Ketvirto laipsnio lygtis
y
4
+
a
y
3
+
b
y
2
+
c
y
+
d
=
0
{\displaystyle y^{4}+ay^{3}+by^{2}+cy+d=0}
redukuojama keitiniu
y
=
x
−
a
4
{\displaystyle y=x-{\frac {a}{4}}}
x
4
+
p
x
2
+
q
x
+
r
=
0.
{\displaystyle x^{4}+px^{2}+qx+r=0.}
Surasime kam lygūs koeficientai p, q ir r.
Iš Niutono Binomo formulės žinome, kad
(
a
−
b
)
2
=
a
2
−
2
a
b
+
b
2
,
{\displaystyle (a-b)^{2}=a^{2}-2ab+b^{2},}
(
a
−
b
)
3
=
a
3
−
3
a
2
b
+
3
a
b
2
−
b
3
,
{\displaystyle (a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3},}
(
a
−
b
)
4
=
a
4
−
4
a
3
b
+
6
a
2
b
2
−
4
a
b
3
+
b
4
.
{\displaystyle (a-b)^{4}=a^{4}-4a^{3}b+6a^{2}b^{2}-4ab^{3}+b^{4}.}
Į lygtį
y
4
+
a
y
3
+
b
y
2
+
c
y
+
d
=
0
{\displaystyle y^{4}+ay^{3}+by^{2}+cy+d=0}
y
{\displaystyle y}
x
−
a
4
{\displaystyle x-{\frac {a}{4}}}
(
x
−
a
4
)
4
+
a
(
x
−
a
4
)
3
+
b
(
x
−
a
4
)
2
+
c
(
x
−
a
4
)
+
d
=
0.
{\displaystyle (x-{\frac {a}{4}})^{4}+a(x-{\frac {a}{4}})^{3}+b(x-{\frac {a}{4}})^{2}+c(x-{\frac {a}{4}})+d=0.}
Surandame kam lygus
(
x
−
a
4
)
2
,
{\displaystyle (x-{\frac {a}{4}})^{2},}
(
x
−
a
4
)
2
=
x
2
−
2
x
⋅
a
4
+
(
a
4
)
2
=
x
2
−
a
2
x
+
a
2
16
.
{\displaystyle (x-{\frac {a}{4}})^{2}=x^{2}-2x\cdot {\frac {a}{4}}+({\frac {a}{4}})^{2}=x^{2}-{\frac {a}{2}}x+{\frac {a^{2}}{16}}.}
Surandame kam lygus
(
x
−
a
4
)
3
,
{\displaystyle (x-{\frac {a}{4}})^{3},}
(
x
−
a
4
)
3
=
x
3
−
3
x
2
⋅
a
4
+
3
x
⋅
(
a
4
)
2
−
(
a
4
)
3
=
x
3
−
3
x
2
⋅
a
4
+
3
x
⋅
a
2
16
−
a
3
64
=
{\displaystyle (x-{\frac {a}{4}})^{3}=x^{3}-3x^{2}\cdot {\frac {a}{4}}+3x\cdot ({\frac {a}{4}})^{2}-({\frac {a}{4}})^{3}=x^{3}-3x^{2}\cdot {\frac {a}{4}}+3x\cdot {\frac {a^{2}}{16}}-{\frac {a^{3}}{64}}=}
=
x
3
−
3
a
4
x
2
+
3
a
2
16
x
−
a
3
64
.
{\displaystyle =x^{3}-{\frac {3a}{4}}x^{2}+{\frac {3a^{2}}{16}}x-{\frac {a^{3}}{64}}.}
Surandame kam lygus
(
x
−
a
4
)
4
,
{\displaystyle (x-{\frac {a}{4}})^{4},}
(
x
−
a
4
)
4
=
x
4
−
4
x
3
⋅
a
4
+
6
x
2
⋅
(
a
4
)
2
−
4
x
⋅
(
a
4
)
3
+
(
a
4
)
4
=
x
4
−
4
x
3
⋅
a
4
+
6
x
2
⋅
a
2
16
−
4
x
⋅
a
3
64
+
a
4
256
=
{\displaystyle (x-{\frac {a}{4}})^{4}=x^{4}-4x^{3}\cdot {\frac {a}{4}}+6x^{2}\cdot ({\frac {a}{4}})^{2}-4x\cdot ({\frac {a}{4}})^{3}+({\frac {a}{4}})^{4}=x^{4}-4x^{3}\cdot {\frac {a}{4}}+6x^{2}\cdot {\frac {a^{2}}{16}}-4x\cdot {\frac {a^{3}}{64}}+{\frac {a^{4}}{256}}=}
=
x
4
−
a
x
3
+
3
a
2
8
x
2
−
a
3
16
x
+
a
4
256
.
{\displaystyle =x^{4}-ax^{3}+{\frac {3a^{2}}{8}}x^{2}-{\frac {a^{3}}{16}}x+{\frac {a^{4}}{256}}.}
y
4
+
a
y
3
+
b
y
2
+
c
y
+
d
=
0
,
{\displaystyle y^{4}+ay^{3}+by^{2}+cy+d=0,}
(
x
−
a
4
)
4
+
a
(
x
−
a
4
)
3
+
b
(
x
−
a
4
)
2
+
c
(
x
−
a
4
)
+
d
=
0
,
{\displaystyle (x-{\frac {a}{4}})^{4}+a(x-{\frac {a}{4}})^{3}+b(x-{\frac {a}{4}})^{2}+c(x-{\frac {a}{4}})+d=0,}
x
4
−
a
x
3
+
3
a
2
8
x
2
−
a
3
16
x
+
a
4
256
+
a
(
x
3
−
3
a
4
x
2
+
3
a
2
16
x
−
a
3
64
)
+
b
(
x
2
−
a
2
x
+
a
2
16
)
+
c
(
x
−
a
4
)
+
d
=
0
,
{\displaystyle x^{4}-ax^{3}+{\frac {3a^{2}}{8}}x^{2}-{\frac {a^{3}}{16}}x+{\frac {a^{4}}{256}}+a(x^{3}-{\frac {3a}{4}}x^{2}+{\frac {3a^{2}}{16}}x-{\frac {a^{3}}{64}})+b(x^{2}-{\frac {a}{2}}x+{\frac {a^{2}}{16}})+c(x-{\frac {a}{4}})+d=0,}
x
4
−
a
x
3
+
3
a
2
8
x
2
−
a
3
16
x
+
a
4
256
+
a
x
3
−
3
a
2
4
x
2
+
3
a
3
16
x
−
a
4
64
+
b
x
2
−
a
b
2
x
+
a
2
b
16
+
c
x
−
a
c
4
+
d
=
0
,
{\displaystyle x^{4}-ax^{3}+{\frac {3a^{2}}{8}}x^{2}-{\frac {a^{3}}{16}}x+{\frac {a^{4}}{256}}+ax^{3}-{\frac {3a^{2}}{4}}x^{2}+{\frac {3a^{3}}{16}}x-{\frac {a^{4}}{64}}+bx^{2}-{\frac {ab}{2}}x+{\frac {a^{2}b}{16}}+cx-{\frac {ac}{4}}+d=0,}
x
4
−
a
x
3
+
a
x
3
+
3
a
2
8
x
2
−
3
a
2
4
x
2
+
b
x
2
−
a
3
16
x
+
3
a
3
16
x
−
a
b
2
x
+
c
x
+
a
4
256
−
a
4
64
+
a
2
b
16
−
a
c
4
+
d
=
0
,
{\displaystyle x^{4}-ax^{3}+ax^{3}+{\frac {3a^{2}}{8}}x^{2}-{\frac {3a^{2}}{4}}x^{2}+bx^{2}-{\frac {a^{3}}{16}}x+{\frac {3a^{3}}{16}}x-{\frac {ab}{2}}x+cx+{\frac {a^{4}}{256}}-{\frac {a^{4}}{64}}+{\frac {a^{2}b}{16}}-{\frac {ac}{4}}+d=0,}
x
4
−
3
a
2
8
x
2
+
b
x
2
+
2
a
3
16
x
−
a
b
2
x
+
c
x
−
3
a
4
256
+
a
2
b
16
−
a
c
4
+
d
=
0
,
{\displaystyle x^{4}-{\frac {3a^{2}}{8}}x^{2}+bx^{2}+{\frac {2a^{3}}{16}}x-{\frac {ab}{2}}x+cx-{\frac {3a^{4}}{256}}+{\frac {a^{2}b}{16}}-{\frac {ac}{4}}+d=0,}
x
4
+
(
−
3
a
2
8
+
b
)
x
2
+
(
2
a
3
16
−
a
b
2
+
c
)
x
−
3
a
4
256
+
a
2
b
16
−
a
c
4
+
d
=
0
,
{\displaystyle x^{4}+(-{\frac {3a^{2}}{8}}+b)x^{2}+({\frac {2a^{3}}{16}}-{\frac {ab}{2}}+c)x-{\frac {3a^{4}}{256}}+{\frac {a^{2}b}{16}}-{\frac {ac}{4}}+d=0,}
x
4
+
p
x
2
+
q
x
+
r
=
0.
