Ha
Δ
≥
0
{\displaystyle \Delta \geq 0}
(
B
=
0
)
{\displaystyle \left(B=0\right)}
(
A
>
0
)
{\displaystyle \left(A>0\right)}
(
C
=
A
2
4
)
{\displaystyle \left(C={\frac {{A}^{2}}{4}}\right)}
x
1
,
2
=
−
b
4
a
+
i
⋅
A
2
x
3
,
4
=
−
b
4
a
−
i
⋅
A
2
{\displaystyle {\begin{aligned}&{{x}_{1,2}}=-{\frac {b}{4a}}+i\cdot {\sqrt {\frac {A}{2}}}\\&{{x}_{3,4}}=-{\frac {b}{4a}}-i\cdot {\sqrt {\frac {A}{2}}}\\\end{aligned}}}
ellenkező esetben:
x
1
,
2
=
−
b
4
a
+
s
i
g
(
−
B
)
−
A
6
+
u
+
v
±
−
A
3
−
(
u
+
v
)
+
(
A
6
+
2
(
u
+
v
)
)
2
−
C
x
3
,
4
=
−
b
4
a
−
s
i
g
(
−
B
)
−
A
6
+
u
+
v
±
i
⋅
A
3
+
(
u
+
v
)
+
(
A
6
+
2
(
u
+
v
)
)
2
−
C
{\displaystyle {\begin{aligned}&{{x}_{1,2}}=-{\frac {b}{4a}}+sig\left(-B\right){\sqrt {-{\frac {A}{6}}+u+v}}\pm {\sqrt {-{\frac {A}{3}}-\left(u+v\right)+{\sqrt {{{\left({\frac {A}{6}}+2\left(u+v\right)\right)}^{2}}-C}}}}\\&{{x}_{3,4}}=-{\frac {b}{4a}}-sig\left(-B\right){\sqrt {-{\frac {A}{6}}+u+v}}\pm i\cdot {\sqrt {{\frac {A}{3}}+\left(u+v\right)+{\sqrt {{{\left({\frac {A}{6}}+2\left(u+v\right)\right)}^{2}}-C}}}}\\\end{aligned}}}
Ha
Δ
<
0
{\displaystyle \Delta <0}
(
C
>
A
2
4
)
{\displaystyle \left(C>{\frac {{A}^{2}}{4}}\right)}
(
A
>
0
)
{\displaystyle \left(A>0\right)}
x
1
,
2
=
−
b
4
a
−
s
i
g
(
−
B
)
Y
1
±
i
⋅
(
−
Y
2
+
−
Y
3
)
x
3
,
4
=
−
b
4
a
+
s
i
g
(
−
B
)
Y
1
±
i
⋅
(
−
Y
2
−
−
Y
3
)
{\displaystyle {\begin{aligned}&{{x}_{1,2}}=-{\frac {b}{4a}}-sig\left(-B\right){\sqrt {{Y}_{1}}}\pm i\cdot \left({\sqrt {-{{Y}_{2}}}}+{\sqrt {-{{Y}_{3}}}}\right)\\&{{x}_{3,4}}=-{\frac {b}{4a}}+sig\left(-B\right){\sqrt {{Y}_{1}}}\pm i\cdot \left({\sqrt {-{{Y}_{2}}}}-{\sqrt {-{{Y}_{3}}}}\right)\\\end{aligned}}}
ellenkező esetben mind a négy gyök valós:
x
1
,
2
=
−
b
4
a
+
s
i
g
(
−
B
)
Y
1
±
(
Y
2
+
Y
3
)
x
3
,
4
=
−
b
4
a
−
s
i
g
(
−
B
)
Y
1
±
(
Y
2
−
Y
3
)
{\displaystyle {\begin{aligned}&{{x}_{1,2}}=-{\frac {b}{4a}}+sig\left(-B\right){\sqrt {{Y}_{1}}}\pm \left({\sqrt {{Y}_{2}}}+{\sqrt {{Y}_{3}}}\right)\\&{{x}_{3,4}}=-{\frac {b}{4a}}-sig\left(-B\right){\sqrt {{Y}_{1}}}\pm \left({\sqrt {{Y}_{2}}}-{\sqrt {{Y}_{3}}}\right)\\\end{aligned}}}
