+
−
3
3
q
2
p
p
−
1
+
(
−
3
3
q
2
p
p
)
2
3
)
=
{\displaystyle ={\sqrt {p/3}}\left({\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}+{\sqrt {1+\left(-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}\right)^{2}}}}}+{\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}-{\sqrt {1+\left(-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}\right)^{2}}}}}\right)=}
=
p
/
3
(
−
3
3
q
2
p
p
+
1
+
9
⋅
3
q
2
4
p
2
⋅
p
3
+
−
3
3
q
2
p
p
−
1
+
9
⋅
3
q
2
4
p
2
⋅
p
3
)
=
{\displaystyle ={\sqrt {p/3}}\left({\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}+{\sqrt {1+{\frac {9\cdot 3q^{2}}{4p^{2}\cdot p}}}}}}+{\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}-{\sqrt {1+{\frac {9\cdot 3q^{2}}{4p^{2}\cdot p}}}}}}\right)=}
=
p
/
3
(
−
3
3
q
2
p
p
+
1
+
27
q
2
4
p
3
3
+
−
3
3
q
2
p
p
−
1
+
27
q
2
4
p
3
3
)
=
{\displaystyle ={\sqrt {p/3}}\left({\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}+{\sqrt {1+{\frac {27q^{2}}{4p^{3}}}}}}}+{\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}-{\sqrt {1+{\frac {27q^{2}}{4p^{3}}}}}}}\right)=}
=
p
/
3
(
−
3
3
q
2
p
p
+
4
p
3
+
27
q
2
4
p
3
3
+
−
3
3
q
2
p
p
−
4
p
3
+
27
q
2
4
p
3
3
)
=
{\displaystyle ={\sqrt {p/3}}\left({\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}+{\sqrt {\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}+{\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}-{\sqrt {\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}\right)=}
=
(
−
p
3
3
3
3
3
q
2
p
p
+
p
3
3
3
4
p
3
+
27
q
2
4
p
3
3
+
−
p
3
3
3
3
3
q
2
p
p
−
p
3
3
3
4
p
3
+
27
q
2
4
p
3
3
)
=
{\displaystyle =\left({\sqrt[{3}]{-{\sqrt {\frac {p^{3}}{3^{3}}}}{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}+{\sqrt {\frac {p^{3}}{3^{3}}}}{\sqrt {\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}+{\sqrt[{3}]{-{\sqrt {\frac {p^{3}}{3^{3}}}}{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}-{\sqrt {\frac {p^{3}}{3^{3}}}}{\sqrt {\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}\right)=}
=
−
p
p
3
3
3
3
q
2
p
p
+
p
p
3
3
4
p
3
+
27
q
2
4
p
3
3
+
−
p
p
3
3
3
3
q
2
p
p
−
p
p
3
3
4
p
3
+
27
q
2
4
p
3
3
=
{\displaystyle ={\sqrt[{3}]{-{\frac {p{\sqrt {p}}}{3{\sqrt {3}}}}{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}+{\frac {p{\sqrt {p}}}{3{\sqrt {3}}}}{\sqrt {\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}+{\sqrt[{3}]{-{\frac {p{\sqrt {p}}}{3{\sqrt {3}}}}{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}-{\frac {p{\sqrt {p}}}{3{\sqrt {3}}}}{\sqrt {\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}=}
=
−
q
2
+
1
3
3
4
p
3
+
27
q
2
4
3
+
−
q
2
−
1
3
3
4
p
3
+
27
q
2
4
3
=
{\displaystyle ={\sqrt[{3}]{-{\frac {q}{2}}+{\frac {1}{3{\sqrt {3}}}}{\sqrt {\frac {4p^{3}+27q^{2}}{4}}}}}+{\sqrt[{3}]{-{\frac {q}{2}}-{\frac {1}{3{\sqrt {3}}}}{\sqrt {\frac {4p^{3}+27q^{2}}{4}}}}}=}
=
−
q
2
+
4
p
3
+
27
q
2
4
⋅
27
3
+
−
q
2
−
4
p
3
+
27
q
2
4
⋅
27
3
=
{\displaystyle ={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {\frac {4p^{3}+27q^{2}}{4\cdot 27}}}}}+{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {\frac {4p^{3}+27q^{2}}{4\cdot 27}}}}}=}
=
−
q
2
+
p
3
27
+
q
2
4
3
+
−
q
2
−
p
3
27
+
q
2
4
3
.
{\displaystyle ={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {p^{3}}{27}}+{\frac {q^{2}}{4}}}}}}+{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {p^{3}}{27}}+{\frac {q^{2}}{4}}}}}}.}
Sunkiau išspręsti (3) lygtį, kai p yra neigiamas. Ir vėl panaudojame keitinį
x
=
k
y
.
{\displaystyle x=ky.}
y
3
+
p
y
/
k
2
+
q
/
k
3
=
0
{\displaystyle y^{3}+py/k^{2}+q/k^{3}=0}
pasirenkame k , kad
p
/
k
2
=
−
3
,
{\displaystyle p/k^{2}=-3,}
−
p
/
3
=
k
2
,
{\displaystyle -p/3=k^{2},}
k
=
−
p
/
3
.
{\displaystyle k={\sqrt {-p/3}}.}
Pošaknyje yra teigiamas skaičius, nes p <0.
q
/
k
3
=
q
/
(
−
p
/
3
)
3
=
q
−
p
3
/
(
3
⋅
9
)
=
3
q
p
−
p
/
3
=
3
3
q
p
−
p
;
{\displaystyle q/k^{3}=q/({\sqrt {-p/3}})^{3}={\frac {q}{\sqrt {-p^{3}/(3\cdot 9)}}}={\frac {3q}{p{\sqrt {-p/3}}}}={\frac {3{\sqrt {3}}q}{p{\sqrt {-p}}}};}
3
3
q
p
−
p
=
q
/
k
3
=
−
2
m
;
{\displaystyle {\frac {3{\sqrt {3}}q}{p{\sqrt {-p}}}}=q/k^{3}=-2m;}
m
=
−
3
3
q
2
p
−
p
.
{\displaystyle m=-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}.}
Pažymėję laisvajį narį per -2m, gauname lygtį
y
3
−
3
y
−
2
m
=
0.
(
6
)
{\displaystyle y^{3}-3y-2m=0.\quad (6)}
Panaudojame keitinį
y
=
z
+
1
/
z
:
{\displaystyle y=z+1/z:}
(
z
+
1
/
z
)
3
−
3
(
z
+
1
/
z
)
−
2
m
=
0
,
{\displaystyle (z+1/z)^{3}-3(z+1/z)-2m=0,}
(
z
3
+
3
z
2
⋅
1
/
z
+
3
z
⋅
1
/
z
2
+
1
/
z
3
)
−
3
(
z
+
1
/
z
)
−
2
m
=
0
,
{\displaystyle (z^{3}+3z^{2}\cdot 1/z+3z\cdot 1/z^{2}+1/z^{3})-3(z+1/z)-2m=0,}
(
z
3
+
3
z
+
3
/
z
+
1
/
z
3
)
−
3
(
z
+
1
/
z
)
−
2
m
=
0
,
{\displaystyle (z^{3}+3z+3/z+1/z^{3})-3(z+1/z)-2m=0,}
(
z
3
+
3
z
+
3
/
z
+
1
/
z
3
)
−
3
z
−
3
/
z
−
2
m
=
0
,
{\displaystyle (z^{3}+3z+3/z+1/z^{3})-3z-3/z-2m=0,}
(
z
3
+
1
/
z
3
)
−
2
m
=
0
,
{\displaystyle (z^{3}+1/z^{3})-2m=0,}
z
3
+
1
/
z
3
−
2
m
=
0
,
|
⋅
z
3
{\displaystyle z^{3}+1/z^{3}-2m=0,\;\;|\cdot z^{3}}
z
6
+
1
−
2
m
z
3
=
0
,
{\displaystyle z^{6}+1-2mz^{3}=0,}
z
6
−
2
m
z
3
=
−
1
,
{\displaystyle z^{6}-2mz^{3}=-1,}
z
6
−
2
m
z
3
+
m
2
=
m
2
−
1
,
{\displaystyle z^{6}-2mz^{3}+m^{2}=m^{2}-1,}
(
z
3
−
m
)
2
=
m
2
−
1.
