Quartic equation
다항 방정식 중 하나로 다음 꼴의 방정식을 말한다.
a x 4 + b x 3 + c x 2 + d x + e = 0 ax^4+bx^3+cx^2+dx+e=0 a x 4 + b x 3 + c x 2 + d x + e = 0 여기서
a ≠ 0 a\neq 0 a = 0 이다.
일반적인 복소계수 4차방정식
a x 4 + b x 3 + c x 2 + d x + e = 0 ( a ≠ 0 ) ax^4+bx^3+cx^2+dx+e=0\ (a \neq 0) a x 4 + b x 3 + c x 2 + d x + e = 0 ( a = 0 ) 을 생각하자.
적당한
t ∈ C t \in \Bbb C t ∈ C 에 대하여
z = x − b 4 a \displaystyle z=x-\frac{b}{4a} z = x − 4 a b 로 치환하고 4차항의 계수로 방정식을 나누면
z 4 + p z 2 + q z + r = 0 z^4+pz^2+qz+r=0 z 4 + p z 2 + q z + r = 0 꼴의 방정식을 얻을 수 있다.
w ∈ C w \in \Bbb C w ∈ C 에 대하여
( z 2 + w ) 2 = ( 2 w − p ) z 2 − q z + w 2 − r (z^2+w)^2 = (2w-p){z}^2-qz+w^2-r ( z 2 + w ) 2 = ( 2 w − p ) z 2 − q z + w 2 − r 가 되는데, 우변의 판별식이 0인
w w w 를 3차방정식을 풀어 구하면 우변을 완전제곱식꼴로 고칠 수 있고, 따라서 두 개의 이차방정식으로 나눠진 4차방정식을 풀어 해를 구할 수 있다.
이를 이용하면
a x 4 + b x 3 + c x 2 + d x + e = 0 ( a ≠ 0 ) ax^4+bx^3+cx^2+dx+e=0\ (a \neq 0) a x 4 + b x 3 + c x 2 + d x + e = 0 ( a = 0 ) 의 해가
x = − a 4 x = {\frac{-a}{4}} x = 4 − a ± 1 2 a 2 4 − 2 b 3 + 2 1 3 ( b 2 − 3 a c + 12 d ) 3 ( 2 b 3 − 9 a b c + 27 c 2 + 27 a 2 d − 72 b d + − 4 ( b 2 − 3 a c + 12 d ) 3 + ( 2 b 3 − 9 a b c + 27 c 2 + 27 a 2 d − 72 b d ) 2 ) 1 3 + ( 2 b 3 − 9 a b c + 27 c 2 + 27 a 2 d − 72 b d + − 4 ( b 2 − 3 a c + 12 d ) 3 + ( 2 b 3 − 9 a b c + 27 c 2 + 27 a 2 d − 72 b d ) 2 54 ) 1 3 \pm\frac{1}{2}{\sqrt{\frac{a^2}{4}-\frac{2b}{3}+\frac{2^{\frac{1}{3}}\left(b^2-3ac+12d\right)}{3{\left(2b^3-9abc+27c^2+27a^2d-72bd+{\sqrt{-4{\left(b^2-3ac+12d\right)}^3+{\left(2b^3-9abc+27c^2+27a^2d-72bd\right)}^2}}\right)}^{\frac{1}{3}}}+\left(\frac{{2b^3-9abc+27c^2+27a^2d-72bd+{\sqrt{-4{\left(b^2-3ac+12d\right)}^3+{\left(2b^3-9abc+27c^2+27a^2d-72bd\right)}^2}}}}{54}\right)^\frac{1}{3}}} ± 2 1 4 a 2 − 3 2 b + 3 ( 2 b 3 − 9 ab c + 27 c 2 + 27 a 2 d − 72 b d + − 4 ( b 2 − 3 a c + 12 d ) 3 + ( 2 b 3 − 9 ab c + 27 c 2 + 27 a 2 d − 72 b d ) 2 ) 3 1 2 3 1 ( b 2 − 3 a c + 12 d ) + ( 54 2 b 3 − 9 ab c + 27 c 2 + 27 a 2 d − 72 b d + − 4 ( b 2 − 3 a c + 12 d ) 3 + ( 2 b 3 − 9 ab c + 27 c 2 + 27 a 2 d − 72 b d ) 2 ) 3 1 ± 1 2 a 2 2 − 4 b 3 − 2 1 3 ( b 2 − 3 a c + 12 d ) 3 ( 2 b 3 − 9 a b c + 27 c 2 + 27 a 2 d − 72 b d + − 4 ( b 2 − 3 a c + 12 d ) 3 + ( 2 b 3 − 9 a b c + 27 c 2 + 27 a 2 d − 72 b d ) 2 ) 1 3 − ( 2 b 3 − 9 a b c + 27 c 2 + 27 a 2 d − 72 b d + − 4 ( b 2 − 3 a c + 12 d ) 3 + ( 2 b 3 − 9 a b c + 27 c 2 + 27 a 2 d − 72 b d ) 2 54 ) 1 3 − − a 3 + 4 a b − 8 c 4 a 2 4 − 2 b 3 + 2 1 3 ( b 2 − 3 a c + 12 d ) 3 ( 2 b 3 − 9 a b c + 27 c 2 + 27 a 2 d − 72 b d + − 4 ( b 2 − 3 a c + 12 d ) 3 + ( 2 b 3 − 9 a b c + 27 c 2 + 27 a 2 d − 72 b d ) 2 ) 1 3 + ( 2 b 3 − 9 a b c + 27 c 2 + 27 a 2 d − 72 b d + − 4 ( b 2 − 3 a c + 12 d ) 3 + ( 2 b 3 − 9 a b c + 27 c 2 + 27 a 2 d − 72 b d ) 2 54 ) 1 3 \pm\frac{1}{2}{\sqrt{\frac{a^2}{2}-\frac{4b}{3}-\frac{2^{\frac{1}{3}}\left(b^2-3ac+12d\right)}{3{\left(2b^3-9abc+27c^2+27a^2d-72bd+{\sqrt{-4{\left(b^2-3ac+12d\right)}^3+{\left(2b^3-9abc+27c^2+27a^2d-72bd\right)}^2}}\right)}^{\frac{1}{3}}}-\left(\frac{{2b^3-9abc+27c^2+27a^2d-72bd+{\sqrt{-4{\left(b^2-3ac+12d\right)}^3+{\left(2b^3-9abc+27c^2+27a^2d-72bd\right)}^2}}}}{54}\right)^\frac{1}{3}-\frac{-a^3+4ab-8c}{4{\sqrt{\frac{a^2}{4}-\frac{2b}{3}+\frac{2^{\frac{1}{3}}\left(b^2-3ac+12d\right)}{3{\left(2b^3-9abc+27c^2+27a^2d-72bd+{\sqrt{-4{\left(b^2-3ac+12d\right)}^3+{\left(2b^3-9abc+27c^2+27a^2d-72bd\right)}^2}}\right)}^{\frac{1}{3}}}+\left(\frac{{2b^3-9abc+27c^2+27a^2d-72bd+{\sqrt{-4{\left(b^2-3ac+12d\right)}^3+{\left(2b^3-9abc+27c^2+27a^2d-72bd\right)}^2}}}}{54}\right)^\frac{1}{3}}}}}} ± 2 1 2 a 2 − 3 4 b − 3 ( 2 b 3 − 9 ab c + 27 c 2 + 27 a 2 d − 72 b d + − 4 ( b 2 − 3 a c + 12 d ) 3 + ( 2 b 3 − 9 ab c + 27 c 2 + 27 a 2 d − 72 b d ) 2 ) 3 1 2 3 1 ( b 2 − 3 a c + 12 d ) − ( 54 2 b 3 − 9 ab c + 27 c 2 + 27 a 2 d − 72 b d + − 4 ( b 2 − 3 a c + 12 d ) 3 + ( 2 b 3 − 9 ab c + 27 c 2 + 27 a 2 d − 72 b d ) 2 ) 3 1 − 4 4 a 2 − 3 2 b + 3 ( 2 b 3 − 9 ab c + 27 c 2 + 27 a 2 d − 72 b d + − 4 ( b 2 − 3 a c + 12 d ) 3 + ( 2 b 3 − 9 ab c + 27 c 2 + 27 a 2 d − 72 b d ) 2 ) 3 1 2 3 1 ( b 2 − 3 a c + 12 d ) + ( 54 2 b 3 − 9 ab c + 27 c 2 + 27 a 2 d − 72 b d + − 4 ( b 2 − 3 a c + 12 d ) 3 + ( 2 b 3 − 9 ab c + 27 c 2 + 27 a 2 d − 72 b d ) 2 ) 3 1 − a 3 + 4 ab − 8 c 임을 알 수 있다.
사차 방정식
a x 4 + b x 3 + c x 2 + d x + e = 0 ( a ≠ 0 ) ax^4 + bx^3 + cx^2 + dx+e=0 (a \neq 0) a x 4 + b x 3 + c x 2 + d x + e = 0 ( a = 0 ) 의 판별식은
D = − 27 b 4 e 2 + 16 c 4 a e − 27 d 4 a 2 + 256 a 3 e 3 + 18 b 3 c d e − 4 b 3 d 3 − 4 c 3 b 2 a − 4 c 3 b 2 e + 18 d 3 a b c − 128 a 2 c 2 e 2 + 144 a 2 d 2 c e − 192 a 2 e 2 b d + b 2 c 2 d 2 − 6 b 2 d 2 a e + 144 b 2 e 2 a c − 80 c 2 a b d e D=-27b^4e^2+16c^4ae-27d^4a^2+256a^3e^3+18b^3cde-4b^3d^3-4c^3b^2a-4c^3b^2e+18d^3abc-128a^2c^2e^2+144a^2d^2ce-192a^2e^2bd+b^2c^2d^2-6b^2d^2ae+144b^2e^2ac-80c^2abde D = − 27 b 4 e 2 + 16 c 4 a e − 27 d 4 a 2 + 256 a 3 e 3 + 18 b 3 c d e − 4 b 3 d 3 − 4 c 3 b 2 a − 4 c 3 b 2 e + 18 d 3 ab c − 128 a 2 c 2 e 2 + 144 a 2 d 2 ce − 192 a 2 e 2 b d + b 2 c 2 d 2 − 6 b 2 d 2 a e + 144 b 2 e 2 a c − 80 c 2 ab d e 이다.