A
=
−
3
b
2
8
a
2
+
c
a
{\displaystyle A=-{\frac {3{{b}^{2}}}{8{{a}^{2}}}}+{\frac {c}{a}}}
B
=
b
3
8
a
3
−
b
c
2
a
2
+
d
a
{\displaystyle B={\frac {{b}^{3}}{8{{a}^{3}}}}-{\frac {bc}{2{{a}^{2}}}}+{\frac {d}{a}}}
C
=
−
3
b
4
256
a
4
+
b
2
c
16
a
3
−
b
d
4
a
2
+
e
a
{\displaystyle C=-{\frac {3{{b}^{4}}}{256{{a}^{4}}}}+{\frac {{{b}^{2}}c}{16{{a}^{3}}}}-{\frac {bd}{4{{a}^{2}}}}+{\frac {e}{a}}}
P
=
−
A
2
48
−
C
4
{\displaystyle P=-{\frac {{A}^{2}}{48}}-{\frac {C}{4}}}
Q
=
−
A
3
864
−
B
2
64
+
A
C
24
{\displaystyle Q=-{\frac {{A}^{3}}{864}}-{\frac {{B}^{2}}{64}}+{\frac {AC}{24}}}
Δ
=
(
Q
2
)
2
+
(
P
3
)
3
{\displaystyle \Delta ={{\left({\frac {Q}{2}}\right)}^{2}}+{{\left({\frac {P}{3}}\right)}^{3}}}
u
,
v
=
−
Q
2
±
Δ
3
{\displaystyle u,v={\sqrt[{3}]{-{\frac {Q}{2}}\pm {\sqrt {\Delta }}}}}
Y
k
=
−
A
6
+
2
−
P
/
3
⋅
cos
(
2
(
k
−
1
)
⋅
π
3
+
1
3
⋅
arccos
−
Q
/
2
−
(
P
/
3
)
3
)
{\displaystyle {{Y}_{k}}=-{\frac {A}{6}}+2{\sqrt {-P/3}}\cdot \cos \left({\frac {2\left(k-1\right)\cdot \pi }{3}}+{\frac {1}{3}}\cdot \arccos {\frac {-Q/2}{\sqrt {-{{\left(P/3\right)}^{3}}}}}\right)}
s
i
g
(
x
)
=
{
+
1
,
x
≥
0
−
1
,
x
<
0
{\displaystyle sig\left(x\right)=\left\{{\begin{aligned}&+1,x\geq 0\\&-1,x<0\\\end{aligned}}\right.}
Mivel
(
y
1
+
y
2
+
y
3
)
4
−
2
(
y
1
2
+
y
2
2
+
y
3
2
)
(
y
1
+
y
2
+
y
3
)
2
−
8
y
1
y
2
y
3
(
y
1
+
y
2
+
y
3
)
+
(
y
1
2
+
y
2
2
+
y
3
2
)
2
−
4
(
y
1
2
y
2
2
+
y
1
2
y
3
2
+
y
2
2
y
3
2
)
=
0
{\displaystyle {{\left({{y}_{1}}+{{y}_{2}}+{{y}_{3}}\right)}^{4}}-2\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right){{\left({{y}_{1}}+{{y}_{2}}+{{y}_{3}}\right)}^{2}}-8{{y}_{1}}{{y}_{2}}{{y}_{3}}\left({{y}_{1}}+{{y}_{2}}+{{y}_{3}}\right)+{{\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)}^{2}}-4\left(y_{1}^{2}y_{2}^{2}+y_{1}^{2}y_{3}^{2}+y_{2}^{2}y_{3}^{2}\right)=0}
X
4
−
2
(
y
1
2
+
y
2
2
+
y
3
2
)
⋅
X
2
−
8
y
1
y
2
y
3
⋅
X
+
(
y
1
2
+
y
2
2
+
y
3
2
)
2
−
4
(
y
1
2
y
2
2
+
y
1
2
y
3
2
+
y
2
2
y
3
2
)
=
0