(
7
)
{\displaystyle (z^{3}-m)^{2}=m^{2}-1.\quad (7)}
Lygtis (7) nevisada turi sprendinius ir reikia išnagrinėti trys atvejus:
m
2
>
1
,
{\displaystyle m^{2}>1,\;}
m
2
=
1
{\displaystyle m^{2}=1\;}
m
2
<
1.
{\displaystyle \;m^{2}<1.}
keisti
Jeigu
m
2
>
1
,
{\displaystyle m^{2}>1,}
[
(
z
3
−
m
)
2
=
m
2
−
1.
(
7
)
{\displaystyle (z^{3}-m)^{2}=m^{2}-1.\quad (7)}
z
3
−
m
=
±
m
2
−
1
,
{\displaystyle z^{3}-m=\pm {\sqrt {m^{2}-1}},}
z
3
=
m
±
m
2
−
1
,
{\displaystyle z^{3}=m\pm {\sqrt {m^{2}-1}},}
z
=
m
±
m
2
−
1
3
,
{\displaystyle z={\sqrt[{3}]{m\pm {\sqrt {m^{2}-1}}}},}
ir abi šios z reikšmės duoda tą patį sprendinį:
y
=
z
+
1
/
z
=
m
+
m
2
−
1
3
+
1
m
+
m
2
−
1
3
=
{\displaystyle y=z+1/z={\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}+{\frac {1}{\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}}=}
=
m
+
m
2
−
1
3
+
m
−
m
2
−
1
3
m
−
m
2
−
1
3
⋅
m
+
m
2
−
1
3
=
m
+
m
2
−
1
3
+
m
−
m
2
−
1
3
m
2
−
(
m
2
−
1
)
2
3
=
{\displaystyle ={\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}+{\frac {\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}{{\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}\cdot {\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}}}={\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}+{\frac {\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}{\sqrt[{3}]{m^{2}-({\sqrt {m^{2}-1}})^{2}}}}=}
=
m
+
m
2
−
1
3
+
m
−
m
2
−
1
3
m
2
−
(
m
2
−
1
)
3
=
m
+
m
2
−
1
3
+
m
−
m
2
−
1
3
1
=
{\displaystyle ={\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}+{\frac {\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}{\sqrt[{3}]{m^{2}-(m^{2}-1)}}}={\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}+{\frac {\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}{1}}=}
=
m
+
m
2
−
1
3
+
m
−
m
2
−
1
3
;
{\displaystyle ={\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}+{\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}};}
y
=
z
+
1
/
z
=
m
−
m
2
−
1
3
+
1
m
−
m
2
−
1
3
=
{\displaystyle y=z+1/z={\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}+{\frac {1}{\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}}=}
=
m
−
m
2
−
1
3
+
m
+
m
2
−
1
3
m
+
m
2
−
1
3
⋅
m
−
m
2
−
1
3
=
m
−
m
2
−
1
3
+
m
+
m
2
−
1
3
m
2
−
(
m
2
−
1
)
2
3
=
{\displaystyle ={\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}+{\frac {\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}{{\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}\cdot {\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}}}={\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}+{\frac {\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}{\sqrt[{3}]{m^{2}-({\sqrt {m^{2}-1}})^{2}}}}=}
=
m
−
m
2
−
1
3
+
m
+
m
2
−
1
3
m
2
−
(
m
2
−
1
)
3
=
m
−
m
2
−
1
3
+
m
+
m
2
−
1
3
1
=
{\displaystyle ={\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}+{\frac {\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}{\sqrt[{3}]{m^{2}-(m^{2}-1)}}}={\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}+{\frac {\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}{1}}=}
=
m
−
m
2
−
1
3
+
m
+
m
2
−
1
3
.
{\displaystyle ={\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}+{\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}.}
Gavome tas pačias y reikšmes. Gausime x reikšmę ((3) lygties sprendinį):
k
=
−
p
/
3
;
{\displaystyle k={\sqrt {-p/3}};}
m
=
−
3
3
q
2
p
−
p
;
{\displaystyle m=-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}};}
x
=
k
y
=
−
p
/
3
y
=
−
p
/
3
(
m
−
m
2
−
1
3
+
m
−
m
2
−
1
3
)
=
{\displaystyle x=ky={\sqrt {-p/3}}\;y={\sqrt {-p/3}}\left({\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}+{\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}\right)=}
=
−
p
/
3
(
−
3
3
q
2
p
−
p
+
(
−
3
3
q
2
p
−
p
)
2
−
1
3
+
−
3
3
q
2
p
−
p
−
(
−
3
3
q
2
p
−
p
)
2
−
1
3
)
=
{\displaystyle ={\sqrt {-p/3}}\left({\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}+{\sqrt {\left(-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}\right)^{2}-1}}}}+{\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}-{\sqrt {\left(-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}\right)^{2}-1}}}}\right)=}
=
−
p
/
3
(
−
3
3
q
2
p
−
p
+
9
⋅
3
q
2
4
p
2
⋅
(
−
p
)
−
1
3
+
3
3
q
2
p
−
p
−
9
⋅
3
q
2
−
4
p
2
⋅
p
−
1
3
)
=
{\displaystyle ={\sqrt {-p/3}}\left({\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}+{\sqrt {{\frac {9\cdot 3q^{2}}{4p^{2}\cdot (-p)}}-1}}}}+{\sqrt[{3}]{{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}-{\sqrt {{\frac {9\cdot 3q^{2}}{-4p^{2}\cdot p}}-1}}}}\right)=}
=
−
p
/
3
(
−
3
3
q
2
p
−
p
+
27
q
2
−
4
p
3
−
1
3
+
3
3
q
2
p
−
p
−
27
q
2
−
4
p
3
−
1
3
)
=
{\displaystyle ={\sqrt {-p/3}}\left({\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}+{\sqrt {{\frac {27q^{2}}{-4p^{3}}}-1}}}}+{\sqrt[{3}]{{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}-{\sqrt {{\frac {27q^{2}}{-4p^{3}}}-1}}}}\right)=}
=
−
p
/
3
(
−
3
3
q
2
p
−
p
+
4
p
3
+
27
q
2
−
4
p
3
3
+
3
3
q
2
p
−
p
−
−
4
p
3
+
27
q
2
4
p
3
3
)
=
{\displaystyle ={\sqrt {-p/3}}\left({\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}+{\sqrt {\frac {4p^{3}+27q^{2}}{-4p^{3}}}}}}+{\sqrt[{3}]{{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}-{\sqrt {-{\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}}\right)=}
=
(
−
−
p
3
3
3
3
3
q
2
p
−
p
+
−
p
3
3
3
4
p
3
+
27
q
2
4
p
3
3
+
−
p
3
3
3
3
3
q
2
p
−
p
−
−
p
3
3
3
−
4
p
3
+
27
q
2
4
p
3
3
)
=
{\displaystyle =\left({\sqrt[{3}]{-{\sqrt {-{\frac {p^{3}}{3^{3}}}}}{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}+{\sqrt {\frac {-p^{3}}{3^{3}}}}{\sqrt {\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}+{\sqrt[{3}]{{\sqrt {\frac {-p^{3}}{3^{3}}}}{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}-{\sqrt {\frac {-p^{3}}{3^{3}}}}{\sqrt {-{\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}}\right)=}
=
−
p
−
p
3
3
3
3
q
2
p
−
p
+
p
−
p
3
3
−
4
p
3
+
27
q
2
4
p
3
3
+
p
−
p
3
3
3
3
q
2
p
−
p
−
p
−
p
3
3
−
4
p
3
+
27
q
2
4
p
3
3
=
{\displaystyle ={\sqrt[{3}]{-{\frac {p{\sqrt {-p}}}{3{\sqrt {3}}}}{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}+{\frac {p{\sqrt {-p}}}{3{\sqrt {3}}}}{\sqrt {-{\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}}+{\sqrt[{3}]{{\frac {p{\sqrt {-p}}}{3{\sqrt {3}}}}{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}-{\frac {p{\sqrt {-p}}}{3{\sqrt {3}}}}{\sqrt {-{\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}}=}
=
−
q
2
+
1
3
3
4
p
3
+
27
q
2
4
3
+
−
q
2
−
1
3
3
4
p
3
+
27
q
2
4
3
=
{\displaystyle ={\sqrt[{3}]{-{\frac {q}{2}}+{\frac {1}{3{\sqrt {3}}}}{\sqrt {\frac {4p^{3}+27q^{2}}{4}}}}}+{\sqrt[{3}]{-{\frac {q}{2}}-{\frac {1}{3{\sqrt {3}}}}{\sqrt {\frac {4p^{3}+27q^{2}}{4}}}}}=}
=
−
q
2
+
4
p
3
+
27
q
2
4
⋅
27
3
+
−
q
2
−
4
p
3
+
27
q
2
4
⋅
27
3
=
{\displaystyle ={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {\frac {4p^{3}+27q^{2}}{4\cdot 27}}}}}+{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {\frac {4p^{3}+27q^{2}}{4\cdot 27}}}}}=}
=
−
q
2
+
p
3
27
+
q
2
4
3
+
−
q
2
−
p
3
27
+
q
2
4
3
.
