- 一元四次方程式解法计算机
阶段一:变形去除三次项[编辑]
1.
以
代入
2.得
,令其四根为
阶段二:变身为三次方程式[编辑]
3.由
可得
4.
设![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/876cf9b61f67734078c5857ba7034ddc13edde4d) | 则 | ![{\displaystyle {\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0908114cc06734bfecdead4792f35aff30bfe5a9) | 注意:
对 而言是 2 次
对 而言是 2 次
对 而言是 4 次
对 而言是 6 次 |
---|
5.故
为
的三根,
此方程式对
而言是 6 次,其四项对
而言分别是 0+6 次、 2+4 次、 4+2 次、 6+0 次。
阶段三:以三次方程式之三根求四次方程式之四根[编辑]
6.
为
之二根,
或
7.设
(此有另一种解,判断方法补充于最后)
8.同理,可得
9.解联立方程式,得
补充一:[编辑]
在步骤 7 中,若假设
则需确认
否则需假设
即满足
补充二:[编辑]
在步骤 4 中,
,两边展开化简后得 ![{\displaystyle 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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5380be4063500a759dc26e9dc757db74e97f3f5b)
且![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1fcfdb6815c339611987b915ab32c974a77906b)
,两边展开化简后得
例题一[编辑]
题目:
则 ![{\displaystyle -2p=44,p^{2}-4r=576,q^{2}=2304}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94266785f59de173ac757f36cd07e45f9b242a3a)
为
的三根
![{\displaystyle \because (y+8)(y+12)(y+24)=0\therefore y_{1}=-8,y_{2}=-12,y_{3}=-24}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d506bf6dc33069d5c3ed7e10d9e3f104ed84d44a)
![{\displaystyle x_{1}={\frac {1}{2}}({\sqrt {8}}+{\sqrt {12}}+{\sqrt {24}})={\sqrt {2}}+{\sqrt {3}}+{\sqrt {6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ace476345fbbf16a802ff2e61cbb5fc1e641eb2)
![{\displaystyle x_{2}={\frac {1}{2}}({\sqrt {8}}-{\sqrt {12}}-{\sqrt {24}})={\sqrt {2}}-{\sqrt {3}}-{\sqrt {6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae8c116d24ba6803f2ac189fdfdcb0e84b70b90d)
![{\displaystyle x_{3}={\frac {1}{2}}(-{\sqrt {8}}+{\sqrt {12}}-{\sqrt {24}})=-{\sqrt {2}}+{\sqrt {3}}-{\sqrt {6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/60e5b08d9216813412f6512d7bb351a436d10bd9)
![{\displaystyle x_{4}={\frac {1}{2}}(-{\sqrt {8}}-{\sqrt {12}}+{\sqrt {24}})=-{\sqrt {2}}-{\sqrt {3}}+{\sqrt {6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08f2e4a9991cdbc1dfa79f4c8b9222a9059d8dc1)
例题二[编辑]
题目:
则 ![{\displaystyle -2p=44,p^{2}-4r=576,q^{2}=2304}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94266785f59de173ac757f36cd07e45f9b242a3a)
为
的三根
![{\displaystyle \because (y+8)(y+12)(y+24)=0\therefore y_{1}=-8,y_{2}=-12,y_{3}=-24}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d506bf6dc33069d5c3ed7e10d9e3f104ed84d44a)
(到此之前除了 q 之外其他与例题一一模一样)
- 但是因为
,因此不可取
,应该取
。
再验证一下:
,所以是正确的。
![{\displaystyle x_{1}={\frac {1}{2}}(-{\sqrt {8}}+{\sqrt {12}}+{\sqrt {24}})=-{\sqrt {2}}+{\sqrt {3}}+{\sqrt {6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d976ce2c3c73c3a64c5c9dde6ae1f79bc49f7959)
![{\displaystyle x_{2}={\frac {1}{2}}(-{\sqrt {8}}-{\sqrt {12}}-{\sqrt {24}})=-{\sqrt {2}}-{\sqrt {3}}-{\sqrt {6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/799e2a8f787453302ec71513e389f166e3e60ce9)
![{\displaystyle x_{3}={\frac {1}{2}}({\sqrt {8}}+{\sqrt {12}}-{\sqrt {24}})={\sqrt {2}}+{\sqrt {3}}-{\sqrt {6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5932956d6ebbfd2b867c24f40e968108a395d3d)
![{\displaystyle x_{4}={\frac {1}{2}}({\sqrt {8}}-{\sqrt {12}}+{\sqrt {24}})={\sqrt {2}}-{\sqrt {3}}+{\sqrt {6}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a26508d4debb85e72c89cfde3160f904de88378a)
例题三[编辑]
题目:
则 ![{\displaystyle -2p=4,p^{2}-4r=0,q^{2}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df44c4d6d17d366b860a3dd4e73ae2dad25c6eee)
为
的三根
![{\displaystyle y_{1}=0,y_{2}=0,y_{3}=-4\therefore {\sqrt {-y_{1}}}=0,{\sqrt {-y_{2}}}=0,{\sqrt {-y_{3}}}=2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7ad15beb2636de57b4f3f681cc3892a65cc2981)
![{\displaystyle x_{1}={\frac {1}{2}}(0+0+2)=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83a4a82e73d66011d1847531270fe0bede0756d0)
![{\displaystyle x_{2}={\frac {1}{2}}(0-0-2)=-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e06a10f7cef4542cff3a0b02a40b1fb5604ebfc2)
