一元四次方程式

解法

階段一：變形去除三次項

1. $at^{4}+bt^{3}+ct^{2}+dt+e=0\Rightarrow$ 以 $t=x-{\frac {b}{4a}}$ 代入

2.得 $x^{4}+px^{2}+qx+r=0$ ，令其四根為 $x_{1},x_{2},x_{3},x_{4}$

階段二：變身為三次方程式

3.由 $(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})=0$ 可得 ${\begin{cases}x_{1}+x_{2}+x_{3}+x_{4}=0\\x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4}=p\\x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4}=-q\\x_{1}x_{2}x_{3}x_{4}=r\end{cases}}$

4.

設 ${\begin{cases}y_{1}=(x_{1}+x_{2})(x_{3}+x_{4})\\y_{2}=(x_{1}+x_{3})(x_{2}+x_{4})\\y_{3}=(x_{1}+x_{4})(x_{2}+x_{3})\end{cases}}$	則	${\begin{cases}y_{1}+y_{2}+y_{3}=2p\\y_{1}y_{2}+y_{1}y_{3}+y_{2}y_{3}=p^{2}-4r\\y_{1}y_{2}y_{3}=-q^{2}\end{cases}}$	注意： $y_{1},y_{2},y_{3}$ 對 $x$ 而言是 2 次 $p$ 對 $x$ 而言是 2 次 $p^{2},r$ 對 $x$ 而言是 4 次 $q^{2}$ 對 $x$ 而言是 6 次

5.故 $y_{1},y_{2},y_{3}$ 為 $y^{3}-2py^{2}+(p^{2}-4r)y+q^{2}=0$ 的三根，
此方程式對 $x$ 而言是 6 次，其四項對 $x$ 而言分別是 0+6 次、 2+4 次、 4+2 次、 6+0 次。

階段三：以三次方程式之三根求四次方程式之四根

6. $\because (x_{1}+x_{2})+(x_{3}+x_{4})=0,(x_{1}+x_{2})(x_{3}+x_{4})=y_{1}\therefore (x_{1}+x_{2}),(x_{3}+x_{4})$ 為 $w^{2}+y_{1}=0$ 之二根， $(x_{1}+x_{2})={\sqrt {-y_{1}}}$ 或 $(x_{1}+x_{2})=-{\sqrt {-y_{1}}}$

7.設 $(x_{1}+x_{2})={\sqrt {-y_{1}}},(x_{3}+x_{4})=-{\sqrt {-y_{1}}}$ (此有另一種解，判斷方法補充於最後)

8.同理，可得 ${\begin{cases}(x_{1}+x_{2})={\sqrt {-y_{1}}},(x_{3}+x_{4})=-{\sqrt {-y_{1}}}\\(x_{1}+x_{3})={\sqrt {-y_{2}}},(x_{2}+x_{4})=-{\sqrt {-y_{2}}}\\(x_{1}+x_{4})={\sqrt {-y_{3}}},(x_{2}+x_{3})=-{\sqrt {-y_{3}}}\end{cases}}$

9.解聯立方程式，得 ${\begin{cases}x_{1}={\frac {1}{2}}({\sqrt {-y_{1}}}+{\sqrt {-y_{2}}}+{\sqrt {-y_{3}}})\\x_{2}={\frac {1}{2}}({\sqrt {-y_{1}}}-{\sqrt {-y_{2}}}-{\sqrt {-y_{3}}})\\x_{3}={\frac {1}{2}}(-{\sqrt {-y_{1}}}+{\sqrt {-y_{2}}}-{\sqrt {-y_{3}}})\\x_{4}={\frac {1}{2}}(-{\sqrt {-y_{1}}}-{\sqrt {-y_{2}}}+{\sqrt {-y_{3}}})\end{cases}}$

補充一：

在步驟 7 中，若假設 $(x_{1}+x_{2})={\sqrt {-y_{1}}},(x_{3}+x_{4})=-{\sqrt {-y_{1}}}$ 則需確認 ${\sqrt {-y_{1}}}\times {\sqrt {-y_{2}}}\times {\sqrt {-y_{3}}}=-q$

否則需假設 $(x_{1}+x_{2})=-{\sqrt {-y_{1}}},(x_{3}+x_{4})={\sqrt {-y_{1}}}$ 即滿足 $-{\sqrt {-y_{1}}}\times {\sqrt {-y_{2}}}\times {\sqrt {-y_{3}}}=-q$