{\displaystyle x^{4}+px^{2}+qx+r=0.}
Iš čia turime, kad
p
=
−
3
a
2
8
+
b
,
{\displaystyle p=-{\frac {3a^{2}}{8}}+b,}
q
=
2
a
3
16
−
a
b
2
+
c
,
{\displaystyle q={\frac {2a^{3}}{16}}-{\frac {ab}{2}}+c,}
r
=
−
3
a
4
256
+
a
2
b
16
−
a
c
4
+
d
.
{\displaystyle r=-{\frac {3a^{4}}{256}}+{\frac {a^{2}b}{16}}-{\frac {ac}{4}}+d.}
Ketvirto laipsnio lygties sprendimas pirmu budu [ keisti ] Imame redukuotą lygtį
f
(
x
)
=
x
4
+
2
p
x
2
+
q
x
+
r
=
0.
{\displaystyle f(x)=x^{4}+2px^{2}+qx+r=0.}
Įvedame, pagalbinį nežinomąjį z , kurio reikšmę vėliau surasime ir užrašome taip:
f
(
x
)
=
(
x
2
+
p
+
z
)
2
+
q
x
+
r
−
z
2
−
2
x
2
z
−
2
p
z
−
p
2
=
(
x
2
+
p
+
z
)
2
−
[
2
z
x
2
−
q
x
+
(
z
2
+
2
p
z
−
r
+
p
2
)
]
,
{\displaystyle f(x)=(x^{2}+p+z)^{2}+qx+r-z^{2}-2x^{2}z-2pz-p^{2}=(x^{2}+p+z)^{2}-[2zx^{2}-qx+(z^{2}+2pz-r+p^{2})],}
Čia
(
x
2
+
p
+
z
)
2
=
x
4
+
x
2
p
+
x
2
z
+
p
x
2
+
p
2
+
p
z
+
z
x
2
+
z
p
+
z
2
=
x
4
+
2
p
x
2
+
2
z
x
2
+
p
2
+
z
2
+
2
p
z
.
{\displaystyle (x^{2}+p+z)^{2}=x^{4}+x^{2}p+x^{2}z+px^{2}+p^{2}+pz+zx^{2}+zp+z^{2}=x^{4}+2px^{2}+2zx^{2}+p^{2}+z^{2}+2pz.}
Kad polinmas
2
z
x
2
−
q
x
+
(
z
2
+
2
p
z
−
r
+
p
2
)
{\displaystyle 2zx^{2}-qx+(z^{2}+2pz-r+p^{2})}
d
2
=
q
2
−
4
⋅
2
z
⋅
(
z
2
+
2
p
z
−
r
+
p
2
)
{\displaystyle d_{2}=q^{2}-4\cdot 2z\cdot (z^{2}+2pz-r+p^{2})}
būtų lygus 0. Tada galėsime pasinaudoti formule
a
2
−
b
2
=
(
a
−
b
)
(
a
+
b
)
,
{\displaystyle a^{2}-b^{2}=(a-b)(a+b),}
2
z
x
2
−
q
x
+
(
z
2
+
2
p
z
−
r
+
p
2
)
{\displaystyle 2zx^{2}-qx+(z^{2}+2pz-r+p^{2})}
a
x
2
+
b
x
+
c
=
a
(
x
−
x
1
)
(
x
−
x
2
)
,
{\displaystyle ax^{2}+bx+c=a(x-x_{1})(x-x_{2}),}
x
1
=
x
2
,
{\displaystyle x_{1}=x_{2},}
a
x
2
+
b
x
+
c
=
a
(
x
−
x
0
)
2
{\displaystyle ax^{2}+bx+c=a(x-x_{0})^{2}}
2
z
x
2
−
q
x
+
(
z
2
+
2
p
z
−
r
+
p
2
)
=
b
2
{\displaystyle 2zx^{2}-qx+(z^{2}+2pz-r+p^{2})=b^{2}}
(
x
2
+
p
+
z
)
2
=
a
2
.
{\displaystyle (x^{2}+p+z)^{2}=a^{2}.}
Taigi,
f
(
x
)
=
(
x
2
+
p
+
z
)
2
−
[
2
z
x
2
−
q
x
+
(
z
2
+
2
p
z
−
r
+
p
2
)
]
=
0.
{\displaystyle f(x)=(x^{2}+p+z)^{2}-[2zx^{2}-qx+(z^{2}+2pz-r+p^{2})]=0.}
[
2
z
x
2
−
q
x
+
(
z
2
+
2
p
z
−
r
+
p
2
)
]
=
0
;
{\displaystyle [2zx^{2}-qx+(z^{2}+2pz-r+p^{2})]=0;}
d
2
=
q
2
−
4
⋅
2
z
⋅
(
z
2
+
2
p
z
−
r
+
p
2
)
=
0
,
{\displaystyle d_{2}=q^{2}-4\cdot 2z\cdot (z^{2}+2pz-r+p^{2})=0,}
d
2
=
q
2
−
8
z
(
z
2
+
2
p
z
−
r
+
p
2
)
=
q
2
−
8
z
3
−
16
p
z
2
+
8
z
r
−
8
z
p
2
=
0
,
{\displaystyle d_{2}=q^{2}-8z(z^{2}+2pz-r+p^{2})=q^{2}-8z^{3}-16pz^{2}+8zr-8zp^{2}=0,}
d
2
=
8
z
3
+
16
p
z
2
+
8
(
p
2
−
r
)
z
−
q
2
=
0.
{\displaystyle d_{2}=8z^{3}+16pz^{2}+8(p^{2}-r)z-q^{2}=0.}
Lygtis
8
z
3
+
16
p
z
2
+
8
(
p
2
−
r
)
z
−
q
2
=
0
{\displaystyle 8z^{3}+16pz^{2}+8(p^{2}-r)z-q^{2}=0}
rezolvente (išsprendėja). Vieną iš jos trijų šaknų (realiają) gausime
z
0
{\displaystyle z_{0}}
z
0
{\displaystyle z_{0}}
d
2
=
8
z
3
+
16
p
z
2
+
8
(
p
2
−
r
)
z
−
q
2
=
0
{\displaystyle d_{2}=8z^{3}+16pz^{2}+8(p^{2}-r)z-q^{2}=0}
z , galėsime apskaičiuoti
d
2
.
{\displaystyle d_{2}.}
[
2
z
x
2
−
q
x
+
(
z
2
+
2
p
z
−
r
+
p
2
)
]
=
0
{\displaystyle [2zx^{2}-qx+(z^{2}+2pz-r+p^{2})]=0}
x
1
=
x
2
{\displaystyle x_{1}=x_{2}}
Taigi, turime:
8
z
3
+
16
p
z
2
+
8
(
p
2
−
r
)
z
−
q
2
=
0
,
{\displaystyle 8z^{3}+16pz^{2}+8(p^{2}-r)z-q^{2}=0,}
z
3
+
2
p
z
2
+
(
p
2
−
r
)
z
−
q
2
8
=
0
,
{\displaystyle z^{3}+2pz^{2}+(p^{2}-r)z-{\frac {q^{2}}{8}}=0,}
pakeičiame
z
=
w
−
2
p
3
.
{\displaystyle z=w-{\frac {2p}{3}}.}
Lygčiai
z
3
+
2
p
z
2
+
(
p
2
−
r
)
z
−
q
2
8
=
0
,
{\displaystyle z^{3}+2pz^{2}+(p^{2}-r)z-{\frac {q^{2}}{8}}=0,}
z
=
w
−
2
p
3
,
{\displaystyle z=w-{\frac {2p}{3}},}
w
3
+
m
w
+
n
=
0.
{\displaystyle w^{3}+mw+n=0.}
Lygties
w
3
+
m
w
+
n
=
0
{\displaystyle w^{3}+mw+n=0}
viena šaknis yra:
w
0
=
−
n
2
+
n
2
4
+
m
3
27
3
+
−
n
2
−
n
2
4
+
m
3
27
3
.
{\displaystyle w_{0}={\sqrt[{3}]{-{\frac {n}{2}}+{\sqrt {{\frac {n^{2}}{4}}+{\frac {m^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {n}{2}}-{\sqrt {{\frac {n^{2}}{4}}+{\frac {m^{3}}{27}}}}}}.}
Tada
z
0
=
−
n
2
+
n
2
4
+
m
3
27
3
+
−
n
2
−
n
2
4
+
m
3
27
3
−
2
p
3
.