![{\displaystyle x_{3}={\frac {1}{2}}(-0+0-2)=-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b6dcc271c7fee7d4c16c325bea63bcdcb1208b7)
![{\displaystyle x_{4}={\frac {1}{2}}(-0-0+2)=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c33e09a5e5cbb5ededb0fdab96bf5e90339c1a7)
例题四[编辑]
题目:
则 ![{\displaystyle -2p=2(h^{2}+k^{2}),p^{2}-4r=(h^{2}-k^{2})^{2},q^{2}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4d1b2d5fdda1131b2312f41f3010d12e9d2a6d50)
为
的三根,![{\displaystyle y_{1}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ff2cde62f4affa983601a506f2740aaa0d808d4)
为
的两根
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/645638c9b6d7782a5201c520fe64883572aa4bc4)
![{\displaystyle x_{1}={\frac {1}{2}}[0+(h+k)+(h-k)]=h}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48b8159f264638d28f4a5f1e54f405159933682b)
![{\displaystyle x_{2}={\frac {1}{2}}[0-(h+k)-(h-k)]=-h}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed8fe98f4c926a388f4f5ef1ec4632374dba08d8)
![{\displaystyle x_{3}={\frac {1}{2}}[-0+(h+k)-(h-k)]=k}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f85fa58eca36a027f244c2d368e69a58e8e1503f)
![{\displaystyle x_{4}={\frac {1}{2}}[-0-(h+k)+(h-k)]=-k}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3aa85d8a014796a8a31014483deef12d53479f0a)
例题五[编辑]
题目:
则 ![{\displaystyle -2p=6,p^{2}-4r=9,q^{2}=4}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84eb8db21025ad23b3ca173dbf8d488fae8c56b1)
为
的三根
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/801206e431edc042170a98b112dcd50d125afcc8)
![{\displaystyle x_{1}={\frac {1}{2}}(2+(1)+(1))=2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/24924f5036c90d63441b085ada365b46f251a448)
![{\displaystyle x_{2}={\frac {1}{2}}(2-(1)-(1))=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49626c71071584caeae99212f2e1f80082f191bd)
![{\displaystyle x_{3}={\frac {1}{2}}(-2+(1)-(1))=-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7b58dcb3431f3c5e78561eabbb1f97f0ed63deba)
![{\displaystyle x_{4}={\frac {1}{2}}(-2-(1)+(1))=-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c30f70c8b4c29da8f4af03f70a773def89f3a1c)
例题六[编辑]
题目:
则 ![{\displaystyle -2p=14,p^{2}-4r=49,q^{2}=36}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7986a3c8689d84a48cd59e3120517c6e9026d70)
为
的三根
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e840a7c0cd1d2644b30f28ad53efc93e78e29cd5)
![{\displaystyle x_{1}={\frac {1}{2}}(3+2+1)=3}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ddfb0084911f8300fa954ce39e4eed546732c41)
![{\displaystyle x_{2}={\frac {1}{2}}(3-2-1)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6348a7828e12b8735fea8cb884e0a6c06e560c2e)
![{\displaystyle x_{3}={\frac {1}{2}}(-3+2-1)=-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/644d796ec54d17f3d632f4cfc8a2267335b5c65a)
![{\displaystyle x_{4}={\frac {1}{2}}(-3-2+1)=-2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c3d15e1c41aff2968dbb241ce00191a0e7c36c2)
例题七[编辑]
题目:
则 ![{\displaystyle -2p=0,p^{2}-4r=4,q^{2}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c9e51f881cc1c1b097e5c56e17cddf9b0957497)
为
的三根
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a08907c5011a3d56483ea10a6127c03bab91683a)
![{\displaystyle x_{1}={\frac {1}{2}}(0+(1+i)+(1-i))=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb951d9ee572faceb4e765aadced535d2f12dfa4)
![{\displaystyle x_{2}={\frac {1}{2}}(0-(1+i)-(1-i))=-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9268d203a0857477f48a63fdff9a4ae2095991af)
![{\displaystyle x_{3}={\frac {1}{2}}(-0+(1+i)-(1-i))=i}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9ecf22519312ee7fa479c157b6abebad137b35d7)
![{\displaystyle x_{4}={\frac {1}{2}}(-0-(1+i)+(1-i))=-i}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e8b10e1f532a4a77cdb07a68552a9d6b25bbf19)