補充二：

在步驟 4 中， $y_{1}y_{2}+y_{1}y_{3}+y_{2}y_{3}+4r=p^{2}+(x_{1}+x_{2}+x_{3}+x_{4})\times (x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4})$ ，兩邊展開化簡後得 $y_{1}y_{2}+y_{1}y_{3}+y_{2}y_{3}+4r=p^{2}\therefore y_{1}y_{2}+y_{1}y_{3}+y_{2}y_{3}=p^{2}-4r$
且 ${\begin{aligned}y_{1}y_{2}y_{3}+q^{2}=&(x_{1}+x_{2}+x_{3}+x_{4})\times [x_{1}x_{2}x_{3}(x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})+x_{1}x_{2}x_{4}(x_{1}x_{2}+x_{1}x_{4}+x_{2}x_{4})\\&+x_{1}x_{3}x_{4}(x_{1}x_{3}+x_{1}x_{4}+x_{3}x_{4})+x_{2}x_{3}x_{4}(x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4})+2x_{1}x_{2}x_{3}x_{4}(x_{1}+x_{2}+x_{3}+x_{4})]\end{aligned}}$
，兩邊展開化簡後得 $y_{1}y_{2}y_{3}+q^{2}=0\therefore y_{1}y_{2}y_{3}=-q^{2}$

例題

例題一

題目： $x^{4}-22x^{2}-48x-23=0$

$p=-22,q=-48,r=-23$ 則 $-2p=44,p^{2}-4r=576,q^{2}=2304$
$y_{1},y_{2},y_{3}$ 為 $y^{3}+44y^{2}+576y+2304=0$ 的三根
$\because (y+8)(y+12)(y+24)=0\therefore y_{1}=-8,y_{2}=-12,y_{3}=-24$
$\therefore$
$x_{1}={\frac {1}{2}}({\sqrt {8}}+{\sqrt {12}}+{\sqrt {24}})={\sqrt {2}}+{\sqrt {3}}+{\sqrt {6}}$

$x_{2}={\frac {1}{2}}({\sqrt {8}}-{\sqrt {12}}-{\sqrt {24}})={\sqrt {2}}-{\sqrt {3}}-{\sqrt {6}}$

$x_{3}={\frac {1}{2}}(-{\sqrt {8}}+{\sqrt {12}}-{\sqrt {24}})=-{\sqrt {2}}+{\sqrt {3}}-{\sqrt {6}}$

$x_{4}={\frac {1}{2}}(-{\sqrt {8}}-{\sqrt {12}}+{\sqrt {24}})=-{\sqrt {2}}-{\sqrt {3}}+{\sqrt {6}}$

例題二

題目： $x^{4}-22x^{2}+48x-23=0$

$p=-22,q=48,r=-23$ 則 $-2p=44,p^{2}-4r=576,q^{2}=2304$
$y_{1},y_{2},y_{3}$ 為 $y^{3}+44y^{2}+576y+2304=0$ 的三根
$\because (y+8)(y+12)(y+24)=0\therefore y_{1}=-8,y_{2}=-12,y_{3}=-24$
(到此之前除了 q 之外其他與例題一一模一樣)
但是因為 ${\sqrt {8}}\times {\sqrt {12}}\times {\sqrt {24}}=48\neq -q$ ，因此不可取 $x_{1}+x_{2}={\sqrt {-y_{1}}}$ ，應該取 $x_{1}+x_{2}=-{\sqrt {-y_{1}}}$ 。
再驗證一下： $-{\sqrt {8}}\times {\sqrt {12}}\times {\sqrt {24}}=-48=-q$ ，所以是正確的。
$\therefore$
$x_{1}={\frac {1}{2}}(-{\sqrt {8}}+{\sqrt {12}}+{\sqrt {24}})=-{\sqrt {2}}+{\sqrt {3}}+{\sqrt {6}}$

$x_{2}={\frac {1}{2}}(-{\sqrt {8}}-{\sqrt {12}}-{\sqrt {24}})=-{\sqrt {2}}-{\sqrt {3}}-{\sqrt {6}}$

$x_{3}={\frac {1}{2}}({\sqrt {8}}+{\sqrt {12}}-{\sqrt {24}})={\sqrt {2}}+{\sqrt {3}}-{\sqrt {6}}$

$x_{4}={\frac {1}{2}}({\sqrt {8}}-{\sqrt {12}}+{\sqrt {24}})={\sqrt {2}}-{\sqrt {3}}+{\sqrt {6}}$

例題三

題目： $x^{4}-2x^{2}+1=0$

$p=-2,q=0,r=1$ 則 $-2p=4,p^{2}-4r=0,q^{2}=0$
$y_{1},y_{2},y_{3}$ 為 $y^{3}+4y^{2}=0$ 的三根
$y_{1}=0,y_{2}=0,y_{3}=-4\therefore {\sqrt {-y_{1}}}=0,{\sqrt {-y_{2}}}=0,{\sqrt {-y_{3}}}=2$
$\therefore$
$x_{1}={\frac {1}{2}}(0+0+2)=1$