{\displaystyle z_{0}={\sqrt[{3}]{-{\frac {n}{2}}+{\sqrt {{\frac {n^{2}}{4}}+{\frac {m^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {n}{2}}-{\sqrt {{\frac {n^{2}}{4}}+{\frac {m^{3}}{27}}}}}}-{\frac {2p}{3}}.}
Dabar galime surasti lygties
[
2
z
0
x
2
−
q
x
+
(
z
0
2
+
2
p
z
0
−
r
+
p
2
)
]
{\displaystyle [2z_{0}x^{2}-qx+(z_{0}^{2}+2pz_{0}-r+p^{2})]}
x
1
=
x
2
=
−
(
−
q
)
±
d
2
2
⋅
2
z
0
=
q
±
0
2
⋅
2
z
0
=
q
4
z
0
.
{\displaystyle x_{1}=x_{2}={\frac {-(-q)\pm {\sqrt {d_{2}}}}{2\cdot 2z_{0}}}={\frac {q\pm 0}{2\cdot 2z_{0}}}={\frac {q}{4z_{0}}}.}
Toliau, žinodami, kad
a
x
2
+
b
x
+
c
=
a
(
x
−
x
1
)
(
x
−
x
2
)
,
{\displaystyle ax^{2}+bx+c=a(x-x_{1})(x-x_{2}),}
[
2
z
0
x
2
−
q
x
+
(
z
0
2
+
2
p
z
0
−
r
+
p
2
)
]
=
2
z
0
(
x
−
q
4
z
0
)
⋅
(
x
−
q
4
z
0
)
=
2
z
0
(
x
−
q
4
z
0
)
2
=
[
2
z
0
(
x
−
q
4
z
0
)
]
2
.
{\displaystyle [2z_{0}x^{2}-qx+(z_{0}^{2}+2pz_{0}-r+p^{2})]=2z_{0}\left(x-{\frac {q}{4z_{0}}}\right)\cdot \left(x-{\frac {q}{4z_{0}}}\right)=2z_{0}\left(x-{\frac {q}{4z_{0}}}\right)^{2}=\left[{\sqrt {2z_{0}}}\left(x-{\frac {q}{4z_{0}}}\right)\right]^{2}.}
Įstatę į lygtį gauname:
f
(
x
)
=
(
x
2
+
p
+
z
0
)
2
−
[
2
z
0
x
2
−
q
x
+
(
z
0
2
+
2
p
z
0
−
r
+
p
2
)
]
=
(
x
2
+
p
+
z
0
)
2
−
[
2
z
0
(
x
−
q
4
z
0
)
]
2
;
{\displaystyle f(x)=(x^{2}+p+z_{0})^{2}-[2z_{0}x^{2}-qx+(z_{0}^{2}+2pz_{0}-r+p^{2})]=(x^{2}+p+z_{0})^{2}-\left[{\sqrt {2z_{0}}}\left(x-{\frac {q}{4z_{0}}}\right)\right]^{2};}
(
x
2
+
p
+
z
0
)
2
−
[
2
z
0
(
x
−
q
4
z
0
)
]
2
=
0
,
{\displaystyle (x^{2}+p+z_{0})^{2}-\left[{\sqrt {2z_{0}}}\left(x-{\frac {q}{4z_{0}}}\right)\right]^{2}=0,}
(
x
2
+
p
+
z
0
−
2
z
0
(
x
−
q
4
z
0
)
)
(
x
2
+
p
+
z
0
+
2
z
0
(
x
−
q
4
z
0
)
)
=
0
,
{\displaystyle \left(x^{2}+p+z_{0}-{\sqrt {2z_{0}}}\left(x-{\frac {q}{4z_{0}}}\right)\right)\left(x^{2}+p+z_{0}+{\sqrt {2z_{0}}}\left(x-{\frac {q}{4z_{0}}}\right)\right)=0,}
(
x
2
+
p
+
z
0
−
2
z
0
⋅
x
+
2
z
0
⋅
q
4
z
0
)
(
x
2
+
p
+
z
0
+
2
z
0
⋅
x
−
2
z
0
⋅
q
4
z
0
)
=
0
,
{\displaystyle \left(x^{2}+p+z_{0}-{\sqrt {2z_{0}}}\cdot x+{\sqrt {2z_{0}}}\cdot {\frac {q}{4z_{0}}}\right)\left(x^{2}+p+z_{0}+{\sqrt {2z_{0}}}\cdot x-{\sqrt {2z_{0}}}\cdot {\frac {q}{4z_{0}}}\right)=0,}
(
x
2
+
p
+
z
0
−
2
z
0
⋅
x
+
2
z
0
⋅
q
16
z
0
2
)
(
x
2
+
p
+
z
0
+
2
z
0
⋅
x
−
2
z
0
⋅
q
16
z
0
2
)
=
0
,
{\displaystyle \left(x^{2}+p+z_{0}-{\sqrt {2z_{0}}}\cdot x+{\sqrt {2z_{0}}}\cdot {\frac {q}{\sqrt {16z_{0}^{2}}}}\right)\left(x^{2}+p+z_{0}+{\sqrt {2z_{0}}}\cdot x-{\sqrt {2z_{0}}}\cdot {\frac {q}{\sqrt {16z_{0}^{2}}}}\right)=0,}
(
x
2
+
p
+
z
0
−
2
z
0
⋅
x
+
q
8
z
0
)
(
x
2
+
p
+
z
0
+
2
z
0
⋅
x
−
q
8
z
0
)
=
0
,
{\displaystyle \left(x^{2}+p+z_{0}-{\sqrt {2z_{0}}}\cdot x+{\frac {q}{\sqrt {8z_{0}}}}\right)\left(x^{2}+p+z_{0}+{\sqrt {2z_{0}}}\cdot x-{\frac {q}{\sqrt {8z_{0}}}}\right)=0,}
(
x
2
+
p
+
z
0
−
2
z
0
⋅
x
+
q
2
2
z
0
)
(
x
2
+
p
+
z
0
+
2
z
0
⋅
x
−
q
2
2
z
0
)
=
0.
{\displaystyle \left(x^{2}+p+z_{0}-{\sqrt {2z_{0}}}\cdot x+{\frac {q}{2{\sqrt {2z_{0}}}}}\right)\left(x^{2}+p+z_{0}+{\sqrt {2z_{0}}}\cdot x-{\frac {q}{2{\sqrt {2z_{0}}}}}\right)=0.}
Iš čia nesunku matyti, kad arba
x
2
−
2
z
0
⋅
x
+
(
p
+
z
0
+
q
2
2
z
0
)
=
0
{\displaystyle x^{2}-{\sqrt {2z_{0}}}\cdot x+\left(p+z_{0}+{\frac {q}{2{\sqrt {2z_{0}}}}}\right)=0}
x
2
+
2
z
0
⋅
x
+
(
p
+
z
0
−
q
2
2
z
0
)
=
0.
{\displaystyle x^{2}+{\sqrt {2z_{0}}}\cdot x+\left(p+z_{0}-{\frac {q}{2{\sqrt {2z_{0}}}}}\right)=0.}
f
(
x
)
=
x
4
+
2
p
x
2
+
q
x
+
r
=
0
{\displaystyle f(x)=x^{4}+2px^{2}+qx+r=0}
Taigi,
x
1
,
2
=
1
2
⋅
[
−
(
−
2
z
0
)
±
2
z
0
−
4
(
p
+
z
0
+
q
2
2
z
0
)
]
=
z
0
2
±
z
0
2
−
(
p
+
z
0
+
q
2
2
z
0
)
;
{\displaystyle x_{1,2}={\frac {1}{2}}\cdot \left[-(-{\sqrt {2z_{0}}})\pm {\sqrt {2z_{0}-4\left(p+z_{0}+{\frac {q}{2{\sqrt {2z_{0}}}}}\right)}}\right]={\frac {\sqrt {z_{0}}}{\sqrt {2}}}\pm {\sqrt {{\frac {z_{0}}{2}}-\left(p+z_{0}+{\frac {q}{2{\sqrt {2z_{0}}}}}\right)}};}
x
3
,
4
=
1
2
⋅
[
−
2
z
0
±
2
z
0
−
4
(
p
+
z
0
−
q
2
2
z
0
)
]
=
−
z
0
2
±
z
0
2
−
p
−
z
0
+
q
2
2
z
0
.