{\displaystyle ={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {p^{3}}{27}}+{\frac {q^{2}}{4}}}}}}+{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {p^{3}}{27}}+{\frac {q^{2}}{4}}}}}}.}
Įsitikinkime, kad šis sprendinys yra unikalus (vienintelis). Šio sprendinio modulis yra didesnis už 2:
y
2
=
(
z
+
1
/
z
)
2
=
z
2
+
2
+
1
/
z
2
=
z
2
−
2
+
1
/
z
2
+
4
=
(
z
−
1
/
z
)
2
+
4
>
4.
{\displaystyle y^{2}=(z+1/z)^{2}=z^{2}+2+1/z^{2}=z^{2}-2+1/z^{2}+4=(z-1/z)^{2}+4>4.}
Vadinasi, |y|>2 ir iš (6) lygties
2
m
=
y
3
−
3
y
=
y
(
y
2
−
3
)
>
y
(
4
−
3
)
=
y
,
{\displaystyle 2m=y^{3}-3y=y(y^{2}-3)>y(4-3)=y,}
2
m
>
y
;
{\displaystyle 2m>y;}
2
m
>
y
>
2
(
kai
y
>
0
)
,
{\displaystyle 2m>y>2\quad ({\text{kai}}\;y>0),}
m
>
1
(
kai
y
>
0
)
.
{\displaystyle m>1\quad ({\text{kai}}\;y>0).}
Jeigu pvz., y=-3, tai
2
m
>
−
3
(
kai
y
<
0
)
,
{\displaystyle 2m>-3\quad ({\text{kai}}\;y<0),}
2
m
>
−
2
>
y
(
kai
y
<
0
)
,
{\displaystyle 2m>-2>y\quad ({\text{kai}}\;y<0),}
m
>
−
1
(
kai
y
<
0
)
.
{\displaystyle m>-1\quad ({\text{kai}}\;y<0).}
Bet mūsų atveju turi būti
m
2
>
1
,
{\displaystyle m^{2}>1,}
m
>
−
1
,
{\displaystyle m>-1,}
m
2
=
(
−
0.9
)
2
=
0.81
<
1
,
{\displaystyle m^{2}=(-0.9)^{2}=0.81<1,}
m
2
>
1.
{\displaystyle m^{2}>1.}
y reikšmės patenka į sąlygą
m
2
>
1.
{\displaystyle m^{2}>1.}
y yra neigiamas, tai jo reikšmės mažesnės už -2 (y<-2, kai y<0) ir 2m gali būti lygu -2 (nes 2m>y). Tada gaunasi, kad 2m=-2, m=-1. Ir
m
2
=
(
−
1
)
2
=
1
,
{\displaystyle m^{2}=(-1)^{2}=1,}
m
2
>
1.
{\displaystyle m^{2}>1.}
Bet jeigu, pavyzdžiui, y=-3, tai iš nelygybės
2
m
>
y
,
{\displaystyle 2m>y,}
2
m
>
−
3
,
{\displaystyle 2m>-3,}
m
>
−
1.5.
{\displaystyle m>-1.5.}
Ir gaunasi, kad sąlyga
m
2
>
1
{\displaystyle m^{2}>1}
(
−
1.49
)
2
=
2.2201
>
1.
{\displaystyle (-1.49)^{2}=2.2201>1.}
Vadinasi, sąlyga
m
2
>
1
{\displaystyle m^{2}>1}
2
m
=
−
2
,
m
=
−
1.
{\displaystyle 2m=-2,\;\;m=-1.}
y reikšmėm, m gali įgyti vienintelę reikšmę
m
=
−
1
,
{\displaystyle m=-1,}
p
<
0
,
m
2
>
1.
{\displaystyle p<0,\;m^{2}>1.}
Pataisymas. Jeigu m =-0.5, tai tada m>-1 (kai y<0). Arba 2m=-1>-2>y (kai y<0). Todėl, kai y neigiamas, m reikšmės gali būti segmente [-1; 0]. Arba
−
1
≤
m
≤
0
,
{\displaystyle -1\leq m\leq 0,}
y >0, tada m reikšmės gali būti intervale
(
1
;
∞
)
.
{\displaystyle (1;\;\infty ).}
m reikšmės (
−
1
≤
m
≤
0
{\displaystyle -1\leq m\leq 0}
a reikšmės iš segmento [-1; 0] visada tenkins nelygybę 2a>y (kai y<0), bet m reikšmės iš šio segmento nepriklauso šiam atvejui (
p
<
1
,
m
2
>
1
{\displaystyle p<1,\;m^{2}>1}
Kiti sprendiniai galėtų būti gauti iš (4) lygties, kuri dabar atrodo taip:
[
x
2
+
x
α
+
α
2
+
p
=
0.
(
4
)
{\displaystyle x^{2}+x\alpha +\alpha ^{2}+p=0.\quad (4)}
[
y
3
−
3
y
−
2
m
=
0.
(
6
)
{\displaystyle y^{3}-3y-2m=0.\quad (6)}
y
2
+
y
α
+
α
2
−
3
=
0
(
kur
|
α
|
>
2
)
.
