# 一元四次方程式

## 解法

### 階段一：變形去除三次項

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

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

### 階段二：變身為三次方程式

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

4.

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

5.故 ${\displaystyle y_{1},y_{2},y_{3}}$ ${\displaystyle y^{3}-2py^{2}+(p^{2}-4r)y+q^{2}=0}$  的三根，

### 階段三：以三次方程式之三根求四次方程式之四根

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

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

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

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

### 補充二：

{\displaystyle {\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 ${\displaystyle y_{1}y_{2}y_{3}+q^{2}=0\therefore y_{1}y_{2}y_{3}=-q^{2}}$

## 例題

### 例題一

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

### 例題二

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

### 例題三

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

### 例題四

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

### 例題五

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

### 例題六

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

### 例題七

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