{\displaystyle {{X}^{4}}-2\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)\cdot {{X}^{2}}-8{{y}_{1}}{{y}_{2}}{{y}_{3}}\cdot X+{{\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)}^{2}}-4\left(y_{1}^{2}y_{2}^{2}+y_{1}^{2}y_{3}^{2}+y_{2}^{2}y_{3}^{2}\right)=0}
X
1
=
y
1
+
y
2
+
y
3
{\displaystyle {{X}_{1}}={{y}_{1}}+{{y}_{2}}+{{y}_{3}}}
Ez igaz marad akkor is ha
X
=
y
1
±
(
y
2
+
y
3
)
{\displaystyle X={{y}_{1}}\pm \left({{y}_{2}}+{{y}_{3}}\right)}
X
=
−
y
1
±
(
y
2
−
y
3
)
{\displaystyle X=-{{y}_{1}}\pm \left({{y}_{2}}-{{y}_{3}}\right)}
X
4
−
2
(
y
1
2
+
y
2
2
+
y
3
2
)
⋅
X
2
−
8
y
1
y
2
y
3
⋅
X
+
(
y
1
2
+
y
2
2
+
y
3
2
)
2
−
4
(
y
1
2
y
2
2
+
y
1
2
y
3
2
+
y
2
2
y
3
2
)
=
0
{\displaystyle {{X}^{4}}-2\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)\cdot {{X}^{2}}-8{{y}_{1}}{{y}_{2}}{{y}_{3}}\cdot X+{{\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)}^{2}}-4\left(y_{1}^{2}y_{2}^{2}+y_{1}^{2}y_{3}^{2}+y_{2}^{2}y_{3}^{2}\right)=0}
X
1
,
2
=
+
y
1
±
(
y
2
+
y
3
)
X
3
,
4
=
−
y
1
±
(
y
2
−
y
3
)
{\displaystyle {\begin{aligned}&{{X}_{1,2}}=+{{y}_{1}}\pm \left({{y}_{2}}+{{y}_{3}}\right)\\&{{X}_{3,4}}=-{{y}_{1}}\pm \left({{y}_{2}}-{{y}_{3}}\right)\\\end{aligned}}}
X
4
+
A
⋅
X
2
+
B
⋅
X
+
C
=
0
{\displaystyle {{X}^{4}}+A\cdot {{X}^{2}}+B\cdot X+C=0}
{
−
2
(
y
1
2
+
y
2
2
+
y
3
2
)
=
A
−
8
y
1
y
2
y
3
=
B
(
y
1
2
+
y
2
2
+
y
3
2
)
2
−
4
(
y
1
2
y
2
2
+
y
1
2
y
3
2
+
y
2
2
y
3
2
)
=
C
{\displaystyle \left\{{\begin{aligned}&-2\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)=A\\&-8{{y}_{1}}{{y}_{2}}{{y}_{3}}=B\\&{{\left(y_{1}^{2}+y_{2}^{2}+y_{3}^{2}\right)}^{2}}-4\left(y_{1}^{2}y_{2}^{2}+y_{1}^{2}y_{3}^{2}+y_{2}^{2}y_{3}^{2}\right)=C\\\end{aligned}}\right.}
y
1
,
2
,
3
{\displaystyle {{y}_{1,2,3}}}
A
,
B
,
C
{\displaystyle A,B,C}
{
y
1
2
+
y
2
2
+
y
3
2
=
−
A
2
y
1
2
y
2
2
+
y
1
2
y
3
2
+
y
2
2
y
3
2
=
A
2
16
−
C
4
y
1
2
y
2
2
y
3
2
=
B
2
64
{\displaystyle \left\{{\begin{aligned}&y_{1}^{2}+y_{2}^{2}+y_{3}^{2}=-{\frac {A}{2}}\\&y_{1}^{2}y_{2}^{2}+y_{1}^{2}y_{3}^{2}+y_{2}^{2}y_{3}^{2}={\frac {{A}^{2}}{16}}-{\frac {C}{4}}\\&y_{1}^{2}y_{2}^{2}y_{3}^{2}={\frac {{B}^{2}}{64}}\\\end{aligned}}\right.}
