一元三次方程求根公式详细逐步推导 | 您所在的位置:网站首页 › 3次方程怎么解 详细过程 › 一元三次方程求根公式详细逐步推导 |
/ / 注 : 颜 色 为 代 数 置 换 标 记 , 推 导 后 期 需 将 前 面 的 设 定 值 回 代 //注:颜色为代数置换标记,推导后期需将前面的设定值回代 //注:颜色为代数置换标记,推导后期需将前面的设定值回代 \quad 已 知 : a x 3 + b x 2 + c x + d = 0 a ≠ 0 a , b , c , d ∈ ℜ 问 题 : x = ? \begin{aligned} &已知:ax^3+bx^2+cx+d=0 \qquad a\neq0 \qquad a,b,c,d\in\Re \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \\ &问题:\mathbf x\mathbf=\mathbf? \end{aligned} 已知:ax3+bx2+cx+d=0a=0a,b,c,d∈ℜ问题:x=? \quad 推 导 开 始 : 推导开始: 推导开始: \quad x 3 + b x 2 a + c x a + d a = 0 a ≠ 0 a , b , c , d ∈ ℜ \begin{aligned} &x^3+\dfrac{bx^2}{a}+\dfrac{cx}{a}+\dfrac{d}{a}=0 \qquad a\neq0 \qquad a,b,c,d\in\Re \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \end{aligned} x3+abx2+acx+ad=0a=0a,b,c,d∈ℜ 设 k 0 = 1 , k 1 = b a , k 2 = c a , k 3 = d a k 0 x 3 + k 1 x 2 + k 2 x + k 3 = 0 设 x = y + η k 0 ( y + η ) 3 + k 1 ( y + η ) 2 + k 2 ( y + η ) + k 3 = 0 ( y + η ) 3 = y 3 + 3 y 2 η + 3 y η 2 + η 3 ( y + η ) 2 = y 2 + 2 y η + η 2 k 0 y 3 + k 0 3 y 2 η + 3 y η 2 k 0 + η 3 k 0 + y 2 k 1 + 2 y η k 1 + η 2 k 1 + y k 2 + η k 2 + k 3 = 0 y 3 k 0 + y 2 ( k 0 3 η + k 1 ) + y 1 ( 3 η 2 k 0 + 2 η k 1 + k 2 ) + y 0 ( η 3 k 0 + η 2 k 1 + η k 2 + k 3 ) = 0 \begin{aligned} &\color{red}{设\quad k_0=1,k_1=\dfrac{b}{a},k_2=\dfrac{c}{a},k_3=\dfrac{d}{a}}\\ &k_0x^3+k_1x^2+k_2x+k_3=0\\ \\ &\color{orange}{设 \quad x=y+\eta}\\ &k_0(y+\eta)^3+k_1(y+\eta)^2+k_2(y+\eta)+k_3=0\\ \\ &(y+\eta)^3=y^3+3y^2\eta+3y\eta^2+\eta^3\\ &(y+\eta)^2=y^2+2y\eta+\eta^2\\ \\ &k_0y^3+k_03y^2\eta+3y\eta^2k_0+\eta^3k_0+y^2k_1+2y\eta k_1+\eta^2k_1+yk_2+\eta k_2+k_3=0 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\\ &y^3k_0+y^2(k_03\eta+k_1)+y^1(3\eta^2k_0+2\eta k_1+k_2)+y^0(\eta^3k_0+\eta^2k_1+\eta k_2+k_3)=0 \\ \end{aligned} 设k0=1,k1=ab,k2=ac,k3=adk0x3+k1x2+k2x+k3=0设x=y+ηk0(y+η)3+k1(y+η)2+k2(y+η)+k3=0(y+η)3=y3+3y2η+3yη2+η3(y+η)2=y2+2yη+η2k0y3+k03y2η+3yη2k0+η3k0+y2k1+2yηk1+η2k1+yk2+ηk2+k3=0y3k0+y2(k03η+k1)+y1(3η2k0+2ηk1+k2)+y0(η3k0+η2k1+ηk2+k3)=0 k 0 = 