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}}}
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}}}
A valós együtthatós negyedfokú egyenlet megoldása Ludovico Ferrari módszere szerint
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)