(
4.6
)
{\displaystyle y^{2}+y\alpha +\alpha ^{2}-3=0\quad ({\text{kur}}\;|\alpha |>2).\quad (4.6)}
Jos diskriminantas
D
=
α
2
−
4
(
α
2
−
3
)
=
−
3
α
2
+
12
=
3
(
4
−
α
2
)
<
0
{\displaystyle D=\alpha ^{2}-4(\alpha ^{2}-3)=-3\alpha ^{2}+12=3(4-\alpha ^{2})<0}
yra neigiamas ir (4.6) lygtis neturi sprendinių (su
α
=
m
−
m
2
−
1
3
+
m
+
m
2
−
1
3
{\displaystyle \alpha ={\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}+{\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}}
p
<
0
,
m
2
>
1
{\displaystyle p<0,\;\;m^{2}>1}
α
{\displaystyle \alpha \;}
α
=
m
+
m
2
−
1
3
+
m
−
m
2
−
1
3
{\displaystyle \alpha ={\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}+{\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}}
keisti
Jeigu
m
2
=
1
,
{\displaystyle m^{2}=1,}
[
(
z
3
−
m
)
2
=
m
2
−
1.
(
7
)
{\displaystyle (z^{3}-m)^{2}=m^{2}-1.\quad (7)}
turime
z
3
=
m
.
{\displaystyle z^{3}=m.}
Kai m=1, tai z=1 ir
y = z + 1/z = 2.
(6) lygtis tranformuojasi į
y
3
−
3
y
−
2
=
0
,
{\displaystyle y^{3}-3y-2=0,}
kuri gali būti išskaidyta nepasinaudojant (4) lygtim:
y
3
−
3
y
−
2
=
0
,
(
y
−
2
)
(
y
2
+
2
y
+
1
)
=
0
,
(
y
−
2
)
(
y
+
1
)
2
=
0.
{\displaystyle y^{3}-3y-2=0,\quad (y-2)(y^{2}+2y+1)=0,\quad (y-2)(y+1)^{2}=0.}
Taigi lygtis (6) (ir (3)) šiuo atveju turi du sprendinius ((6) lygties sprendiniai yra 2 ir -1, kai p<0, m*m=1). Grįžtame prie x.
x
=
k
y
=
−
p
/
3
⋅
y
.
{\displaystyle x=ky={\sqrt {-p/3}}\cdot y.}
x
1
=
k
y
1
=
−
p
/
3
⋅
2
=
2
−
p
/
3
;
{\displaystyle x_{1}=ky_{1}={\sqrt {-p/3}}\cdot 2=2{\sqrt {-p/3}};}
x
2
=
k
y
2
=
−
p
/
3
⋅
(
−
1
)
=
−
−
p
/
3
.
{\displaystyle x_{2}=ky_{2}={\sqrt {-p/3}}\cdot (-1)=-{\sqrt {-p/3}}.}
Kai m=-1, tai z=-1 ir
y = z + 1/z = -1 + 1/(-1) = -2.
(6) lygtis tampa
y
3
−
3
y
+
2
=
0
,
(
y
+
2
)
(
y
2
−
2
y
+
1
)
=
0
,
(
y
+
2
)
(
y
−
1
)
2
=
0.
{\displaystyle y^{3}-3y+2=0,\quad (y+2)(y^{2}-2y+1)=0,\quad (y+2)(y-1)^{2}=0.}
Ir taip pat turi du sprendinius (kaip ir (3)). Grįžtame prie x.
x
=
k
y
=
−
p
/
3
⋅
y
.
{\displaystyle x=ky={\sqrt {-p/3}}\cdot y.}
x
1
=
k
y
1
=
−
p
/
3
⋅
(
−
2
)
=
−
2
−
p
/
3
;
{\displaystyle x_{1}=ky_{1}={\sqrt {-p/3}}\cdot (-2)=-2{\sqrt {-p/3}};}
x
2
=
k
y
2
=
−
p
/
3
⋅
1
=
−
p
/
3
.
{\displaystyle x_{2}=ky_{2}={\sqrt {-p/3}}\cdot 1={\sqrt {-p/3}}.}
p<0.
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
x
3
−
12
x
+
16
=
0
{\displaystyle x^{3}-12x+16=0}
pavyzdys, kurios p=-12<0.
Tada
x
1
=
−
2
−
p
/
3
=
−
2
−
(
−
12
)
/
3
=
−
2
12
/
3
=
−
2
4
=
−
2
⋅
2
=
−
4
;
{\displaystyle x_{1}=-2{\sqrt {-p/3}}=-2{\sqrt {-(-12)/3}}=-2{\sqrt {12/3}}=-2{\sqrt {4}}=-2\cdot 2=-4;}
x
2
=
−
p
/
3
=
−
(
−
12
)
/
3
=
12
/
3
=
4
=
2.
{\displaystyle x_{2}={\sqrt {-p/3}}={\sqrt {-(-12)/3}}={\sqrt {12/3}}={\sqrt {4}}=2.}
Šie sprendiniai yra tokie patys kaip ir pateiktoje nuorodoje.
Patikriname:
x
1
3
−
12
x
1
+
16
=
0
,
{\displaystyle x_{1}^{3}-12x_{1}+16=0,}
(
−
4
)
3
−
12
⋅
(
−
4
)
+
16
=
0
,
{\displaystyle (-4)^{3}-12\cdot (-4)+16=0,}
−
64
+
48
+
16
=
0
;
{\displaystyle -64+48+16=0;}
x
2
3
−
12
x
2
+
16
=
0
,
{\displaystyle x_{2}^{3}-12x_{2}+16=0,}
2
3
−
12
⋅
2
+
16
=
0
,
{\displaystyle 2^{3}-12\cdot 2+16=0,}
8
−
24
+
16
=
0.
{\displaystyle 8-24+16=0.}
Vadinasi sprendiniai teisingi.
m
=
−
3
3
q
2
p
−
p
=
−
3
3
⋅
16
2
⋅
(
−
12
)
−
(
−
12
)
=
−
3
3
⋅
16
−
24
12
=
−
3
3
⋅
16
−
24
3
⋅
4
=
{\displaystyle m=-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}=-{\frac {3{\sqrt {3}}\cdot 16}{2\cdot (-12){\sqrt {-(-12)}}}}=-{\frac {3{\sqrt {3}}\cdot 16}{-24{\sqrt {12}}}}=-{\frac {3{\sqrt {3}}\cdot 16}{-24{\sqrt {3\cdot 4}}}}=}
=
−
3
⋅
16
−
24
4
=
−
3
⋅
16
−
3
⋅
8
4
=
−
16
−
8
4
=
−
16
−
16
=
1.
{\displaystyle =-{\frac {3\cdot 16}{-24{\sqrt {4}}}}=-{\frac {3\cdot 16}{-3\cdot 8{\sqrt {4}}}}=-{\frac {16}{-8{\sqrt {4}}}}=-{\frac {16}{-16}}=1.}
Kažkodėl šitame pavyzdyje m=1, bet
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
[
y
3
+
p
y
/
k
2
+
q
/
k
3
=
0
{\displaystyle y^{3}+py/k^{2}+q/k^{3}=0}
pasirenkame k , kad
p
/
k
2
=
−
3
,
{\displaystyle p/k^{2}=-3,}
−
p
/
3
=
k
2
,
{\displaystyle -p/3=k^{2},}
k
=
−
p
/
3
.
{\displaystyle k={\sqrt {-p/3}}.}
Pošaknyje yra teigiamas skaičius, nes p <0.
q
/
k
3
=
q
/
(
−
p
/
3
)
3
=
q
−
p
3
/
(
3
⋅
9
)
=
3
q
p
−
p
/
3
=
3
3
q
p
−
p
;
{\displaystyle q/k^{3}=q/({\sqrt {-p/3}})^{3}={\frac {q}{\sqrt {-p^{3}/(3\cdot 9)}}}={\frac {3q}{p{\sqrt {-p/3}}}}={\frac {3{\sqrt {3}}q}{p{\sqrt {-p}}}};}
3
3
q
p
−
p
=
q
/
k
3
=
−
2
m
;
{\displaystyle {\frac {3{\sqrt {3}}q}{p{\sqrt {-p}}}}=q/k^{3}=-2m;}
m
=
−
3
3
q
2
p
−
p
.