1 3 k 0 η + k 1 = 0 3 η 2 k 0 + 2 η k 1 + k 2 = p η 3 k 0 + η 2 k 1 + η k 2 + k 3 = q \begin{aligned} k_0=&1 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ 3k_0\eta+k_1=&0\\ 3\eta^2k_0+2\eta k_1+k_2=&p\\ \eta^3k_0+\eta^2k_1+\eta k_2+k_3=&q \end{aligned} k0=3k0η+k1=3η2k0+2ηk1+k2=η3k0+η2k1+ηk2+k3=10pq 设 y 3 + p y + q = 0 3 η + k 1 = 0 3 η 2 + 2 η k 1 + k 2 = p η 3 + η 2 k 1 + η k 2 + k 3 = q \begin{aligned} 设 \quad y^3+py+q=&0 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ 3\eta+k_1=&0\\ 3\eta^2+2\eta k_1+k_2=&p\\ \eta^3+\eta^2k_1+\eta k_2+k_3=&q\\ \end{aligned} 设y3+py+q=3η+k1=3η2+2ηk1+k2=η3+η2k1+ηk2+k3=00pq η = − k 1 3 p = 3 ( − k 1 3 ) 2 + 2 ( − k 1 3 ) k 1 + k 2 p = − ( k 1 ) 2 3 + k 2 q = ( − k 1 3 ) 3 + ( − k 1 3 ) 2 k 1 + ( − k 1 3 ) k 2 + k 3 q = − k 1 3 27 + 3 k 1 3 27 − k 1 k 2 3 + k 3 q = 2 k 1 3 27 − k 1 k 2 3 + k 3 \begin{aligned} &\color{orange}{\eta=-\dfrac{k_1}{3}}\\ &p=3\left(-\dfrac{k_1}{3}\right)^2+2\left(-\dfrac{k_1}{3}\right)k_1+k_2\\ &\color{green}{p=-\dfrac{(k_1)^2}{3}+k_2}\\ &q=\left(-\dfrac{k_1}{3}\right)^3+\left(-\dfrac{k_1}{3}\right)^2k_1+\left(-\dfrac{k_1}{3}\right)k_2+k_3 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\\ &q=-\dfrac{k_1^3}{27}+\dfrac{3k_1^3}{27}-\dfrac{k_1k_2}{3}+k_3\\ &\color{green}{q=\dfrac{2k_1^3}{27}-\dfrac{k_1k_2}{3}+k_3}\\ \end{aligned} η=−3k1p=3(−3k1)2+2(−3k1)k1+k2p=−3(k1)2+k2q=(−3k1)3+(−3k1)2k1+(−3k1)k2+k3q=−27k13+273k13−3k1k2+k3q=272k13−3k1k2+k3 设 y = A 1 3 + B 1 3 A ′ = A 1 3 A 0 ′ ∈ ℜ , A 1 ′ = ω A 0 ′ , A 2 ′ = ω 2 A 0 ′ B ′ = B 1 3 B 0 ′ ∈ ℜ , B 1 ′ = ω B 0 ′ , B 2 ′ = ω 2 B 0 ′ ∵ y 3 = A + B + ( A B ) 1 3 ( A 1 3 + B 1 3 ) y 3 = A + B + ( A B ) 1 3 y y 3 − ( A + B ) − ( A B ) 1 3 y = 0 y 3 + p y + q = 0 \begin{aligned} 设 &\color{pink}{y=A^\frac13+B^\frac13\qquad A'=A^\frac13\quad A'_0\in\Re,A'_1=\omega A'_0,A'_2=\omega^2 A'_0 \quad B'=B^\frac13 \quad B'_0\in\Re,B'_1=\omega B'_0,B'_2=\omega^2 B'_0}\\ \because &y^3=A+B+(AB)^\frac13(A^\frac13+B^\frac13) \quad \quad \quad \quad \quad \quad \quad\\ &y^3=A+B+(AB)^\frac13y\\ &y^3-(A+B)-(AB)^\frac13y=0\\ &y^3+py+q=0\\ \end{aligned} 设∵y=A31+B31A′=A31A0′∈ℜ,A1′=ωA0′,A2′=ω2A0′B′=B31B0′∈ℜ,B1′=ωB0′,B2′=ω2B0′y3=A+B+(AB)31(A31+B31)y3=A+B+(AB)31yy3−(A+B)−(AB)31y=0y3+py+q=0 ∴ p = − 3 ( A B ) 1 3 q = − ( A + B ) A + B = − q A B = ( − p 3 ) 3 \begin{aligned} \therefore &p=-3(AB)^\frac13 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\\ &q=-(A+B)\\ &A+B=-q \qquad AB=(-\frac{p}{3})^3 \\ \end{aligned} ∴p=−3(AB)31q=−(A+B)A+B=−qAB=(−3p)3 / / 分 界 线 中 的 x , a , b , c 为 展 示 定 理 用 , 与 上 下 文 中 的 x , a , b , c 无 关 根 据 韦 达 定 理 : ∵ x 1 + x 2 = − b a x 1 x 2 = c a a x 2 + b x + c = 0 ∴ A = x 1 , B = x 2 , q = b a , ( − p 3 ) 3 = c a 根 据 Q u a d r a t i c F o r m u l a : ∵ a x 2 + b x + c = 0 x 1 = − b + b 2 − 4 a c 2 a x 2 = − b − b 2 − 4 a c 2 a x 1 = − b + b 2 − 4 a c 2 a = − b 2 a + b 2 2 2 a 2 − c a x 2 = − b + b 2 − 4 a c 2 a = − b 2 a − b 2 2 2 a 2 − c a \begin{aligned} &//分界线中的x,a,b,c为展示定理用,与上下文中的x,a,b,c无关 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\\ \\ &根据韦达定理:\\ &\because x_1+x_2=-\frac{b}{a} \qquad x_1x_2=\frac{c}{a}\qquad ax^2+bx+c=0\\ &\therefore A=x_1,B=x_2,q=\frac{b}{a},(-\frac{p}{3})^3=\frac{c}{a}\\ &根据Quadratic Formula: \\ & \because ax^2+bx+c=0 \\ & x_1=\frac{-b+\sqrt{b^2-4ac}}{2a} \qquad x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}\\ &x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}=-\frac{b}{2a}+\sqrt{\frac{b^2}{2^2a^2}-\frac{c}{a}}\\ &x_2=\frac{-b+\sqrt{b^2-4ac}}{2a}=-\frac{b}{2a}-\sqrt{\frac{b^2}{2^2a^2}-\frac{c}{a}}\\ \end{aligned} //分界线中的x,a,b,c为展示定理用,与上下文中的x,a,b,c无关根据韦达定理:∵x1+x2=−abx1x2=acax2+bx+c=0∴A=x1,B=x2,q=ab,(−3p)3=ac根据QuadraticFormula:∵ax2+bx+c=0x1=2a−b+b2−4ac x2=2a−b−b2−4ac x1=2a−b+b2−4ac =−2ab+22a2b2−ac x2=2a−b+b2−4ac =−2ab−22a2b2−ac ∴ A 0 = − q 2 + q 2 4 + p 3 27 B 0 = − q 2 − q 2 4 + p 3 27 ∵ A ′ = A 1 3 A 0 ′ ∈ ℜ , A 1 ′ = ω A 0 ′ , A 2 ′ = ω 2 A 0 ′ B ′ = B 1 3 B 0 ′ ∈ ℜ , B 1 ′ = ω B 0 ′ , B 2 ′ = ω 2 B 0 ′ ∵ ω = − 1 ± 3 i 2 A B = ( − q 