$x_{2}={\frac {1}{2}}(0-0-2)=-1$

$x_{3}={\frac {1}{2}}(-0+0-2)=-1$

$x_{4}={\frac {1}{2}}(-0-0+2)=1$

例題四

題目： $x^{4}-(h^{2}+k^{2})x^{2}+h^{2}k^{2}=0$

$p=-(h^{2}+k^{2}),q=0,r=h^{2}k^{2}$ 則 $-2p=2(h^{2}+k^{2}),p^{2}-4r=(h^{2}-k^{2})^{2},q^{2}=0$
$y_{1},y_{2},y_{3}$ 為 $y^{3}+2(h^{2}+k^{2})y^{2}+(h^{2}-k^{2})^{2}y=0$ 的三根， $y_{1}=0$
$y_{2},y_{3}$ 為 $y^{2}+2(h^{2}+k^{2})y+(h^{2}-k^{2})^{2}=0$ 的兩根
$y_{2}=-(h+k)^{2},y_{3}=-(h-k)^{2}\therefore {\sqrt {-y_{1}}}=0,{\sqrt {-y_{2}}}=h+k,{\sqrt {-y_{3}}}=h-k$
$\therefore$
$x_{1}={\frac {1}{2}}[0+(h+k)+(h-k)]=h$

$x_{2}={\frac {1}{2}}[0-(h+k)-(h-k)]=-h$

$x_{3}={\frac {1}{2}}[-0+(h+k)-(h-k)]=k$

$x_{4}={\frac {1}{2}}[-0-(h+k)+(h-k)]=-k$

例題五

題目： $x^{4}-3x^{2}-2x=0$

$p=-3,q=-2,r=0$ 則 $-2p=6,p^{2}-4r=9,q^{2}=4$
$y_{1},y_{2},y_{3}$ 為 $y^{3}+6y^{2}+9y+4=0$ 的三根
$\because (y+4)(y+1)(y+1)=0\therefore y_{1}=-4,y_{2}=-1,y_{3}=-1\therefore {\sqrt {-y_{1}}}=2,{\sqrt {-y_{2}}}=1,{\sqrt {-y_{3}}}=1$
$\therefore$
$x_{1}={\frac {1}{2}}(2+(1)+(1))=2$

$x_{2}={\frac {1}{2}}(2-(1)-(1))=0$

$x_{3}={\frac {1}{2}}(-2+(1)-(1))=-1$

$x_{4}={\frac {1}{2}}(-2-(1)+(1))=-1$

例題六

題目： $x^{4}-7x^{2}-6x=0$

$p=-7,q=-6,r=0$ 則 $-2p=14,p^{2}-4r=49,q^{2}=36$
$y_{1},y_{2},y_{3}$ 為 $y^{3}+14y^{2}+49y+36=0$ 的三根
$\because (y+9)(y+4)(y+1)=0\therefore y_{1}=-9,y_{2}=-4,y_{3}=-1\therefore {\sqrt {-y_{1}}}=3,{\sqrt {-y_{2}}}=2,{\sqrt {-y_{3}}}=1$
$\therefore$
$x_{1}={\frac {1}{2}}(3+2+1)=3$

$x_{2}={\frac {1}{2}}(3-2-1)=0$

$x_{3}={\frac {1}{2}}(-3+2-1)=-1$

$x_{4}={\frac {1}{2}}(-3-2+1)=-2$

例題七

題目： $x^{4}-1=0$

$p=0,q=0,r=-1$ 則 $-2p=0,p^{2}-4r=4,q^{2}=0$
$y_{1},y_{2},y_{3}$ 為 $y^{3}+4y=0$ 的三根
$\because (y)(y+2i)(y-2i)=0\therefore y_{1}=0,y_{2}=-2i,y_{3}=2i\therefore {\sqrt {-y_{1}}}=0,{\sqrt {-y_{2}}}=1+i,{\sqrt {-y_{3}}}=1-i$
$\therefore$
$x_{1}={\frac {1}{2}}(0+(1+i)+(1-i))=1$

$x_{2}={\frac {1}{2}}(0-(1+i)-(1-i))=-1$

$x_{3}={\frac {1}{2}}(-0+(1+i)-(1-i))=i$

$x_{4}={\frac {1}{2}}(-0-(1+i)+(1-i))=-i$

参见

一元三次方程式