{\displaystyle m=-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}.}
Gali būti, kad gavome m=1, o ne m=-1, nes
q
/
k
3
=
q
/
(
−
p
/
3
)
3
=
q
−
p
3
/
(
3
⋅
9
)
=
3
q
−
p
2
⋅
p
/
3
=
3
3
q
p
2
−
p
;
{\displaystyle q/k^{3}=q/({\sqrt {-p/3}})^{3}={\frac {q}{\sqrt {-p^{3}/(3\cdot 9)}}}={\frac {3q}{\sqrt {-p^{2}\cdot p/3}}}={\frac {3{\sqrt {3}}q}{{\sqrt {p^{2}}}{\sqrt {-p}}}};}
tada
3
3
q
p
2
−
p
=
q
/
k
3
=
−
2
m
;
{\displaystyle {\frac {3{\sqrt {3}}q}{{\sqrt {p^{2}}}{\sqrt {-p}}}}=q/k^{3}=-2m;}
m
=
−
3
3
q
2
p
2
−
p
=
{\displaystyle m=-{\frac {3{\sqrt {3}}q}{2{\sqrt {p^{2}}}{\sqrt {-p}}}}=}
=
−
3
3
⋅
16
2
(
−
12
)
2
−
(
−
12
)
=
{\displaystyle =-{\frac {3{\sqrt {3}}\cdot 16}{2{\sqrt {(-12)^{2}}}{\sqrt {-(-12)}}}}=}
=
−
3
3
⋅
16
2
144
12
=
{\displaystyle =-{\frac {3{\sqrt {3}}\cdot 16}{2{\sqrt {144}}{\sqrt {12}}}}=}
=
−
3
3
⋅
16
2
⋅
12
3
⋅
4
=
{\displaystyle =-{\frac {3{\sqrt {3}}\cdot 16}{2\cdot 12{\sqrt {3\cdot 4}}}}=}
=
−
3
⋅
16
2
⋅
12
4
=
{\displaystyle =-{\frac {3\cdot 16}{2\cdot 12{\sqrt {4}}}}=}
=
−
3
⋅
16
2
⋅
12
⋅
2
=
{\displaystyle =-{\frac {3\cdot 16}{2\cdot 12\cdot 2}}=}
=
−
3
⋅
16
4
⋅
3
⋅
4
=
−
1.
{\displaystyle =-{\frac {3\cdot 16}{4\cdot 3\cdot 4}}=-1.}
Dabar ir gavome
m
=
−
1.
{\displaystyle m=-1.}
keisti
Jeigu
m
2
<
1
,
{\displaystyle m^{2}<1,}
y
=
z
+
1
/
z
{\displaystyle y=z+1/z}
y
=
2
z
{\displaystyle y=2z}
[
y
3
−
3
y
−
2
m
=
0.
(
6
)
{\displaystyle y^{3}-3y-2m=0.\quad (6)}
(
2
z
)
3
−
3
⋅
2
z
−
2
m
=
0
,
{\displaystyle (2z)^{3}-3\cdot 2z-2m=0,}
8
z
3
−
6
z
−
2
m
=
0
,
{\displaystyle 8z^{3}-6z-2m=0,}
4
z
3
−
3
z
−
m
=
0.
{\displaystyle 4z^{3}-3z-m=0.}
Darome keitinį
z
=
cos
φ
:
{\displaystyle z=\cos \varphi :}
4
cos
3
φ
−
3
cos
φ
−
m
=
0
,
cos
(
3
φ
)
−
m
=
0
,
cos
3
φ
=
m
.
{\displaystyle 4\cos ^{3}\varphi -3\cos \varphi -m=0,\quad \cos(3\varphi )-m=0,\quad \cos 3\varphi =m.}
Ši lygtis turi sprendinius, kai |m|<1 (arba
m
2
<
1
{\displaystyle m^{2}<1}
φ
{\displaystyle \varphi }
cos
3
φ
=
m
,
{\displaystyle \cos 3\varphi =m,}
cos
(
±
3
φ
)
=
m
,
{\displaystyle \cos(\pm 3\varphi )=m,}
±
3
φ
=
arccos
m
,
{\displaystyle \pm 3\varphi =\arccos m,}
±
(
±
3
φ
)
=
3
φ
=
±
arccos
m
,
{\displaystyle \pm (\pm 3\varphi )=3\varphi =\pm \arccos m,}
3
φ
=
±
arccos
m
,
{\displaystyle 3\varphi =\pm \arccos m,}
3
φ
−
2
k
π
=
±
arccos
m
,
{\displaystyle 3\varphi -2k\pi =\pm \arccos m,}
3
φ
=
2
k
π
±
arccos
m
,
{\displaystyle 3\varphi =2k\pi \pm \arccos m,}
φ
=
2
k
π
3
±
1
3
arccos
m
;
{\displaystyle \varphi ={\frac {2k\pi }{3}}\pm {\frac {1}{3}}\arccos m;}
z
=
cos
φ
=
cos
(
2
3
k
π
±
1
3
arccos
m
)
,
{\displaystyle z=\cos \varphi =\cos \left({\frac {2}{3}}k\pi \pm {\frac {1}{3}}\arccos m\right),}
y
=
2
z
=
2
cos
φ
=
2
cos
(
2
3
k
π
±
1
3
arccos
m
)
,
{\displaystyle y=2z=2\cos \varphi =2\cos \left({\frac {2}{3}}k\pi \pm {\frac {1}{3}}\arccos m\right),}
k - sveikasis skaičius.Remdamiesi redukcijos formulėmis, suprantame, kad su daugiau nei trim skaičiaus k reikšmėm z reikšmės kartosis, nes kubinė lygtis negali turėti daugiau nei trys sprendinius. Todėl pasirenkame
k
=
0
,
1
,
2.
{\displaystyle k=0,\;1,\;2.}
y
1
=
2
z
1
=
2
cos
(
1
3
arccos
m
)
,
{\displaystyle y_{1}=2z_{1}=2\cos \left({\frac {1}{3}}\arccos m\right),}
y
2
=
2
z
2
=
2
cos
(
2
3
π
±
1
3
arccos
m
)
,
{\displaystyle y_{2}=2z_{2}=2\cos \left({\frac {2}{3}}\pi \pm {\frac {1}{3}}\arccos m\right),}
y
3
=
2
z
3
=
2
cos
(
4
3
π
±
1
3
arccos
m
)
.
{\displaystyle y_{3}=2z_{3}=2\cos \left({\frac {4}{3}}\pi \pm {\frac {1}{3}}\arccos m\right).}
Grįžtame prie x .
x
=
k
y
=
−
p
/
3
y
;
{\displaystyle x=ky={\sqrt {-p/3}}\;y;}
x
1
=
y
1
−
p
/
3
=
2
−
p
/
3
cos
(
1
3
arccos
m
)
,
{\displaystyle x_{1}=y_{1}{\sqrt {-p/3}}=2{\sqrt {-p/3}}\cos \left({\frac {1}{3}}\arccos m\right),}
x
2
=
y
2
−
p
/
3
=
2
−
p
/
3
cos
(
2
3
π
±
1
3
arccos
m
)
,
{\displaystyle x_{2}=y_{2}{\sqrt {-p/3}}=2{\sqrt {-p/3}}\cos \left({\frac {2}{3}}\pi \pm {\frac {1}{3}}\arccos m\right),}
x
3
=
y
3
−
p
/
3
=
2
−
p
/
3
cos
(
4
3
π
±
1
3
arccos
m
)
.