{\displaystyle x_{3,4}={\frac {1}{2}}\cdot \left[-{\sqrt {2z_{0}}}\pm {\sqrt {2z_{0}-4\left(p+z_{0}-{\frac {q}{2{\sqrt {2z_{0}}}}}\right)}}\right]=-{\frac {\sqrt {z_{0}}}{\sqrt {2}}}\pm {\sqrt {{\frac {z_{0}}{2}}-p-z_{0}+{\frac {q}{2{\sqrt {2z_{0}}}}}}}.}
[ keisti ] Pagalbinę kubinę lygtį (ketvirto laipsnio lygties rezolventę)
8
z
3
+
16
p
z
2
+
8
(
p
2
−
r
)
z
−
q
2
=
0
,
{\displaystyle 8z^{3}+16pz^{2}+8(p^{2}-r)z-q^{2}=0,}
z
3
+
2
p
z
2
+
(
p
2
−
r
)
z
−
q
2
8
=
0
{\displaystyle z^{3}+2pz^{2}+(p^{2}-r)z-{\frac {q^{2}}{8}}=0}
sutvarkysime padarę keitinį
z
=
w
−
2
p
3
.
{\displaystyle z=w-{\frac {2p}{3}}.}
z
3
+
2
p
z
2
+
(
p
2
−
r
)
z
−
q
2
8
=
0
{\displaystyle z^{3}+2pz^{2}+(p^{2}-r)z-{\frac {q^{2}}{8}}=0}
(
w
−
2
p
3
)
3
+
2
p
(
w
−
2
p
3
)
2
+
(
p
2
−
r
)
(
w
−
2
p
3
)
−
q
2
8
=
0
,
{\displaystyle \left(w-{\frac {2p}{3}}\right)^{3}+2p\left(w-{\frac {2p}{3}}\right)^{2}+(p^{2}-r)\left(w-{\frac {2p}{3}}\right)-{\frac {q^{2}}{8}}=0,}
(
w
3
−
3
w
2
2
p
3
+
3
w
(
2
p
3
)
2
−
(
2
p
3
)
3
)
+
2
p
(
w
2
−
2
w
2
p
3
+
(
2
p
3
)
2
)
+
(
p
2
−
r
)
(
w
−
2
p
3
)
−
q
2
8
=
0
,
{\displaystyle \left(w^{3}-3w^{2}{\frac {2p}{3}}+3w({\frac {2p}{3}})^{2}-({\frac {2p}{3}})^{3}\right)+2p\left(w^{2}-2w{\frac {2p}{3}}+({\frac {2p}{3}})^{2}\right)+(p^{2}-r)\left(w-{\frac {2p}{3}}\right)-{\frac {q^{2}}{8}}=0,}
(
w
3
−
w
2
2
p
+
3
w
4
p
2
9
−
8
p
3
27
)
+
2
p
(
w
2
−
4
w
p
3
+
4
p
2
9
)
+
(
p
2
−
r
)
w
−
(
p
2
−
r
)
2
p
3
−
q
2
8
=
0
,
{\displaystyle \left(w^{3}-w^{2}2p+3w{\frac {4p^{2}}{9}}-{\frac {8p^{3}}{27}}\right)+2p\left(w^{2}-{\frac {4wp}{3}}+{\frac {4p^{2}}{9}}\right)+(p^{2}-r)w-(p^{2}-r){\frac {2p}{3}}-{\frac {q^{2}}{8}}=0,}
w
3
−
w
2
2
p
+
4
w
p
2
3
−
8
p
3
27
+
2
p
w
2
−
8
w
p
2
3
+
8
p
3
9
+
w
p
2
−
w
r
−
2
p
3
3
+
2
p
r
3
−
q
2
8
=
0
,
{\displaystyle w^{3}-w^{2}2p+{\frac {4wp^{2}}{3}}-{\frac {8p^{3}}{27}}+2pw^{2}-{\frac {8wp^{2}}{3}}+{\frac {8p^{3}}{9}}+wp^{2}-wr-{\frac {2p^{3}}{3}}+{\frac {2pr}{3}}-{\frac {q^{2}}{8}}=0,}
w
3
+
4
w
p
2
3
−
8
w
p
2
3
+
w
p
2
−
w
r
−
8
p
3
27
+
8
p
3
9
−
2
p
3
3
+
2
p
r
3
−
q
2
8
=
0
,
{\displaystyle w^{3}+{\frac {4wp^{2}}{3}}-{\frac {8wp^{2}}{3}}+wp^{2}-wr-{\frac {8p^{3}}{27}}+{\frac {8p^{3}}{9}}-{\frac {2p^{3}}{3}}+{\frac {2pr}{3}}-{\frac {q^{2}}{8}}=0,}
w
3
−
4
w
p
2
3
+
w
(
p
2
−
r
)
+
−
8
p
3
+
3
⋅
8
p
3
−
9
⋅
2
p
3
27
+
2
p
r
3
−
q
2
8
=
0
,
{\displaystyle w^{3}-{\frac {4wp^{2}}{3}}+w(p^{2}-r)+{\frac {-8p^{3}+3\cdot 8p^{3}-9\cdot 2p^{3}}{27}}+{\frac {2pr}{3}}-{\frac {q^{2}}{8}}=0,}
w
3
+
−
4
w
p
2
+
3
w
(
p
2
−
r
)
3
+
−
8
p
3
+
24
p
3
−
18
p
3
27
+
2
p
r
3
−
q
2
8
=
0
,
{\displaystyle w^{3}+{\frac {-4wp^{2}+3w(p^{2}-r)}{3}}+{\frac {-8p^{3}+24p^{3}-18p^{3}}{27}}+{\frac {2pr}{3}}-{\frac {q^{2}}{8}}=0,}
w
3
+
−
4
w
p
2
+
3
w
p
2
−
3
w
r
3
+
−
2
p
3
27
+
2
p
r
3
−
q
2
8
=
0
,
{\displaystyle w^{3}+{\frac {-4wp^{2}+3wp^{2}-3wr}{3}}+{\frac {-2p^{3}}{27}}+{\frac {2pr}{3}}-{\frac {q^{2}}{8}}=0,}
w
3
+
−
w
p
2
−
3
w
r
3
−
2
p
3
27
+
2
p
r
3
−
q
2
8
=
0
,
{\displaystyle w^{3}+{\frac {-wp^{2}-3wr}{3}}-{\frac {2p^{3}}{27}}+{\frac {2pr}{3}}-{\frac {q^{2}}{8}}=0,}
w
3
−
p
2
+
3
r
3
w
−
2
p
3
27
+
2
p
r
3
−
q
2
8
=
0.
{\displaystyle w^{3}-{\frac {p^{2}+3r}{3}}w-{\frac {2p^{3}}{27}}+{\frac {2pr}{3}}-{\frac {q^{2}}{8}}=0.}
Gavome redukuotą kubinę lygtį
w
3
+
m
w
+
n
=
0
,
{\displaystyle w^{3}+mw+n=0,}
kur
m
=
−
p
2
+
3
r
3
,
{\displaystyle m=-{\frac {p^{2}+3r}{3}},}
n
=
−
2
p
3
27
+
2
p
r
3
−
q
2
8
.
{\displaystyle n=-{\frac {2p^{3}}{27}}+{\frac {2pr}{3}}-{\frac {q^{2}}{8}}.}
Ketvirto laipsnio lygties sprendimas antru budu [ keisti ] Imame redukuota lygtį:
f
(
x
)
=
x
4
+
2
p
x
2
+
q
x
+
r
=
0.
{\displaystyle f(x)=x^{4}+2px^{2}+qx+r=0.}
Į lygtį
x
4
+
2
p
x
2
+
q
x
+
r
=
0
{\displaystyle x^{4}+2px^{2}+qx+r=0}
x įvedame tris nežinomuosius, kuriuos vėliau susiesime dviem lygtim. Imame
2
x
=
u
+
v
+
w
.