(
y
2
)
3
+
A
2
⋅
(
y
2
)
2
+
(
A
2
16
−
C
4
)
⋅
(
y
2
)
−
(
B
8
)
2
=
0
{\displaystyle {{\left({{y}^{2}}\right)}^{3}}+{\frac {A}{2}}\cdot {{\left({{y}^{2}}\right)}^{2}}+\left({\frac {{A}^{2}}{16}}-{\frac {C}{4}}\right)\cdot \left({{y}^{2}}\right)-{{\left({\frac {B}{8}}\right)}^{2}}=0}
y
1
y
2
y
3
=
−
B
8
{\displaystyle {{y}_{1}}{{y}_{2}}{{y}_{3}}=-{\frac {B}{8}}}
P
=
−
A
2
48
−
C
4
Q
=
−
A
3
864
−
B
2
64
+
A
C
24
{\displaystyle {\begin{aligned}&P=-{\frac {{A}^{2}}{48}}-{\frac {C}{4}}\\&Q=-{\frac {{A}^{3}}{864}}-{\frac {{B}^{2}}{64}}+{\frac {AC}{24}}\\\end{aligned}}}
Δ
=
(
Q
2
)
2
+
(
P
3
)
3
u
,
v
=
−
Q
2
±
Δ
3
{\displaystyle {\begin{aligned}&\Delta ={{\left({\frac {Q}{2}}\right)}^{2}}+{{\left({\frac {P}{3}}\right)}^{3}}\\&u,v={\sqrt[{3}]{-{\frac {Q}{2}}\pm {\sqrt {\Delta }}}}\\\end{aligned}}}
Δ
≥
0
{\displaystyle \Delta \geq 0}
y
1
2
=
−
A
6
+
u
+
v
y
2
,
3
2
=
−
A
6
−
u
+
v
2
±
i
(
u
−
v
)
3
2
{\displaystyle {\begin{aligned}&y_{1}^{2}=-{\frac {A}{6}}+u+v\\&y_{2,3}^{2}=-{\frac {A}{6}}-{\frac {u+v}{2}}\pm i{\frac {\left(u-v\right){\sqrt {3}}}{2}}\\\end{aligned}}}
y
1
=
−
A
6
+
u
+
v
{\displaystyle {{y}_{1}}={\sqrt {-{\frac {A}{6}}+u+v}}}
y
2
,
3
{\displaystyle {{y}_{2,3}}}
α
±
i
⋅
β
=
±
(
α
+
α
2
+
β
2
2
±
i
⋅
s
i
g
(
b
)
⋅
−
α
+
α
2
+
β
2
2
)
{\displaystyle {\sqrt {\alpha \pm i\cdot \beta }}=\pm \left({\sqrt {\frac {\alpha +{\sqrt {{{\alpha }^{2}}+{{\beta }^{2}}}}}{2}}}\pm i\cdot sig\left(b\right)\cdot {\sqrt {\frac {-\alpha +{\sqrt {{{\alpha }^{2}}+{{\beta }^{2}}}}}{2}}}\right)}
ennek eredményeként:
y
2
,
3
=
1
2
−
A
3
−
(
u
+
v
)
+
(
A
6
+
2
(
u
+
v
)
)
2
−
C
±
i
2
⋅
A
3
+
(
u
+
v
)
+
(
A
6
+
2
(
u
+
v
)
)
2
−
C
{\displaystyle {{y}_{2,3}}={\frac {1}{2}}{\sqrt {-{\frac {A}{3}}-\left(u+v\right)+{\sqrt {{{\left({\frac {A}{6}}+2\left(u+v\right)\right)}^{2}}-C}}}}\pm {\frac {i}{2}}\cdot {\sqrt {{\frac {A}{3}}+\left(u+v\right)+{\sqrt {{{\left({\frac {A}{6}}+2\left(u+v\right)\right)}^{2}}-C}}}}}
y
2
⋅
y
3
=
1
2
(
A
6
+
2
(
u
+
v
)
)
2
−
C
≥
0