{\displaystyle x_{3}=y_{3}{\sqrt {-p/3}}=2{\sqrt {-p/3}}\cos \left({\frac {4}{3}}\pi \pm {\frac {1}{3}}\arccos m\right).}
Kubinė lygtis (3) yra išspręsta.
m
=
−
3
3
q
2
p
−
p
.
{\displaystyle m=-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}.}
x
3
−
19
x
+
30
=
0
,
{\displaystyle x^{3}-19x+30=0,}
kurios
p
=
−
19
{\displaystyle p=-19}
q
=
30.
{\displaystyle q=30.}
Jos sprendiniai yra 2, 3 ir
−
5.
{\displaystyle -5.}
Apskaičiuosime juos pagal ką tik išvestas formules.
m
=
−
3
3
q
2
p
−
p
=
−
3
3
⋅
30
2
⋅
(
−
19
)
−
(
−
19
)
=
−
90
3
2
⋅
(
−
19
)
19
=
−
45
3
(
−
19
)
19
=
{\displaystyle m=-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}=-{\frac {3{\sqrt {3}}\cdot 30}{2\cdot (-19){\sqrt {-(-19)}}}}=-{\frac {90{\sqrt {3}}}{2\cdot (-19){\sqrt {19}}}}=-{\frac {45{\sqrt {3}}}{(-19){\sqrt {19}}}}=}
= 45*3^0.5 /(19*19^0.5) = 0.9411150958093732309574814856304.
Arba kitas variantas:
m
=
−
3
3
q
2
p
2
−
p
=
−
3
3
⋅
30
2
(
−
19
)
2
−
(
−
19
)
=
−
90
3
2
361
19
=
−
45
3
19
19
=
{\displaystyle m=-{\frac {3{\sqrt {3}}q}{2{\sqrt {p^{2}}}{\sqrt {-p}}}}=-{\frac {3{\sqrt {3}}\cdot 30}{2{\sqrt {(-19)^{2}}}{\sqrt {-(-19)}}}}=-{\frac {90{\sqrt {3}}}{2{\sqrt {361}}{\sqrt {19}}}}=-{\frac {45{\sqrt {3}}}{19{\sqrt {19}}}}=}
= -45*3^0.5 /(19*19^0.5) = -0.9411150958093732309574814856304.
Tada
arccos
m
=
arccos
(
0.9411150958093732309574814856304
)
=
{\displaystyle \arccos m=\arccos(0.9411150958093732309574814856304)=}
Arba
arccos
m
=
arccos
(
−
0.9411150958093732309574814856304
)
=
{\displaystyle \arccos m=\arccos(-0.9411150958093732309574814856304)=}
Apskaičiuosime
x
1
.
{\displaystyle x_{1}.}
Su pirma m reikšme:
x
1
=
2
−
p
/
3
cos
(
1
3
arccos
m
)
=
2
−
(
−
19
)
/
3
cos
(
1
3
arccos
(
0.94111509580937323
)
)
=
{\displaystyle x_{1}=2{\sqrt {-p/3}}\cos \left({\frac {1}{3}}\arccos m\right)=2{\sqrt {-(-19)/3}}\cos \left({\frac {1}{3}}\arccos(0.94111509580937323)\right)=}
=
2
19
/
3
cos
(
1
3
⋅
0.3448827615021193521
)
=
2
19
/
3
cos
(
0.11496092050070645
)
=
{\displaystyle =2{\sqrt {19/3}}\cos \left({\frac {1}{3}}\cdot 0.3448827615021193521\right)=2{\sqrt {19/3}}\cos(0.11496092050070645)=}
= 2*(19/3)^0.5 * cos(0.11496092050070645) = 2*(19/3)^0.5 *0.99339926779878285489956379038765 = 5 (Windows 10 kalkuliatoriaus duoda tokią tikslią reikšmę "5", įstačius pajuodintą reikšmę). Su antra m reikšme:
x
1
=
2
−
p
/
3
cos
(
1
3
arccos
m
)
=
2
−
(
−
19
)
/
3
cos
(
1
3
arccos
(
−
0.94111509580937323
)
)
=
{\displaystyle x_{1}=2{\sqrt {-p/3}}\cos \left({\frac {1}{3}}\arccos m\right)=2{\sqrt {-(-19)/3}}\cos \left({\frac {1}{3}}\arccos(-0.94111509580937323)\right)=}
=
2
19
/
3
cos
(
1
3
⋅
2.796709892087673886353886095715
)
=
2
19
/
3
cos
(
0.93223663069589129545
)
=
{\displaystyle =2{\sqrt {19/3}}\cos \left({\frac {1}{3}}\cdot 2.796709892087673886353886095715\right)=2{\sqrt {19/3}}\cos(0.93223663069589129545)=}
= 2*(19/3)^0.5 * cos(0.93223663069589129545) = 2*(19/3)^0.5 *0.59603956067926971293973827423259 = 3 (Windows 10 kalkuliatoriaus duoda tokią tikslią reikšmę "3", įstačius pajuodintą reikšmę).
Su pirma m reikšme, atsakymas 5 yra be minuso ženklo. O su antra m reikšme, atsakymas 3 yra duotos lygties sprendinys.
Toliau apskaičiuosime
x
2
.
{\displaystyle x_{2}.}
Su pirma m reikšme:
x
2
=
2
−
p
/
3
cos
(
2
3
π
±
1
3
arccos
m
)
=
2
−
(
−
19
)
/
3
cos
(
2
3
π
±
1
3
arccos
(
0.3448827615021193521
)
)
=
{\displaystyle x_{2}=2{\sqrt {-p/3}}\cos \left({\frac {2}{3}}\pi \pm {\frac {1}{3}}\arccos m\right)=2{\sqrt {-(-19)/3}}\cos \left({\frac {2}{3}}\pi \pm {\frac {1}{3}}\arccos(0.3448827615021193521)\right)=}
=
2
19
/
3
cos
(
2
3
π
±
0.11496092050070645
)
=
{\displaystyle =2{\sqrt {19/3}}\cos \left({\frac {2}{3}}\pi \pm 0.11496092050070645\right)=}
= 2*(19/3)^0.5 * cos(2*pi /3 +0.11496092050070645070291909585483) =
= 2*(19/3)^0.5 * cos(2.0943951023931954923084289221863 +0.11496092050070645070291909585483) =
= 2*(19/3)^0.5 * cos(2.2093560228939019430113480180411) = 2*(19/3)^0.5 * (-0.59603956067926971293973827423256) =
= -2.9999999999999999999999999999999.
O jeigu po pi paimsime ne pliuso , bet minuso ženklą, tai gausime:
= 2*(19/3)^0.5 * cos(2.0943951023931954923084289221863 -0.11496092050070645070291909585483) =
= 2*(19/3)^0.5 * cos(1.9794341818924890416055098263315) = 2*(19/3)^0.5 * (-0.39735970711951314195982551615502) =
= -1.9999999999999999999999999999998.