{\displaystyle 2x=u+v+w.}
Išskaičiuojame:
(
2
x
)
2
=
4
x
2
=
(
u
+
v
+
w
)
(
u
+
v
+
w
)
=
u
2
+
u
v
+
u
w
+
v
u
+
v
2
+
v
w
+
w
u
+
w
v
+
w
2
=
u
2
+
v
2
+
w
2
+
2
(
u
v
+
u
w
+
v
w
)
;
{\displaystyle (2x)^{2}=4x^{2}=(u+v+w)(u+v+w)=u^{2}+uv+uw+vu+v^{2}+vw+wu+wv+w^{2}=u^{2}+v^{2}+w^{2}+2(uv+uw+vw);}
(
2
x
)
4
=
(
4
x
2
)
2
=
16
x
4
=
(
u
2
+
v
2
+
w
2
+
2
(
u
v
+
u
w
+
v
w
)
)
2
=
{\displaystyle (2x)^{4}=(4x^{2})^{2}=16x^{4}=(u^{2}+v^{2}+w^{2}+2(uv+uw+vw))^{2}=}
=
u
4
+
u
2
v
2
+
u
2
w
2
+
2
u
2
(
u
v
+
u
w
+
v
w
)
+
v
2
u
2
+
v
4
+
v
2
w
2
+
2
v
2
(
u
v
+
u
w
+
v
w
)
+
w
2
u
2
+
w
2
v
2
+
w
4
+
2
w
2
(
u
v
+
u
w
+
v
w
)
+
2
(
u
v
+
u
w
+
v
w
)
(
u
2
+
v
2
+
w
2
)
+
4
(
u
v
+
u
w
+
v
w
)
2
=
{\displaystyle =u^{4}+u^{2}v^{2}+u^{2}w^{2}+2u^{2}(uv+uw+vw)+v^{2}u^{2}+v^{4}+v^{2}w^{2}+2v^{2}(uv+uw+vw)+w^{2}u^{2}+w^{2}v^{2}+w^{4}+2w^{2}(uv+uw+vw)+2(uv+uw+vw)(u^{2}+v^{2}+w^{2})+4(uv+uw+vw)^{2}=}
=
u
4
+
v
4
+
w
4
+
2
(
u
2
v
2
+
u
2
w
2
+
v
2
w
2
)
+
2
(
u
2
+
v
2
+
w
2
)
(
u
v
+
u
w
+
v
w
)
+
2
(
u
v
+
u
w
+
v
w
)
(
u
2
+
v
2
+
w
2
)
+
4
(
u
2
v
2
+
u
2
v
w
+
u
v
2
w
+
u
2
w
v
+
u
2
w
2
+
u
v
w
2
+
u
v
2
w
+
u
v
w
2
+
v
2
w
2
)
=
{\displaystyle =u^{4}+v^{4}+w^{4}+2(u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2})+2(u^{2}+v^{2}+w^{2})(uv+uw+vw)+2(uv+uw+vw)(u^{2}+v^{2}+w^{2})+4(u^{2}v^{2}+u^{2}vw+uv^{2}w+u^{2}wv+u^{2}w^{2}+uvw^{2}+uv^{2}w+uvw^{2}+v^{2}w^{2})=}
=
(
u
2
+
v
2
+
w
2
)
2
+
4
(
u
2
+
v
2
+
w
2
)
(
u
v
+
u
w
+
v
w
)
+
4
(
u
2
v
2
+
u
2
w
2
+
v
2
w
2
+
2
(
u
2
v
w
+
u
v
2
w
+
u
v
w
2
)
)
=
{\displaystyle =(u^{2}+v^{2}+w^{2})^{2}+4(u^{2}+v^{2}+w^{2})(uv+uw+vw)+4(u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2}+2(u^{2}vw+uv^{2}w+uvw^{2}))=}
=
(
u
2
+
v
2
+
w
2
)
2
+
4
(
u
2
+
v
2
+
w
2
)
(
u
v
+
u
w
+
v
w
)
+
4
(
u
2
v
2
+
u
2
w
2
+
v
2
w
2
)
+
8
u
v
w
(
u
+
v
+
w
)
.
{\displaystyle =(u^{2}+v^{2}+w^{2})^{2}+4(u^{2}+v^{2}+w^{2})(uv+uw+vw)+4(u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2})+8uvw(u+v+w).}
Įstatę šias reikšmes į lygtį
x
4
+
2
p
x
2
+
q
x
+
r
=
0
{\displaystyle x^{4}+2px^{2}+qx+r=0}
16
x
4
+
32
p
x
2
+
16
q
x
+
16
r
=
0
,
{\displaystyle 16x^{4}+32px^{2}+16qx+16r=0,}
[
(
u
2
+
v
2
+
w
2
)
2
+
4
(
u
2
+
v
2
+
w
2
)
(
u
v
+
u
w
+
v
w
)
+
4
(
u
2
v
2
+
u
2
w
2
+
v
2
w
2
)
+
8
u
v
w
(
u
+
v
+
w
)
]
+
8
p
[
u
2
+
v
2
+
w
2
+
2
(
u
v
+
u
w
+
v
w
)
]
+
8
q
[
u
+
v
+
w
]
+
16
r
=
0
,
{\displaystyle [(u^{2}+v^{2}+w^{2})^{2}+4(u^{2}+v^{2}+w^{2})(uv+uw+vw)+4(u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2})+8uvw(u+v+w)]+8p[u^{2}+v^{2}+w^{2}+2(uv+uw+vw)]+8q[u+v+w]+16r=0,}
(
u
2
+
v
2
+
w
2
)
2
+
4
(
u
2
+
v
2
+
w
2
)
(
u
v
+
u
w
+
v
w
)
+
4
(
u
2
v
2
+
u
2
w
2
+
v
2
w
2
)
+
8
u
v
w
(
u
+
v
+
w
)
+
8
p
(
u
2
+
v
2
+
w
2
)
+
16
p
(
u
v
+
u
w
+
v
w
)
+
8
q
(
u
+
v
+
w
)
+
16
r
=
0
,
{\displaystyle (u^{2}+v^{2}+w^{2})^{2}+4(u^{2}+v^{2}+w^{2})(uv+uw+vw)+4(u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2})+8uvw(u+v+w)+8p(u^{2}+v^{2}+w^{2})+16p(uv+uw+vw)+8q(u+v+w)+16r=0,}
(
u
2
+
v
2
+
w
2
)
2
+
4
(
u
2
+
v
2
+
w
2
+
4
p
)
(
u
v
+
u
w
+
v
w
)
+
4
(
u
2
v
2
+
u
2
w
2
+
v
2
w
2
)
+
8
(
u
v
w
+
q
)
(
u
+
v
+
w
)
+
8
p
(
u
2
+
v
2
+
w
2
)
+
16
r
=
0.
{\displaystyle (u^{2}+v^{2}+w^{2})^{2}+4(u^{2}+v^{2}+w^{2}+4p)(uv+uw+vw)+4(u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2})+8(uvw+q)(u+v+w)+8p(u^{2}+v^{2}+w^{2})+16r=0.}
Dabar reikalaujame, kad
u
2
+
v
2
+
w
2
+
4
p
=
0
,
{\displaystyle u^{2}+v^{2}+w^{2}+4p=0,}
u
v
w
+
q
=
0.
{\displaystyle uvw+q=0.}
Įvedę šias sąlygas turėsime lygtį:
(
u
2
+
v
2
+
w
2
)
2
+
4
(
u
2
v
2
+
u
2
w
2
+
v
2
w
2
)
+
8
p
(
u
2
+
v
2
+
w
2
)
+
16
r
=
0
,
{\displaystyle (u^{2}+v^{2}+w^{2})^{2}+4(u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2})+8p(u^{2}+v^{2}+w^{2})+16r=0,}
(
−
4
p
)
2
+
4
(
u
2
v
2
+
u
2
w
2
+
v
2
w
2
)
+
8
p
⋅
(
−
4
p
)
+
16
r
=
0
,
{\displaystyle (-4p)^{2}+4(u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2})+8p\cdot (-4p)+16r=0,}
16
p
2
+
4
(
u
2
v
2
+
u
2
w
2
+
v
2
w
2
)
−
32
p
2
+
16
r
=
0
,
{\displaystyle 16p^{2}+4(u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2})-32p^{2}+16r=0,}
4
(
u
2
v
2
+
u
2
w
2
+
v
2
w
2
)
−
16
p
2
+
16
r
=
0
,
{\displaystyle 4(u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2})-16p^{2}+16r=0,}
4
(
u
2
v
2
+
u
2
w
2
+
v
2
w
2
)
=
16
p
2
−
16
r
,
{\displaystyle 4(u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2})=16p^{2}-16r,}
u
2
v
2
+
u
2
w
2
+
v
2
w
2
=
4
(
p
2
−
r
)
.
{\displaystyle u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2}=4(p^{2}-r).}
Pagaliau, vietoje lygties
x
4
+
2
p
x
2
+
q
x
+
r
=
0
{\displaystyle x^{4}+2px^{2}+qx+r=0}
u
2
+
v
2
+
w
2
=
−
4
p
,
{\displaystyle u^{2}+v^{2}+w^{2}=-4p,}
u
2
v
2
+
u
2
w
2
+
v
2
w
2
=
4
(
p
2
−
r
)
,
{\displaystyle u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2}=4(p^{2}-r),}
u
v
w
=
−
q
.