3 ) 3 q ∈ ℜ ∴ A 0 = ( A 0 ′ ) 3 B 0 = ( B 0 ′ ) 3 A 1 = ( A 1 ′ ) 3 B 1 = ( B 2 ′ ) 3 A 2 = ( A 2 ′ ) 3 B 2 = ( B 1 ′ ) 3 ∵ y = A 1 3 + B 1 3 ∴ y 0 = A 0 ′ + B 0 ′ y 1 = A 1 ′ + B 2 ′ y 2 = A 2 ′ + B 1 ′ \begin{aligned} &\therefore A_0=-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}\qquad B_0=-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}\\ &\because \color{pink}{A'=A^\frac13\quad A'_0\in\Re,A'_1=\omega A'_0,A'_2=\omega^2 A'_0 \quad B'=B^\frac13 \quad B'_0\in\Re,B'_1=\omega B'_0,B'_2=\omega^2 B'_0}\quad \quad \quad \quad \quad \quad \quad \quad\\ &\because \omega=\dfrac{-1\pm\sqrt{3}i}{2} \quad AB=(-\frac{q}{3})^3 \quad q\in\Re\\ &\therefore A_0=(A'_0)^3 \quad B_0=(B'_0)^3\\ &\quad A_1=(A'_1)^3 \quad B_1=(B'_2)^3\\ &\quad A_2=(A'_2)^3 \quad B_2=(B'_1)^3\\ &\because y=A^\frac13+B^\frac13 \quad \\ &\therefore y_0=A'_0+B'_0\\ &\quad y_1=A'_1+B'_2\\ &\quad y_2=A'_2+B'_1\\ \end{aligned} ∴A0=−2q+4q2+27p3 B0=−2q−4q2+27p3 ∵A′=A31A0′∈ℜ,A1′=ωA0′,A2′=ω2A0′B′=B31B0′∈ℜ,B1′=ωB0′,B2′=ω2B0′∵ω=2−1±3 iAB=(−3q)3q∈ℜ∴A0=(A0′)3B0=(B0′)3A1=(A1′)3B1=(B2′)3A2=(A2′)3B2=(B1′)3∵y=A31+B31∴y0=A0′+B0′y1=A1′+B2′y2=A2′+B1′ y 0 = − q 2 + ( q 2 4 + p 3 27 ) 1 2 3 + − q 2 − ( q 2 4 + p 3 27 ) 1 2 3 y 1 = − 1 + 3 i 2 − q 2 + ( q 2 4 + p 3 27 ) 1 2 3 + − 1 − 3 i 2 − q 2 − ( q 2 4 + p 3 27 ) 1 2 3 y 2 = − 1 − 3 i 2 − q 2 + ( q 2 4 + p 3 27 ) 1 2 3 + − 1 + 3 i 2 − q 2 − ( q 2 4 + p 3 27 ) 1 2 3 \begin{aligned} \\ \quad y_0=\sqrt[3]{-\frac{q}{2}+(\frac{q^2}{4}+\frac{p^3}{27})^{\frac{1}{2}}}+\sqrt[3]{-\frac{q}{2}-(\frac{q^2}{4}+\frac{p^3}{27})^{\frac{1}{2}}}\quad \quad \quad \quad \quad \quad \quad \quad \quad\\ \\ \quad y_1=\dfrac{-1+\sqrt{3}i}{2}\sqrt[3]{-\frac{q}{2}+(\frac{q^2}{4}+\frac{p^3}{27})^{\frac{1}{2}}}+\dfrac{-1-\sqrt{3}i}{2}\sqrt[3]{-\frac{q}{2}-(\frac{q^2}{4}+\frac{p^3}{27})^{\frac{1}{2}}}\\ \\ \quad y_2=\dfrac{-1-\sqrt{3}i}{2}\sqrt[3]{-\frac{q}{2}+(\frac{q^2}{4}+\frac{p^3}{27})^{\frac{1}{2}}}+\dfrac{-1+\sqrt{3}i}{2}\sqrt[3]{-\frac{q}{2}-(\frac{q^2}{4}+\frac{p^3}{27})^{\frac{1}{2}}}\\ \\ \end{aligned} y0=3−2q+(4q2+27p3)21 +3−2q−(4q2+27p3)21 y1=2−1+3 i3−2q+(4q2+27p3)21 +2−1−3 i3−2q−(4q2+27p3)21 y2=2−1−3 i3−2q+(4q2+27p3)21 +2−1+3 i3−2q−(4q2+27p3)21 ∵ x = y + η \because \color{orange}{x=y+\eta} ∵x=y+η η = − k 1 3 \quad \color{orange}{\eta=-\dfrac{k_1}{3}} η=−3k1 k 1 = b a \color{red}{\quad k_1=\dfrac{b}{a}} k1=ab ∴ y = x + b 3 a \therefore y=x+\frac{b}{3a} ∴y=x+3ab ∵ p = − ( k 1 ) 2 3 + k 2 \because \color{green}{p=-\dfrac{(k_1)^2}{3}+k_2} ∵p=−3(k1)2+k2 q = 2 k 1 3 27 − k 1 k 2 3 + k 3 \quad \color{green}{q=\dfrac{2k_1^3}{27}-\dfrac{k_1k_2}{3}+k_3} q=272k13−3k1k2+k3 ∴ x 0 + b 3 a = − k 1 3 27 + k 1 k 2 6 − k 3 2 + 1 4 ( k 1 3 27 − k 1 k 2 6 − d 2 a ) 2 + ( k 2 − k 1 2 3 ) 3 27 3 + − k 1 3 27 + k 1 k 2 6 − k 3 2 − 1 4 ( 2 k 1 3 27 − k 1 k 2 3 + k 3 ) 2 + ( k 2 − k 1 2 3 ) 3 27 3 x 1 + b 3 a = − 1 + 3 i 2 − k 1 3 27 + k 1 k 2 6 − k 3 2 + 1 4 ( 2 k 1 3 27 − k 1 k 2 3 + k 3 ) 2 + ( k 2 − k 1 2 3 ) 3 27 3 + − 1 − 3 i 2 − k 1 3 27 + k 1 k 2 6 − k 3 2 − 1 4 ( 2 k 1 3 27 − k 1 k 2 3 + k 3 ) 2 + ( k 2 − k 1 2 3 ) 3 27 3 x 2 + b 3 a = − 1 − 3 i 2 − k 1 3 27 + k 1 k 2 6 − k 3 2 + 1 4 ( 2 k 1 3 27 − k 1 k 2 3 + k 3 ) 2 + ( k 2 − k 1 2 3 ) 3 27 3 + − 1 + 3 i 2 − k 1 3 27 + k 1 k 2 6 − k 3 2 − 1 4 ( 2 k 1 3 27 − k 1 k 2 3 + k 3 ) 2 + ( k 2 − k 1 2 3 ) 3 27 3 \begin{aligned} \\ \therefore \quad x_0+\frac{b}{3a}=\sqrt[3]{-\dfrac{k_1^3}{27}+\dfrac{k_1k_2}{6}-\frac{k_3}{2}+\sqrt{\frac{1}{4}(\dfrac{k_1^3}{27}-\dfrac{k_1k_2}{6}-\frac{d}{2a})^2+\frac{(k_2-\frac{k_1^2}{3})^3}{27}}}&\\ \qquad+\sqrt[3]{-\dfrac{k_1^3}{27}+\dfrac{k_1k_2}{6}-\frac{k_3}{2}-\sqrt{\frac{1}{4}(\dfrac{2k_1^3}{27}-\dfrac{k_1k_2}{3}+k_3)^2+\frac{(k_2-\frac{k_1^2}{3})^3}{27}}}\\ \\ \quad x_1+\frac{b}{3a}=\dfrac{-1+\sqrt{3}i}{2}\sqrt[3]{-\dfrac{k_1^3}{27}+\dfrac{k_1k_2}{6}-\frac{k_3}{2}+\sqrt{\frac{1}{4}(\dfrac{2k_1^3}{27}-\dfrac{k_1k_2}{3}+k_3)^2+\frac{(k_2-\frac{k_1^2}{3})^3}{27}}}&\\ \qquad+\dfrac{-1-\sqrt{3}i}{2}\sqrt[3]{-\dfrac{k_1^3}{27}+\dfrac{k_1k_2}{6}-\frac{k_3}{2}-\sqrt{\frac{1}{4}(\dfrac{2k_1^3}{27}-\dfrac{k_1k_2}{3}+k_3)^2+\frac{(k_2-\frac{k_1^2}{3})^3}{27}}}\\ \\ \quad x_2+\frac{b}{3a}=\dfrac{-1-\sqrt{3}i}{2}\sqrt[3]{-\dfrac{k_1^3}{27}+\dfrac{k_1k_2}{6}-\frac{k_3}{2}+\sqrt{\frac{1}{4}(\dfrac{2k_1^3}{27}-\dfrac{k_1k_2}{3}+k_3)^2+\frac{(k_2-\frac{k_1^2}{3})^3}{27}}}&\\ \qquad+\dfrac{-1+\sqrt{3}i}{2}\sqrt[3]{-\dfrac{k_1^3}{27}+\dfrac{k_1k_2}{6}-\frac{k_3}{2}-\sqrt{\frac{1}{4}(\dfrac{2k_1^3}{27}-\dfrac{k_1k_2}{3}+k_3)^2+\frac{(k_2-\frac{k_1^2}{3})^3}{27}}}\\ \\ \end{aligned} ∴x0+3ab=3−27k13+6k1k2−2k3+41(27k13−6k1k2−2ad)2+27(k2−3k12)3 +3−27k13+6k1k2−2k3−41(272k13−3k1k2+k3)2+27(k2−3k12)3 x1+3ab=2−1+3 i3−27k13+6k1k2−2k3+41(272k13−3k1k2+k3)2+27(k2−3k12)3 +2−1−3 i3−27k13+6k1k2−2k3−41(272k13−3k1k2+k3)2+27(k2−3k12)3 x2+3ab=2−1−3 i3−27k13+6k1k2−2k3+41(272k13−3k1k2+k3)2+27(k2−3k12)3 +2−1+3 i3−27k13+6k1k2−2k3−41(272k13−3k1k2+k3)2+27(k2−3k12)3 ∵ k 1 = b a , k 2 = c a , k 3 = d a \because \color{red}{k_1=\dfrac{b}{a},k_2=\dfrac{c}{a},k_3=\dfrac{d}{a}} ∵k1=ab,k2=ac,k3=ad ∴ x 0 = − b 3 a + − b 3 27 a 3 + b c 6 a 2 − d 2 a + 1 4 ( 2 d 3 27 a 3 − b c 3 a 2 + d a ) 2 + ( c 3 a − b 2 9 a 2 ) 3 3 + − b 3 27 a 3 + b c 6 a 2 − d 2 a − 1 4 ( 2 d 3 27 a 3 − b c 3 a 2 + d a ) 2 + ( c 3 a − b 2 9 a 2 ) 3 3 x 1 = − b 3 a + − 1 + 3 i 2 − b 3 27 a 3 + b c 6 a 2 − d 2 a + 1 4 ( 2 d 3 27 a 3 − b c 3 a 2 + d a ) 2 + ( c 3 a − b 2 9 a 2 ) 3 3 + − 1 − 3 i 2 − b 3 27 a 3 + b c 6 a 2 − d 2 a − 1 4 ( 2 d 3 27 a 3 − b c 3 a 2 + d a ) 2 + ( c 3 a − b 2 9 a 2 ) 3 3 x 2 = − b 3 a + − 1 − 3 i 2 − b 3 27 a 3 + b c 6 a 2 − d 2 a + 1 4 ( 2 d 3 27 a 3 − b c 3 a 2 + d a ) 2 + ( c 3 a − b 2 9 a 2 ) 3 3 + − 1 + 3 i 2 − b 3 27 a 3 + b c 6 a 2 − d 2 a − 1 4 ( 2 d 3 27 a 3 − b c 3 a 2 + d a ) 2 + ( c 3 a − b 2 9 a 2 ) 3 3 \\ \\ \begin{aligned} \\ \\ \therefore \quad x_0=-\frac{b}{3a}+\sqrt[3]{-\dfrac{b^3}{27a^3}+\dfrac{bc}{6a^2}-\frac{d}{2a}+\sqrt{\frac{1}{4}(\dfrac{2d^3}{27a^3}-\dfrac{bc}{3a^2}+\frac{d}{a})^2+(\frac{c}{3a}-\frac{b^2}{9a^2})^3}}&\\ +\sqrt[3]{-\dfrac{b^3}{27a^3}+\dfrac{bc}{6a^2}-\frac{d}{2a}-\sqrt{\frac{1}{4}(\dfrac{2d^3}{27a^3}-\dfrac{bc}{3a^2}+\frac{d}{a})^2+(\frac{c}{3a}-\frac{b^2}{9a^2})^3}}\\ \\ \\ \quad x_1=-\frac{b}{3a}+\dfrac{-1+\sqrt{3}i}{2}\sqrt[3]{-\dfrac{b^3}{27a^3}+\dfrac{bc}{6a^2}-\frac{d}{2a}+\sqrt{\frac{1}{4}(\dfrac{2d^3}{27a^3}-\dfrac{bc}{3a^2}+\frac{d}{a})^2+(\frac{c}{3a}-\frac{b^2}{9a^2})^3}}&\\ +\dfrac{-1-\sqrt{3}i}{2}\sqrt[3]{-\dfrac{b^3}{27a^3}+\dfrac{bc}{6a^2}-\frac{d}{2a}-\sqrt{\frac{1}{4}(\dfrac{2d^3}{27a^3}-\dfrac{bc}{3a^2}+\frac{d}{a})^2+(\frac{c}{3a}-\frac{b^2}{9a^2})^3}}\\ \\ \\ \quad x_2=-\frac{b}{3a}+\dfrac{-1-\sqrt{3}i}{2}\sqrt[3]{-\dfrac{b^3}{27a^3}+\dfrac{bc}{6a^2}-\frac{d}{2a}+\sqrt{\frac{1}{4}(\dfrac{2d^3}{27a^3}-\dfrac{bc}{3a^2}+\frac{d}{a})^2+(\frac{c}{3a}-\frac{b^2}{9a^2})^3}}&\\ +\dfrac{-1+\sqrt{3}i}{2}\sqrt[3]{-\dfrac{b^3}{27a^3}+\dfrac{bc}{6a^2}-\frac{d}{2a}-\sqrt{\frac{1}{4}(\dfrac{2d^3}{27a^3}-\dfrac{bc}{3a^2}+\frac{d}{a})^2+(\frac{c}{3a}-\frac{b^2}{9a^2})^3}}\\ \end{aligned} ∴x0=−3ab+3−27a3b3+6a2bc−2ad+41(27a32d3−3a2bc+ad)2+(3ac−9a2b2)3 +3−27a3b3+6a2bc−2ad−41(27a32d3−3a2bc+ad)2+(3ac−9a2b2)3 x1=−3ab+2−1+3 i3−27a3b3+6a2bc−2ad+41(27a32d3−3a2bc+ad)2+(3ac−9a2b2)3 +2−1−3 i3−27a3b3+6a2bc−2ad−41(27a32d3−3a2bc+ad)2+(3ac−9a2b2)3 x2=−3ab+2−1−3 i3−27a3b3+6a2bc−2ad+41(27a32d3−3a2bc+ad)2+(3ac−9a2b2)3 +2−1+3 i3−27a3b3+6a2bc−2ad−41(27a32d3−3a2bc+ad)2+(3ac−9a2b2)3
附:一元三次方程的故事 塔尔塔利亚是意大利人,出生于1500年。他12岁那年,被入侵的法国兵砍伤了头部和舌头,从此说话结结巴巴,人们就给他一个绰号“塔尔塔利亚”(在意大利语中,这是口吃的意思),真名反倒少有人叫了,他自学成才,成了数学家,宣布自己找到了三次方程的的解法。这时,意大利数学家卡丹出场,请求塔尔塔利把解方程的方法告诉他,可是遭到了拒绝。后来卡丹对塔尔塔利假装说要推荐他去当西班牙炮兵顾问,称自己因为无法解三次方程而内心痛苦并发誓永远不泄漏塔尔塔利亚解一元三次方程式的秘密。塔尔塔利亚这才把解一元三次方程的秘密告诉了卡丹。六年以后,卡丹不顾原来的信约,在他的著作《关于代数的大法》中,将经过改进的三次方程的解法公开发表。后人就把这个方法叫作卡丹公式,塔尔塔利亚的名字反而被湮没了,正如他的真名在口吃以后被埋没了一样。 塔尔塔利亚对卡丹的背信行为非常恼怒,互相写信指骂对方。最终在一个不明的夜晚,卡丹派人秘密刺杀了塔尔塔利亚。 一元三次方程应有三个根。塔塔利亚公式给出的只是一个实根。又过了大约200年后,随着人们对虚数认识的加深,到了1732年,才由瑞士数学家欧拉找到了一元三次方程三个根的完整的表达式。 维护日志: 2019-12-16:review 2020-10-24:排版调整 |
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