Čia kosinusas skaičiuojamas su reikšmėm didesnėm nei pi /2 =~ 1.57, todėl galimos kai kurios klaidos. Bendrai, negalimos klaidos, nes čia kosinusas, o ne arkkosinusas. Su antra m reikšme:
x
2
=
2
−
p
/
3
cos
(
2
3
π
±
1
3
arccos
m
)
=
2
−
(
−
19
)
/
3
cos
(
2
3
π
±
1
3
arccos
(
−
0.94111509580937323
)
)
=
{\displaystyle x_{2}=2{\sqrt {-p/3}}\cos \left({\frac {2}{3}}\pi \pm {\frac {1}{3}}\arccos m\right)=2{\sqrt {-(-19)/3}}\cos \left({\frac {2}{3}}\pi \pm {\frac {1}{3}}\arccos(-0.94111509580937323)\right)=}
=
2
19
/
3
cos
(
2
3
π
±
0.932236630695891295451
)
=
{\displaystyle =2{\sqrt {19/3}}\cos \left({\frac {2}{3}}\pi \pm 0.932236630695891295451\right)=}
= 2*(19/3)^0.5 * cos(2*pi /3 +0.93223663069589129545129536523833) =
= 2*(19/3)^0.5 * cos(2.0943951023931954923084289221863 + 0.93223663069589129545129536523833) =
= 2*(19/3)^0.5 * cos(3.0266317330890867877597242874246) = 2*(19/3)^0.5 * (-0.99339926779878285489956379038764) =
= -5 (įstačius į Windows 10 kalkuliatorių pajuodintą tekstą).
O jeigu po pi paimsime ne pliuso , bet minuso ženklą, tai gausime:
= 2*(19/3)^0.5 * cos(2.0943951023931954923084289221863 - 0.93223663069589129545129536523833) =
= 2*(19/3)^0.5 * cos(1.162158471697304196857133556948) = 2*(19/3)^0.5 * 0.39735970711951314195982551615505 =
= 2 (įstačius į Windows 10 kalkuliatorių pajuodintą tekstą).
Su antra m reikšme iškart gavome du teisingus tos lygties sprendinius.
Toliau apskaičiuosime
x
3
.
{\displaystyle x_{3}.}
Su pirma m reikšme:
x
3
=
2
−
p
/
3
cos
(
4
3
π
±
1
3
arccos
m
)
=
2
−
(
−
19
)
/
3
cos
(
4
3
π
±
1
3
arccos
(
0.3448827615021193521
)
)
=
{\displaystyle x_{3}=2{\sqrt {-p/3}}\cos \left({\frac {4}{3}}\pi \pm {\frac {1}{3}}\arccos m\right)=2{\sqrt {-(-19)/3}}\cos \left({\frac {4}{3}}\pi \pm {\frac {1}{3}}\arccos(0.3448827615021193521)\right)=}
=
2
19
/
3
cos
(
4
3
π
±
0.11496092050070645
)
=
{\displaystyle =2{\sqrt {19/3}}\cos \left({\frac {4}{3}}\pi \pm 0.11496092050070645\right)=}
= 2*(19/3)^0.5 * cos(4*pi /3 +0.11496092050070645070291909585483) =
= 2*(19/3)^0.5 * cos(4.1887902047863909846168578443727 +0.11496092050070645070291909585483) =
= 2*(19/3)^0.5 * cos(4.3037511252870974353197769402275) = 2*(19/3)^0.5 * (-0.39735970711951314195982551615506) =
= -2.
O jeigu po pi paimsime ne pliuso , bet minuso ženklą, tai gausime:
= 2*(19/3)^0.5 * cos(4.1887902047863909846168578443727 - 0.11496092050070645070291909585483) =
= 2*(19/3)^0.5 * cos(4.0738292842856845339139387485179) = 2*(19/3)^0.5 * (-0.59603956067926971293973827423257) =
= -2.9999999999999999999999999999999. Su antra m reikšme:
x
3
=
2
−
p
/
3
cos
(
4
3
π
±
1
3
arccos
m
)
=
2
−
(
−
19
)
/
3
cos
(
4
3
π
±
1
3
arccos
(
−
0.94111509580937323
)
)
=
{\displaystyle x_{3}=2{\sqrt {-p/3}}\cos \left({\frac {4}{3}}\pi \pm {\frac {1}{3}}\arccos m\right)=2{\sqrt {-(-19)/3}}\cos \left({\frac {4}{3}}\pi \pm {\frac {1}{3}}\arccos(-0.94111509580937323)\right)=}
=
2
19
/
3
cos
(
4
3
π
±
0.932236630695891295451
)
=
{\displaystyle =2{\sqrt {19/3}}\cos \left({\frac {4}{3}}\pi \pm 0.932236630695891295451\right)=}
= 2*(19/3)^0.5 * cos(4*pi /3 +0.93223663069589129545129536523833) =
= 2*(19/3)^0.5 * cos(4.1887902047863909846168578443727 + 0.93223663069589129545129536523833) =
= 2*(19/3)^0.5 * cos(5.121026835482282280068153209611) = 2*(19/3)^0.5 * 0.39735970711951314195982551615505 =
= 2 (įstačius į Windows 10 kalkuliatorių pajuodintą tekstą).
O jeigu po pi paimsime ne pliuso , bet minuso ženklą, tai gausime:
= 2*(19/3)^0.5 * cos(4.1887902047863909846168578443727 - 0.93223663069589129545129536523833) =
= 2*(19/3)^0.5 * cos(3.2565535740904996891655624791344) = 2*(19/3)^0.5 * (-0.99339926779878285489956379038764) =
= -5 (įstačius į Windows 10 kalkuliatorių pajuodintą tekstą).
Su antra m reikšme iškart gavome du teisingus tos lygties sprendinius.
Su antra m reikšmę gavome visus teisingus lygties
x
3
−
19
x
+
30
=
0
{\displaystyle x^{3}-19x+30=0}
m reikšme gavome sprendinius su priešingais ženklais (-2, -3, 5; dėl to visi jie neteisingi).