{\displaystyle uvw=-q.}
Šią sistemą spręsime panašiai, kaip sprendžiama trečio laipsnio lygčių sistema. Pakėlę lygtį
u
v
w
=
−
q
{\displaystyle uvw=-q}
u
2
v
2
w
2
=
q
2
.
{\displaystyle u^{2}v^{2}w^{2}=q^{2}.}
Pagal Vijeto teoremą, iš sistemos lygčių nesunku pastebėti, kad
u
2
{\displaystyle u^{2}}
v
2
{\displaystyle v^{2}}
w
2
{\displaystyle w^{2}}
y
3
+
4
p
y
2
+
4
(
p
2
−
r
)
y
−
q
2
=
0
{\displaystyle y^{3}+4py^{2}+4(p^{2}-r)y-q^{2}=0}
šaknys. Ši lygtis taip pat vadinama ketvirto laipsnio lygties
x
4
+
2
p
x
2
+
q
x
+
r
=
0
{\displaystyle x^{4}+2px^{2}+qx+r=0}
y
1
{\displaystyle y_{1}}
y
2
{\displaystyle y_{2}}
y
3
{\displaystyle y_{3}}
u
2
{\displaystyle u^{2}}
v
2
{\displaystyle v^{2}}
w
2
{\displaystyle w^{2}}
u
2
+
v
2
+
w
2
=
−
4
p
,
{\displaystyle u^{2}+v^{2}+w^{2}=-4p,}
u
2
v
2
+
u
2
w
2
+
v
2
w
2
=
4
(
p
2
−
r
)
{\displaystyle u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2}=4(p^{2}-r)}
u
v
w
=
−
q
{\displaystyle uvw=-q}
u , v ir w atžvilgiu, tai kurią lygčių
y
3
+
4
p
y
2
+
4
(
p
2
−
r
)
y
−
q
2
=
0
{\displaystyle y^{3}+4py^{2}+4(p^{2}-r)y-q^{2}=0}
u
2
{\displaystyle u^{2}}
v
2
{\displaystyle v^{2}}
w
2
{\displaystyle w^{2}}
x
4
+
2
p
x
2
+
q
x
+
r
=
0
{\displaystyle x^{4}+2px^{2}+qx+r=0}
u , v ir w , nes reikia tik ištraukti kvadratines šaknis iš
y
1
{\displaystyle y_{1}}
y
2
{\displaystyle y_{2}}
y
3
{\displaystyle y_{3}}
u
=
y
1
,
v
=
y
2
,
w
=
y
3
.
{\displaystyle u={\sqrt {y_{1}}},\quad v={\sqrt {y_{2}}},\quad w={\sqrt {y_{3}}}.}
Pagaliau įstatę u , v ir w reikšmes į lygybę
2
x
=
u
+
v
+
w
{\displaystyle 2x=u+v+w}
x
4
+
2
p
x
2
+
q
x
+
r
=
0
{\displaystyle x^{4}+2px^{2}+qx+r=0}
y
1
{\displaystyle {\sqrt {y_{1}}}}
u
0
{\displaystyle u_{0}}
y
2
,
{\displaystyle {\sqrt {y_{2}}},}
y
3
{\displaystyle {\sqrt {y_{3}}}}
v
0
{\displaystyle v_{0}}
w
0
{\displaystyle w_{0}}
u
0
v
0
w
0
=
−
q
,
{\displaystyle u_{0}v_{0}w_{0}=-q,}
x
1
=
1
2
(
u
0
+
v
0
+
w
0
)
,
{\displaystyle x_{1}={\frac {1}{2}}(u_{0}+v_{0}+w_{0}),}
x
2
=
1
2
(
u
0
−
v
0
−
w
0
)
,
{\displaystyle x_{2}={\frac {1}{2}}(u_{0}-v_{0}-w_{0}),}
x
3
=
1
2
(
−
u
0
+
v
0
−
w
0
)
,
{\displaystyle x_{3}={\frac {1}{2}}(-u_{0}+v_{0}-w_{0}),}
x
4
=
1
2
(
−
u
0
−
v
0
+
w
0
)
,
{\displaystyle x_{4}={\frac {1}{2}}(-u_{0}-v_{0}+w_{0}),}
arba, pavyzdžiui,
x
1
=
1
2
(
u
0
+
v
0
−
w
0
)
,
{\displaystyle x_{1}={\frac {1}{2}}(u_{0}+v_{0}-w_{0}),}
x
2
=
1
2
(
u
0
−
v
0
+
w
0
)
,
{\displaystyle x_{2}={\frac {1}{2}}(u_{0}-v_{0}+w_{0}),}
x
3
=
1
2
(
−
u
0
+
v
0
+
w
0
)
,
{\displaystyle x_{3}={\frac {1}{2}}(-u_{0}+v_{0}+w_{0}),}
x
4
=
1
2
(
−
u
0
−
v
0
−
w
0
)
.
{\displaystyle x_{4}={\frac {1}{2}}(-u_{0}-v_{0}-w_{0}).}
Abi šios sistemos yra lygiavertės (tapatingos), nes jos gaunamos viena iš kitos, pakeitus visų u , v ir w ženklus priešingais. Tai nepakeičia jų reikšmių, bet tik pačius u , v ir w pažymėjimus. Rasime lygties
x
4
+
x
2
+
x
−
1
=
0
{\displaystyle x^{4}+x^{2}+x-1=0}
Turime, kad
2
p
=
1
{\displaystyle 2p=1}
p
=
1
2
,
{\displaystyle p={\frac {1}{2}},}
q
=
1
{\displaystyle q=1}
r
=
−
1
{\displaystyle r=-1}
Turime
2
x
=
u
+
v
+
w
{\displaystyle 2x=u+v+w}
ir lygčių sistemą
{
u
2
+
v
2
+
w
2
=
−
4
p
,
u
2
v
2
+
u
2
w
2
+
v
2
w
2
=
4
(
p
2
−
r
)
,
u
v
w
=
−
q
;
{\displaystyle {\begin{cases}u^{2}+v^{2}+w^{2}=-4p,&\\u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2}=4(p^{2}-r),&\\uvw=-q;&\end{cases}}}
pakeliame trečią eilutę kvadratu, kad atitiktų Vijeto teoremą:
{
u
2
+
v
2
+
w
2
=
−
4
p
,
u
2
v
2
+
u
2
w
2
+
v
2
w
2
=
4
(
p
2
−
r
)
,
u
2
v
2
w
2
=
q
2
.
{\displaystyle {\begin{cases}u^{2}+v^{2}+w^{2}=-4p,&\\u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2}=4(p^{2}-r),&\\u^{2}v^{2}w^{2}=q^{2}.&\end{cases}}}
Tada įstatę į lygčių sistemą p , q ir r reikšmes, gauname:
u
2
+
v
2
+
w
2
=
−
4
⋅
1
2
=
−
2
;
{\displaystyle u^{2}+v^{2}+w^{2}=-4\cdot {\frac {1}{2}}=-2;}
u
2
v
2
+
u
2
w
2
+
v
2
w
2
=
4
(
(
1
2
)
2
−
(
−
1
)
)
=
4
(
1
4
+
1
)
=
1
+
4
=
5
;
{\displaystyle u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2}=4(\left({\frac {1}{2}}\right)^{2}-(-1))=4({\frac {1}{4}}+1)=1+4=5;}
u
2
v
2
w
2
=
(
−
1
)
2
=
1.
{\displaystyle u^{2}v^{2}w^{2}=(-1)^{2}=1.}
Sudarome kubinę pagalbinę lygtį, pritaikę Vijeto teoremą:
y
3
−
(
u
2
+
v
2
+
w
2
)
y
2
+
(
u
2
v
2
+
u
2
w
2
+
v
2
w
2
)
y
−
u
2
v
2
w
2
=
0
,
{\displaystyle y^{3}-(u^{2}+v^{2}+w^{2})y^{2}+(u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2})y-u^{2}v^{2}w^{2}=0,}
y
3
−
(
−
2
)
y
2
+
5
y
−
1
=
0
,
{\displaystyle y^{3}-(-2)y^{2}+5y-1=0,}
y
3
+
2
y
2
+
5
y
−
1
=
0.
{\displaystyle y^{3}+2y^{2}+5y-1=0.}
Iš kubinės lygties kalkuliatoriaus https://www.calculatorsoup.com/calculators/algebra/cubicequation.php internete, randame, kad
y
1
=
0.1850373752486395
;
{\displaystyle y_{1}=0.1850373752486395;}
y
2
=
−
1.0925186876243198
+
2.0520030453017895
i
;
{\displaystyle y_{2}=-1.0925186876243198+2.0520030453017895i;}
y
3
=
−
1.0925186876243198
−
2.0520030453017895
i
.