{\displaystyle {{y}_{2}}\cdot {{y}_{3}}={\frac {1}{2}}{\sqrt {{{\left({\frac {A}{6}}+2\left(u+v\right)\right)}^{2}}-C}}\geq 0}
y
1
y
2
y
3
=
−
B
8
{\displaystyle {{y}_{1}}{{y}_{2}}{{y}_{3}}=-{\frac {B}{8}}}
y
1
=
s
i
g
(
−
B
)
−
A
6
+
u
+
v
{\displaystyle {{y}_{1}}=sig\left(-B\right){\sqrt {-{\frac {A}{6}}+u+v}}}
Tehát pozitív delta esetén a gyökok:
X
1
,
2
=
+
s
i
g
(
−
B
)
−
A
6
+
u
+
v
±
−
A
3
−
(
u
+
v
)
+
(
A
6
+
2
(
u
+
v
)
)
2
−
C
X
3
,
4
=
−
s
i
g
(
−
B
)
−
A
6
+
u
+
v
±
i
⋅
A
3
+
(
u
+
v
)
+
(
A
6
+
2
(
u
+
v
)
)
2
−
C
{\displaystyle {\begin{aligned}&{{X}_{1,2}}=+sig\left(-B\right){\sqrt {-{\frac {A}{6}}+u+v}}\pm {\sqrt {-{\frac {A}{3}}-\left(u+v\right)+{\sqrt {{{\left({\frac {A}{6}}+2\left(u+v\right)\right)}^{2}}-C}}}}\\&{{X}_{3,4}}=-sig\left(-B\right){\sqrt {-{\frac {A}{6}}+u+v}}\pm i\cdot {\sqrt {{\frac {A}{3}}+\left(u+v\right)+{\sqrt {{{\left({\frac {A}{6}}+2\left(u+v\right)\right)}^{2}}-C}}}}\\\end{aligned}}}
B
=
0
{\displaystyle B=0}
A
>
0
{\displaystyle A>0}
C
=
A
2
4
{\displaystyle C={\frac {{A}^{2}}{4}}}
−
A
6
+
u
+
v
<
0
{\displaystyle -{\frac {A}{6}}+u+v<0}
y
1
{\displaystyle {{y}_{1}}}
komplex szám és ebben az esetben a gyökök:
X
1
,
2
=
+
i
⋅
A
2
X
3
,
4
=
−
i
⋅
A
2
{\displaystyle {\begin{aligned}&{{X}_{1,2}}=+i\cdot {\sqrt {\frac {A}{2}}}\\&{{X}_{3,4}}=-i\cdot {\sqrt {\frac {A}{2}}}\\\end{aligned}}}
Δ
<
0
{\displaystyle \Delta <0}
y
k
=
±
−
A
6
+
2
−
P
/
3
⋅
cos
(
2
(
k
−
1
)
⋅
π
3
+
1
3
⋅
arccos
−
Q
/
2
−
(
P
/
3
)
3
)
{\displaystyle {{y}_{k}}=\pm {\sqrt {-{\frac {A}{6}}+2{\sqrt {-P/3}}\cdot \cos \left({\frac {2\left(k-1\right)\cdot \pi }{3}}+{\frac {1}{3}}\cdot \arccos {\frac {-Q/2}{\sqrt {-{{\left(P/3\right)}^{3}}}}}\right)}}}
(
4
⋅
C
>
A
2
)
{\displaystyle \left(4\cdot C>{{A}^{2}}\right)}
(
A
>
0
)
{\displaystyle \left(A>0\right)}
y
2
,
3
{\displaystyle {{y}_{2,3}}}
y
1
y
2
y
3
=
−
B
8
{\displaystyle {{y}_{1}}{{y}_{2}}{{y}_{3}}=-{\frac {B}{8}}}
s
i
g
(
−
B
)
{\displaystyle sig\left(-B\right)}
X
1
,
2
=
−
s
i
g
(
−
B
)
⋅
y
1
±
i
⋅
(
−
y
2
2
+
−
y
3
2
)
X
3
,
4
=
+
s
i
g
(
−
B
)
⋅
y
1
±
i
⋅
(
−
y
2