![{\displaystyle (y^{3}+3yk^{2}+k^{3})+[3y^{2}k+by^{2}]+b(2yk+k^{2})+c(y+k)+d=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b54db9d0cc00a02ac4d7f3346a889ce48053f5b6)
![{\displaystyle [3y^{2}k+by^{2}]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f408576958133c25e065f4c3f14a88cd3afa532d)
![{\displaystyle x=-{\sqrt[{3}]{q}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/209fe3e430f32aaa9cf167afc6b3f673fc09c622)
![{\displaystyle z={\sqrt[{3}]{m\pm {\sqrt {1+m^{2}}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc7dc9bc54e21d510258a3ef6e1c852594032007)
![{\displaystyle y=z-1/z={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}-{\frac {1}{\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7077b03cb7e996e1cf096e154be72c5fcd298258)
![{\displaystyle ={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}-{\frac {\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}{{\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}\cdot {\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}}}={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}-{\frac {\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}{\sqrt[{3}]{m^{2}-({\sqrt {1+m^{2}}})^{2}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8832c05257e7ba1d1194189fa2b357531ecbb5e)
![{\displaystyle ={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}-{\frac {\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}{\sqrt[{3}]{m^{2}-(1+m^{2})}}}={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}-{\frac {\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}{-1}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/213a6813646e6fc66b1cd7da388ce7c9ded0fd7d)
![{\displaystyle ={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}+{\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}};}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b14b6b0e4ba4a1180e0fe415d2ce25cbf5130e69)
![{\displaystyle y=z-1/z={\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}-{\frac {1}{\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f5af5d93327597b02e8b4593597c7090d100faa)
![{\displaystyle ={\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}-{\frac {\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}{{\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}\cdot {\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}}}={\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}-{\frac {\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}{\sqrt[{3}]{m^{2}-({\sqrt {1+m^{2}}})^{2}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2e170d81d737f1224214604cf6aeca1eddc12c9)
![{\displaystyle ={\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}-{\frac {\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}{\sqrt[{3}]{m^{2}-(1+m^{2})}}}={\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}-{\frac {\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}{-1}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c3c6080564fdf0fc0d4290b76805efb2213e356d)
![{\displaystyle ={\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}+{\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/908b59d58069505a2389c6766ac2df99028c3a26)
![{\displaystyle y={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}+{\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/171b49e42da1b0b3ee1f9d149db792512a6c5dff)
![{\displaystyle y={\sqrt[{3}]{{\sqrt {m^{2}+1}}+m}}+{\sqrt[{3}]{{\sqrt {m^{2}+1}}-m}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c86def769b50883371b90f39aa37a6334b43a132)
![{\displaystyle \alpha ={\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}+{\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bb8d2be93a4da4b705a78b17ce2507b107fa718)
![{\displaystyle x=ky={\sqrt {p/3}}\;y={\sqrt {p/3}}\left({\sqrt[{3}]{m+{\sqrt {1+m^{2}}}}}+{\sqrt[{3}]{m-{\sqrt {1+m^{2}}}}}\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3fe332f1cbef8cf9ef5ee696352f650b9dfdfb6e)
![{\displaystyle ={\sqrt {p/3}}\left({\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}+{\sqrt {1+\left(-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}\right)^{2}}}}}+{\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}-{\sqrt {1+\left(-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}\right)^{2}}}}}\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/41044631f734e9874d0d7a67bea7c5c8cdd6737d)
![{\displaystyle ={\sqrt {p/3}}\left({\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}+{\sqrt {1+{\frac {9\cdot 3q^{2}}{4p^{2}\cdot p}}}}}}+{\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}-{\sqrt {1+{\frac {9\cdot 3q^{2}}{4p^{2}\cdot p}}}}}}\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7319cf5616c469a2b88d253ca5a4d426df998c1)
![{\displaystyle ={\sqrt {p/3}}\left({\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}+{\sqrt {1+{\frac {27q^{2}}{4p^{3}}}}}}}+{\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}-{\sqrt {1+{\frac {27q^{2}}{4p^{3}}}}}}}\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd208b1d56444d6c452a4b2738fd4fb417dbd88b)
![{\displaystyle ={\sqrt {p/3}}\left({\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}+{\sqrt {\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}+{\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}-{\sqrt {\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6e8adde97c575bf07d96d2866d9d714ac234f90c)
![{\displaystyle =\left({\sqrt[{3}]{-{\sqrt {\frac {p^{3}}{3^{3}}}}{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}+{\sqrt {\frac {p^{3}}{3^{3}}}}{\sqrt {\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}+{\sqrt[{3}]{-{\sqrt {\frac {p^{3}}{3^{3}}}}{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}-{\sqrt {\frac {p^{3}}{3^{3}}}}{\sqrt {\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e6a41244dc37f7f1208126f758c9dd9d93459728)
![{\displaystyle ={\sqrt[{3}]{-{\frac {p{\sqrt {p}}}{3{\sqrt {3}}}}{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}+{\frac {p{\sqrt {p}}}{3{\sqrt {3}}}}{\sqrt {\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}+{\sqrt[{3}]{-{\frac {p{\sqrt {p}}}{3{\sqrt {3}}}}{\frac {3{\sqrt {3}}q}{2p{\sqrt {p}}}}-{\frac {p{\sqrt {p}}}{3{\sqrt {3}}}}{\sqrt {\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/87b4240dfc55906e784cb8075c5100c9db783245)
![{\displaystyle ={\sqrt[{3}]{-{\frac {q}{2}}+{\frac {1}{3{\sqrt {3}}}}{\sqrt {\frac {4p^{3}+27q^{2}}{4}}}}}+{\sqrt[{3}]{-{\frac {q}{2}}-{\frac {1}{3{\sqrt {3}}}}{\sqrt {\frac {4p^{3}+27q^{2}}{4}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2486e41a4f5b532eeb4c4602c6c07b3c54b864f8)
![{\displaystyle ={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {\frac {4p^{3}+27q^{2}}{4\cdot 27}}}}}+{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {\frac {4p^{3}+27q^{2}}{4\cdot 27}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6170524033d7e019447365af2a20c6fc7851800a)
![{\displaystyle ={\sqrt[{3}]{-{\frac {q}{2}}+{\sqrt {{\frac {p^{3}}{27}}+{\frac {q^{2}}{4}}}}}}+{\sqrt[{3}]{-{\frac {q}{2}}-{\sqrt {{\frac {p^{3}}{27}}+{\frac {q^{2}}{4}}}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08dfed3e85f23d7b8567872cd6bb2cd5055c0461)
![{\displaystyle z={\sqrt[{3}]{m\pm {\sqrt {m^{2}-1}}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b38ad54377bc6c106fc6531d397172b0c0f07a0)
![{\displaystyle y=z+1/z={\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}+{\frac {1}{\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1c76d77230d4f28935d650b4430d6ab2f3d1e4e)
![{\displaystyle ={\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}+{\frac {\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}{{\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}\cdot {\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}}}={\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}+{\frac {\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}{\sqrt[{3}]{m^{2}-({\sqrt {m^{2}-1}})^{2}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a37d3153e6bef7588a115e1386cfdf11375d7589)
![{\displaystyle ={\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}+{\frac {\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}{\sqrt[{3}]{m^{2}-(m^{2}-1)}}}={\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}+{\frac {\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}{1}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac0418d78d64576389ebabb2f0ed1e4bdf79daa3)
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![{\displaystyle ={\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}+{\frac {\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}{{\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}\cdot {\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}}}={\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}+{\frac {\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}{\sqrt[{3}]{m^{2}-({\sqrt {m^{2}-1}})^{2}}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/78b1a70eebd631b362c51c9caa4e66ee42c98da0)
![{\displaystyle ={\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}+{\frac {\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}{\sqrt[{3}]{m^{2}-(m^{2}-1)}}}={\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}+{\frac {\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}{1}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4958ee965e603a13dc27064c857b39f80e0ef9b)
![{\displaystyle ={\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}+{\sqrt[{3}]{m+{\sqrt {m^{2}-1}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7729260b68792ecdfaf8e23751a9f793ba0e1915)
![{\displaystyle x=ky={\sqrt {-p/3}}\;y={\sqrt {-p/3}}\left({\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}+{\sqrt[{3}]{m-{\sqrt {m^{2}-1}}}}\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b6fadd818605a4dc5dfb0fe190be90ddcacd965a)
![{\displaystyle ={\sqrt {-p/3}}\left({\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}+{\sqrt {\left(-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}\right)^{2}-1}}}}+{\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}-{\sqrt {\left(-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}\right)^{2}-1}}}}\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84cf060b5ed53f9d9597ada847a5c1bc5133dd11)
![{\displaystyle ={\sqrt {-p/3}}\left({\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}+{\sqrt {{\frac {9\cdot 3q^{2}}{4p^{2}\cdot (-p)}}-1}}}}+{\sqrt[{3}]{{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}-{\sqrt {{\frac {9\cdot 3q^{2}}{-4p^{2}\cdot p}}-1}}}}\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba89e395fcd07b77f603489f6af47299d655079c)
![{\displaystyle ={\sqrt {-p/3}}\left({\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}+{\sqrt {{\frac {27q^{2}}{-4p^{3}}}-1}}}}+{\sqrt[{3}]{{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}-{\sqrt {{\frac {27q^{2}}{-4p^{3}}}-1}}}}\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0de27da54141416140d6b182733eb639b10c8e5)
![{\displaystyle ={\sqrt {-p/3}}\left({\sqrt[{3}]{-{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}+{\sqrt {\frac {4p^{3}+27q^{2}}{-4p^{3}}}}}}+{\sqrt[{3}]{{\frac {3{\sqrt {3}}q}{2p{\sqrt {-p}}}}-{\sqrt {-{\frac {4p^{3}+27q^{2}}{4p^{3}}}}}}}\right)=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5274822af57b5365082423d57ab04b18b7912016)
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