{\displaystyle y_{3}=-1.0925186876243198-2.0520030453017895i.}
Tada:
2
x
=
u
+
v
+
w
=
y
1
+
y
2
+
y
3
.
{\displaystyle 2x=u+v+w={\sqrt {y_{1}}}+{\sqrt {y_{2}}}+{\sqrt {y_{3}}}.}
Kad ištrauktume šaknį iš kompleksinio skaičiaus pasinaudosime formulėmis:
a
1
+
a
2
i
=
±
(
a
1
+
a
1
2
+
a
2
2
2
+
i
−
a
1
+
a
1
2
+
a
2
2
2
)
,
k
a
i
a
2
>
0.
{\displaystyle {\sqrt {a_{1}+a_{2}i}}=\pm \left({\sqrt {\frac {a_{1}+{\sqrt {a_{1}^{2}+a_{2}^{2}}}}{2}}}+i{\sqrt {\frac {-a_{1}+{\sqrt {a_{1}^{2}+a_{2}^{2}}}}{2}}}\right),\quad kai\quad a_{2}>0.}
a
1
+
a
2
i
=
±
(
a
1
+
a
1
2
+
a
2
2
2
−
i
−
a
1
+
a
1
2
+
a
2
2
2
)
,
k
a
i
a
2
<
0.
{\displaystyle {\sqrt {a_{1}+a_{2}i}}=\pm \left({\sqrt {\frac {a_{1}+{\sqrt {a_{1}^{2}+a_{2}^{2}}}}{2}}}-i{\sqrt {\frac {-a_{1}+{\sqrt {a_{1}^{2}+a_{2}^{2}}}}{2}}}\right),\quad kai\;\;\;\;a_{2}<0.}
Taigi, gauname:
u
=
y
1
=
0.185037375
=
0.430159708
;
{\displaystyle u={\sqrt {y_{1}}}={\sqrt {0.185037375}}=0.430159708;}
v
=
y
2
=
−
1.092518688
+
2.052003045
i
=
{\displaystyle v={\sqrt {y_{2}}}={\sqrt {-1.092518688+2.052003045i}}=}
=
±
(
−
1.092518688
+
(
−
1.092518688
)
2
+
2.052003045
2
2
+
i
−
(
−
1.092518688
)
+
(
−
1.092518688
)
2
+
2.052003045
2
2
)
=
{\displaystyle =\pm \left({\sqrt {\frac {-1.092518688+{\sqrt {(-1.092518688)^{2}+2.052003045^{2}}}}{2}}}+i{\sqrt {\frac {-(-1.092518688)+{\sqrt {(-1.092518688)^{2}+2.052003045^{2}}}}{2}}}\right)=}
=
±
(
−
1.092518688
+
1.193597084
+
4.210716497
2
+
i
1.092518688
+
1.193597084
+
4.210716497
2
)
=
{\displaystyle =\pm \left({\sqrt {\frac {-1.092518688+{\sqrt {1.193597084+4.210716497}}}{2}}}+i{\sqrt {\frac {1.092518688+{\sqrt {1.193597084+4.210716497}}}{2}}}\right)=}
=
±
(
−
1.092518688
+
2.324717957
2
+
i
1.092518688
+
2.324717957
2
)
=
{\displaystyle =\pm \left({\sqrt {\frac {-1.092518688+2.324717957}{2}}}+i{\sqrt {\frac {1.092518688+2.324717957}{2}}}\right)=}
=
±
(
0.616099634
+
i
1.708618323
)
=
±
(
0.784920145
+
1.307141279
i
)
;
{\displaystyle =\pm \left({\sqrt {0.616099634}}+i{\sqrt {1.708618323}}\right)=\pm \left(0.784920145+1.307141279i\right);}
w
=
y
3
=
−
1.092518688
−
2.052003045
i
=
±
(
0.784920145
−
1.307141279
i
)
.
{\displaystyle w={\sqrt {y_{3}}}={\sqrt {-1.092518688-2.052003045i}}=\pm (0.784920145-1.307141279i).}
Toliau gauname lygties
x
4
+
x
2
+
x
−
1
=
0
{\displaystyle x^{4}+x^{2}+x-1=0}
x
0
=
1
2
(
±
0.430159708
±
(
0.784920145
+
1.307141279
i
)
±
(
0.784920145
−
1.307141279
i
)
)
;
{\displaystyle x_{0}={\frac {1}{2}}(\pm 0.430159708\pm \left(0.784920145+1.307141279i\right)\pm (0.784920145-1.307141279i));}
x
1
=
1
2
(
u
0
+
v
0
+
w
0
)
=
1
2
(
0.430159708
+
(
0.784920145
+
1.307141279
i
)
+
(
0.784920145
−
1.307141279
i
)
)
=
{\displaystyle x_{1}={\frac {1}{2}}(u_{0}+v_{0}+w_{0})={\frac {1}{2}}(0.430159708+\left(0.784920145+1.307141279i\right)+(0.784920145-1.307141279i))=}
=
1
2
(
0.430159708
+
2
⋅
0.784920145
)
=
1.999999998
2
=
1
;
{\displaystyle ={\frac {1}{2}}(0.430159708+2\cdot 0.784920145)={\frac {1.999999998}{2}}=1;}
x
2
=
1
2
(
u
0
−
v
0
−
w
0
)
=
1
2
(
0.430159708
−
(
0.784920145
+
1.307141279
i
)
−
(
0.784920145
−
1.307141279
i
)
)
=
{\displaystyle x_{2}={\frac {1}{2}}(u_{0}-v_{0}-w_{0})={\frac {1}{2}}(0.430159708-\left(0.784920145+1.307141279i\right)-(0.784920145-1.307141279i))=}
=
1
2
(
0.430159708
−
2
⋅
0.784920145
)
=
−
0.569840291
;
{\displaystyle ={\frac {1}{2}}(0.430159708-2\cdot 0.784920145)=-0.569840291;}
x
3
=
1
2
(
−
u
0
+
v
0
−
w
0
)
=
1
2
(
−
0.430159708
+
(
0.784920145
+
1.307141279
i
)
−
(
0.784920145
−
1.307141279
i
)
)
=
{\displaystyle x_{3}={\frac {1}{2}}(-u_{0}+v_{0}-w_{0})={\frac {1}{2}}(-0.430159708+\left(0.784920145+1.307141279i\right)-(0.784920145-1.307141279i))=}
=
1
2
(
−
0.430159708
+
1.307141279
i
+
1.307141279
i
)
=
−
0.215079854
+
1.307141279
i
;
{\displaystyle ={\frac {1}{2}}(-0.430159708+1.307141279i+1.307141279i)=-0.215079854+1.307141279i;}
x
4
=
1
2
(
−
u
0
−
v
0
+
w
0
)
=
1
2
(
−
0.430159708
−
(
0.784920145
+
1.307141279
i
)
+
(
0.784920145
−
1.307141279
i
)
)
=
{\displaystyle x_{4}={\frac {1}{2}}(-u_{0}-v_{0}+w_{0})={\frac {1}{2}}(-0.430159708-\left(0.784920145+1.307141279i\right)+(0.784920145-1.307141279i))=}
=
1
2
(
−
0.430159708
−
1.307141279
i
−
1.307141279
i
)
=
−
0.215079854
−
1.307141279
i
;
{\displaystyle ={\frac {1}{2}}(-0.430159708-1.307141279i-1.307141279i)=-0.215079854-1.307141279i;}
x
5
=
1
2
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0
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w
0
)
=
1
2
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0.430159708
+
(
0.784920145
+
1.307141279
i
)
−
(
0.784920145
−
1.307141279
i
)
)
=
{\displaystyle x_{5}={\frac {1}{2}}(u_{0}+v_{0}-w_{0})={\frac {1}{2}}(0.430159708+\left(0.784920145+1.307141279i\right)-(0.784920145-1.307141279i))=}
=
1
2
(
0.430159708
+
2
⋅
1.307141279
i
)
=
0.215079854
+
1.307141279
i
;
{\displaystyle ={\frac {1}{2}}(0.430159708+2\cdot 1.307141279i)=0.215079854+1.307141279i;}
x
6
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1
2
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2
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0.430159708
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0.784920145
+
1.307141279
i
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+
(
0.784920145
−
1.307141279
i
)
)
=
{\displaystyle x_{6}={\frac {1}{2}}(u_{0}-v_{0}+w_{0})={\frac {1}{2}}(0.430159708-\left(0.784920145+1.307141279i\right)+(0.784920145-1.307141279i))=}
=
1
2
(
0.430159708
+
2
⋅
(
−
1.307141279
i
)
)
=
0.215079854
−
1.307141279
i
;
{\displaystyle ={\frac {1}{2}}(0.430159708+2\cdot (-1.307141279i))=0.215079854-1.307141279i;}
x
7
=
1
2
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−
u
0
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v
0
+
w
0
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1
2
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−
0.430159708
+
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0.784920145
+
1.307141279
i
)
+
(
0.784920145
−
1.307141279
i
)
)
=
{\displaystyle x_{7}={\frac {1}{2}}(-u_{0}+v_{0}+w_{0})={\frac {1}{2}}(-0.430159708+\left(0.784920145+1.307141279i\right)+(0.784920145-1.307141279i))=}
=
1
2
(
−
0.430159708
+
2
⋅
0.784920145
)
=
−
0.215079854
+
0.784920145
=
0.569840291
;
{\displaystyle ={\frac {1}{2}}(-0.430159708+2\cdot 0.784920145)=-0.215079854+0.784920145=0.569840291;}
x
8
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2
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v
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w
0
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=
1
2
(
−
0.430159708
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(
0.784920145
+
1.307141279
i
)
−
(
0.784920145
−
1.307141279
i
)
)
=
{\displaystyle x_{8}={\frac {1}{2}}(-u_{0}-v_{0}-w_{0})={\frac {1}{2}}(-0.430159708-\left(0.784920145+1.307141279i\right)-(0.784920145-1.307141279i))=}
=
1
2
(
−
0.430159708
−
2
⋅
0.784920145
)
=
−
0.215079854
−
0.784920145
=
−
1.