2
−
−
y
3
2
)
{\displaystyle {\begin{aligned}&{{X}_{1,2}}=-sig\left(-B\right)\cdot {{y}_{1}}\pm i\cdot \left({\sqrt {-y_{2}^{2}}}+{\sqrt {-y_{3}^{2}}}\right)\\&{{X}_{3,4}}=+sig\left(-B\right)\cdot {{y}_{1}}\pm i\cdot \left({\sqrt {-y_{2}^{2}}}-{\sqrt {-y_{3}^{2}}}\right)\\\end{aligned}}}
X
1
,
2
=
+
s
i
g
(
−
B
)
⋅
y
1
±
(
y
2
+
y
3
)
X
3
,
4
=
−
s
i
g
(
−
B
)
⋅
y
1
±
(
y
2
−
y
3
)
{\displaystyle {\begin{aligned}&{{X}_{1,2}}=+sig\left(-B\right)\cdot {{y}_{1}}\pm \left({{y}_{2}}+{{y}_{3}}\right)\\&{{X}_{3,4}}=-sig\left(-B\right)\cdot {{y}_{1}}\pm \left({{y}_{2}}-{{y}_{3}}\right)\\\end{aligned}}}
a
x
4
+
b
x
3
+
c
x
2
+
d
x
+
e
=
0
{\displaystyle a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e=0}
x
=
−
b
4
a
+
X
{\displaystyle x=-{\frac {b}{4a}}+X}
X
4
+
(
−
3
b
2
8
a
2
+
c
a
)
⏞
A
⋅
X
2
+
(
b
3
8
a
3
−
b
c
2
a
2
+
d
a
)
⏞
B
⋅
X
+
(
−
3
b
4
256
a
4
+
b
2
c
16
a
3
−
b
d
4
a
2
+
e
a
)
⏞
C
=
0
{\displaystyle {{X}^{4}}+\overbrace {\left(-{\frac {3{{b}^{2}}}{8{{a}^{2}}}}+{\frac {c}{a}}\right)} ^{A}\cdot {{X}^{2}}+\overbrace {\left({\frac {{b}^{3}}{8{{a}^{3}}}}-{\frac {bc}{2{{a}^{2}}}}+{\frac {d}{a}}\right)} ^{B}\cdot X+\overbrace {\left(-{\frac {3{{b}^{4}}}{256{{a}^{4}}}}+{\frac {{{b}^{2}}c}{16{{a}^{3}}}}-{\frac {bd}{4{{a}^{2}}}}+{\frac {e}{a}}\right)} ^{C}=0}
x
1
,
2
,
3
,
4
=
−
b
4
a
+
X
1
,
2
,
3
,
4
{\displaystyle {{x}_{1,2,3,4}}=-{\frac {b}{4a}}+{{X}_{1,2,3,4}}}
szerkesztés
Az
x
4
+
a
⋅
x
3
+
b
⋅
x
2
+
c
⋅
x
+
d
=
0
{\displaystyle x^{4}+a\cdot x^{3}+b\cdot x^{2}+c\cdot x+d=0}
Ludovico Ferraritól (1522-1565) származó módszer szerinti megoldása két másodfokú egyenlet megoldására vezethető vissza. Előbb azonban meg kell oldani egy harmadfokú egyenletet, melynek eredményét a másodfokú egyenletek együtthatóinak képzésekor fogjuk felhasználni. A harmadfokú egyenlet:
y
3
+
3
⋅
p
⋅
y
+
2
⋅
q
=
0
,
{\displaystyle y^{3}+3\cdot p\cdot y+2\cdot q=0,}
3
⋅
p
=
a
⋅
c
/
4
−
b
⋅
b
/
12
−
d
{\displaystyle 3\cdot p=a\cdot c/4-b\cdot b/12-d}
2
⋅
q
=
a
⋅
b
⋅
c
/
24
−
a
⋅
a
⋅
d
/
8
−
b
⋅
b
⋅
b
/
108
+
b
⋅
d
/
3
−
c
⋅
c
/
8.