{\displaystyle ={\frac {1}{2}}(-0.430159708-2\cdot 0.784920145)=-0.215079854-0.784920145=-1.}
Pusė iš šių sprendinių neteisingi. Iš realiųjų sprendinių neteisingi sprendiniai yra
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
u
0
v
0
w
0
=
−
q
.
{\displaystyle u_{0}v_{0}w_{0}=-q.}
x
1
=
−
x
8
,
x
2
=
−
x
7
,
x
3
=
−
x
6
,
x
4
=
−
x
5
.
{\displaystyle x_{1}=-x_{8},\;\;x_{2}=-x_{7},\;\;x_{3}=-x_{6},\;\;x_{4}=-x_{5}.}
Visada arba pirmi keturi sprendiniai teisingi, arba paskutiniai keturi sprendiniai teisingi. Nes pirmies keturiems sprendiniams:
u
0
v
0
w
0
=
u
0
v
0
w
0
,
u
0
(
−
v
0
)
(
−
w
0
)
=
u
0
v
0
w
0
,
(
−
u
0
)
v
0
(
−
w
0
)
=
u
0
v
0
w
0
,
(
−
u
0
)
(
−
v
0
)
w
0
=
u
0
v
0
w
0
;
{\displaystyle u_{0}v_{0}w_{0}=u_{0}v_{0}w_{0},\;u_{0}(-v_{0})(-w_{0})=u_{0}v_{0}w_{0},\;(-u_{0})v_{0}(-w_{0})=u_{0}v_{0}w_{0},\;(-u_{0})(-v_{0})w_{0}=u_{0}v_{0}w_{0};}
o keturiems paskutiniams sprendiniams:
u
0
v
0
(
−
w
0
)
=
−
u
0
v
0
w
0
,
u
0
(
−
v
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)
w
0
=
−
u
0
v
0
w
0
,
(
−
u
0
)
v
0
w
0
=
−
u
0
v
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w
0
,
(
−
u
0
)
(
−
v
0
)
(
−
w
0
)
=
−
u
0
v
0
w
0
.
{\displaystyle u_{0}v_{0}(-w_{0})=-u_{0}v_{0}w_{0},\;u_{0}(-v_{0})w_{0}=-u_{0}v_{0}w_{0},\;(-u_{0})v_{0}w_{0}=-u_{0}v_{0}w_{0},\;(-u_{0})(-v_{0})(-w_{0})=-u_{0}v_{0}w_{0}.}
![{\displaystyle f(x)=(x^{2}+p+z)^{2}+qx+r-z^{2}-2x^{2}z-2pz-p^{2}=(x^{2}+p+z)^{2}-[2zx^{2}-qx+(z^{2}+2pz-r+p^{2})],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/435b51a14990e7259f266143dff4833c08e4816d)
![{\displaystyle f(x)=(x^{2}+p+z)^{2}-[2zx^{2}-qx+(z^{2}+2pz-r+p^{2})]=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/44cef80e6bad000d1721a5585877379eeff76ff8)
![{\displaystyle [2zx^{2}-qx+(z^{2}+2pz-r+p^{2})]=0;}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ef2fc683ce63929bb5f404e9cf4455847293e8e)
![{\displaystyle [2zx^{2}-qx+(z^{2}+2pz-r+p^{2})]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9276c0a23bd169de248a8274865c19fa56200f01)
![{\displaystyle w_{0}={\sqrt[{3}]{-{\frac {n}{2}}+{\sqrt {{\frac {n^{2}}{4}}+{\frac {m^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {n}{2}}-{\sqrt {{\frac {n^{2}}{4}}+{\frac {m^{3}}{27}}}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f4fc5d31d7e59b0a93ccc73360fad4315298ad5)
![{\displaystyle z_{0}={\sqrt[{3}]{-{\frac {n}{2}}+{\sqrt {{\frac {n^{2}}{4}}+{\frac {m^{3}}{27}}}}}}+{\sqrt[{3}]{-{\frac {n}{2}}-{\sqrt {{\frac {n^{2}}{4}}+{\frac {m^{3}}{27}}}}}}-{\frac {2p}{3}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e447682a278f4d88bea48911b3a71dfe4f1785e)
![{\displaystyle [2z_{0}x^{2}-qx+(z_{0}^{2}+2pz_{0}-r+p^{2})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d970e1abf57f0405958d2be2969d9e2f476c7b0)
![{\displaystyle [2z_{0}x^{2}-qx+(z_{0}^{2}+2pz_{0}-r+p^{2})]=2z_{0}\left(x-{\frac {q}{4z_{0}}}\right)\cdot \left(x-{\frac {q}{4z_{0}}}\right)=2z_{0}\left(x-{\frac {q}{4z_{0}}}\right)^{2}=\left[{\sqrt {2z_{0}}}\left(x-{\frac {q}{4z_{0}}}\right)\right]^{2}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb683447087ae43566984cdc09f9b83afa64a533)
![{\displaystyle f(x)=(x^{2}+p+z_{0})^{2}-[2z_{0}x^{2}-qx+(z_{0}^{2}+2pz_{0}-r+p^{2})]=(x^{2}+p+z_{0})^{2}-\left[{\sqrt {2z_{0}}}\left(x-{\frac {q}{4z_{0}}}\right)\right]^{2};}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f6bfc09662d44e0044350f24305daeaffebc4b0)
![{\displaystyle (x^{2}+p+z_{0})^{2}-\left[{\sqrt {2z_{0}}}\left(x-{\frac {q}{4z_{0}}}\right)\right]^{2}=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d6d86961d820422e7e46ea950a178b7261743e8)
![{\displaystyle x_{1,2}={\frac {1}{2}}\cdot \left[-(-{\sqrt {2z_{0}}})\pm {\sqrt {2z_{0}-4\left(p+z_{0}+{\frac {q}{2{\sqrt {2z_{0}}}}}\right)}}\right]={\frac {\sqrt {z_{0}}}{\sqrt {2}}}\pm {\sqrt {{\frac {z_{0}}{2}}-\left(p+z_{0}+{\frac {q}{2{\sqrt {2z_{0}}}}}\right)}};}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6bc3ce5fa77df14e4420ad8942b45f98b8e9a0e)
![{\displaystyle x_{3,4}={\frac {1}{2}}\cdot \left[-{\sqrt {2z_{0}}}\pm {\sqrt {2z_{0}-4\left(p+z_{0}-{\frac {q}{2{\sqrt {2z_{0}}}}}\right)}}\right]=-{\frac {\sqrt {z_{0}}}{\sqrt {2}}}\pm {\sqrt {{\frac {z_{0}}{2}}-p-z_{0}+{\frac {q}{2{\sqrt {2z_{0}}}}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a082f17af5320eba67224c8d7ddb6eeafaaa5ebd)
![{\displaystyle [(u^{2}+v^{2}+w^{2})^{2}+4(u^{2}+v^{2}+w^{2})(uv+uw+vw)+4(u^{2}v^{2}+u^{2}w^{2}+v^{2}w^{2})+8uvw(u+v+w)]+8p[u^{2}+v^{2}+w^{2}+2(uv+uw+vw)]+8q[u+v+w]+16r=0,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96c4b3a4749d197b25371c5a4f92b01fb80f384b)