{\displaystyle 2\cdot q=a\cdot b\cdot c/24-a\cdot a\cdot d/8-b\cdot b\cdot b/108+b\cdot d/3-c\cdot c/8.}
Megoldása a Cardano képlettel történik.
z
{\displaystyle z}
y
{\displaystyle y}
b
6
{\displaystyle {\frac {b}{6}}}
z
=
y
+
b
/
6
{\displaystyle z=y+b/6}
x
2
+
(
a
/
2
+
a
⋅
a
/
4
−
b
+
2
⋅
z
)
⋅
x
+
z
(
+
/
−
)
z
⋅
z
−
d
=
0
{\displaystyle x^{2}+(a/2+{\sqrt {a\cdot a/4-b+2\cdot z}})\cdot x+z(+/-){\sqrt {z\cdot z-d}}=0}
x
2
+
(
a
/
2
−
a
⋅
a
/
4
−
b
+
2
⋅
z
)
⋅
x
+
z
(
−
/
+
)
z
⋅
z
−
d
=
0
{\displaystyle x^{2}+(a/2-{\sqrt {a\cdot a/4-b+2\cdot z}})\cdot x+z(-/+){\sqrt {z\cdot z-d}}=0}
Kettős műveleti jelnél az alsót akkor kell használni, ha
a
⋅
z
−
c
<
0
{\displaystyle a\cdot z-c<0}
p
{\displaystyle p}
q
{\displaystyle q}
PASCAL nyelven megírva) így néz ki:
PROCEDURE negyedfoku ( a , b , c , d : REAL ) ;
VAR p , q , z , z2 , z3 , m , n , w1 , w2 , w3 : REAL ;
BEGIN
p := ( a * c / 4 - b * b / 12 - d ) / 3 ;
q := ( a * b * c / 24 - a * a * d / 8 - b * b * b / 108 + b * d / 3 - c * c / 8 ) / 2 ;
harmadfoku ( p , q , b / 6 , z , w1 , z2 , w2 , z3 , w3 ) ;
IF ( w2 = 0 ) AND ( z2 = z3 ) THEN IF z2 > z THEN z := z2 ;
m := ngyok ( a * a / 4 - b + 2 * z ) ;
n := ngyok ( z * z - d ) ;
IF a * z - c < - 1 . e - 7 THEN n := - n ;
masodfoku ( a / 2 + m , z + n , x [ 1 ] , y [ 1 ] , x [ 2 ] , y [ 2 ]) ;
masodfoku ( a / 2 - m , z - n , x [ 3 ] , y [ 3 ] , x [ 4 ] , y [ 4 ])
END ;
Látható, hogy semmi mást nem csinál, minthogy meghívja a "harmadfokú" eljárást (egyszer), majd a "másodfokú" eljárást (kétszer egymás után), miután kiszámította azok "bemeneti" együtthatóit. [ 1]
![{\displaystyle u,v={\sqrt[{3}]{-{\frac {Q}{2}}\pm {\sqrt {\Delta }}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6ee066f9be3662f902364792649e323e2b83d39)
![{\displaystyle {\begin{aligned}&\Delta ={{\left({\frac {Q}{2}}\right)}^{2}}+{{\left({\frac {P}{3}}\right)}^{3}}\\&u,v={\sqrt[{3}]{-{\frac {Q}{2}}\pm {\sqrt {\Delta }}}}\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3997427636e6fd3d417a0dfd2153dd5a270ded9d)
