求下列五次方程的根式解：x(x+1)(x+2)(x+3)(x+4)=120

drc2000再来

求教:请写出下列五次方程的根式解.
x(x+1)(x+2)(x+3)(x+4)=120
(数值方法得
有实根x=1,
虚根
x=-1.1083+-2.1194i
x= 4.4147+-1.2944i,
但非所愿)

ysr

输入1:  a=1,  b=-2,  c=3,  d=4，e=5;  输出结果1:    x1,2=-0.5795978599+ -0.7253449538i,  x3,4=1.5795978599+ -1.8179487826i
输入2:  a=1,  b=2,  c=-3,  d=4，e=5;  输出结果2:    x1,2=-1.9556023509+ -1.2268754722,  x3,4=0.9556023509+ -1.1148003552i  
输入3:  a=1,  b=2,  c=3,  d=4，e=5;  输出结果3:    x1,2=-0.7790981610+ -1.3540192270,  x3,4=-0.2209018390+ -1.6929245011i  
输入4:  a=1,  b=11,  c=46,  d=96，e=120;  
输出结果4:    x1,2=-4.4147314911+ -1.2945834215i,  x3,4=-1.0852685089+ -2.1193668017i

这是用天山草老师的方法编程计算的结果，代码如下：
Private Sub Command1_Click()
Dim a, b, c, d, f, g
Dim ja, jb, jc
k = DeleteSpace(Text1.Text)
ja = DeleteSpace(Text2.Text)
jb = DeleteSpace(Text3.Text)
jc = DeleteSpace(Text4.Text)
jd = DeleteSpace(Text5.Text)
sd1 = DeleteSpace(Text15.Text)
If Abs(Val(sd1)) <= 10 Then
sd = 10 + 2
Else
If Len(sd1) > 3 And Abs(Val(sd1)) > 300 Then
sd = 300 + 2
Else
sd = Int(Abs(Val(sd1))) + 2
End If
End If

a2 = zhengliys2(Trim(ja), Val(sd))
b2 = zhengliys2(Trim(jb), Val(sd))

c2 = zhengliys2(Trim(jc), Val(sd))
D2 = zhengliys2(Trim(jd), Val(sd))
k2 = zhengliys2(Trim(k), Val(sd))
If MBJC(Trim(k2), 0) = 0 Then
Text4.Text = "a 不能为 0"
   Text5.Text = "a 为 0可能已不是1元4次方程"
  Else
   
a3 = mcc2(Trim(a2), Trim(k2), Val(sd))
b3 = mcc2(Trim(b2), Trim(k2), Val(sd))
c3 = mcc2(Trim(c2), Trim(k2), Val(sd))
d3 = mcc2(Trim(D2), Trim(k2), Val(sd))
jk = 8 & String(sd, "0")
ja1 = mbc2("-4" & String(sd, "0"), Trim(b3), Val(sd))
jb1 = mpc2(mbc2(2 & String(sd, "0"), mbc2(Trim(a3), Trim(c3), Val(sd)), Val(sd)), mbc2(8 & String(sd, "0"), Trim(d3), Val(sd)))
jc1 = mpc2(mbc2(Trim(d3), mpc2(mbc2(4 & String(sd, "0"), Trim(b3), Val(sd)), mbc2(Trim(a3), Trim(a3), Val(sd))), Val(sd)), mbc2(Trim(c3), Trim(c3), Val(sd)))
y = jie3cifc(Trim(ja1), Trim(jb1), Trim(jc1), Trim(jk), Val(sd))
Y1 = zhengliys2(Trim(y), Val(sd))
End If
Text6 = zhengliys2(Trim(y), Val(sd))
Text8 = y
za = 1 & String(sd, "0")
z2 = mpc2(mpc3(mbc2(8 & String(sd, "0"), Trim(Y1), Val(sd)), mbc2(Trim(a3), Trim(a3), Val(sd))), mbc2(4 & String(sd, "0"), Trim(b3), Val(sd)))
z2 = mbbc2(qdfh(Trim(z2)), Val(sd))

zb1 = mcc2(mpc3(Trim(a3), Trim(z2)), 2 & String(sd, "0"), Val(sd))
zb2 = mcc2(mpc2(Trim(a3), Trim(z2)), 2 & String(sd, "0"), Val(sd))
zc1 = mpc3(Trim(Y1), mcc2(mpc2(mbc2(Trim(a3), Trim(Y1), Val(sd)), Trim(c3)), Trim(z2), Val(sd)))
zc2 = mpc2(Trim(Y1), mcc2(mpc2(mbc2(Trim(a3), Trim(Y1), Val(sd)), Trim(c3)), Trim(z2), Val(sd)))
y3 = jie2cifc(Trim(za), Trim(zb1), Trim(zc1), Val(sd))
y4 = jie2cifc(Trim(za), Trim(zb2), Trim(zc2), Val(sd))
Text9 = y3
Text10 = y4
Label1.Caption = Val(Label1.Caption) + 1
Text14.Text = Text14.Text & "  输入" & Label1.Caption & ":  " & "a=" & k & ",  b=" & ja & ",  c=" & jb & ",  d=" & jc _
  & "，e=" & jd & ";  输出结果" & Label1.Caption & ":    " & "x1,2=" & Text9.Text & ",  x3,4=" & Text10.Text
End Sub

ysr

主楼的方程是有根式解的，如下是朋友给出的根式解：（看上去有点复杂）
解 0
x = 1


解 1
x = (-sqrt(225/(2*sqrt((36*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(2/3)-15*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3)+1552)/((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3)))-((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3)-388/(9*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3))-5/6)/2)-sqrt((36*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(2/3)-15*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3)+1552)/((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3))/12-11/4

     
解 2
x = sqrt(225/(2*sqrt((36*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(2/3)-15*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3)+1552)/((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3)))-((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3)-388/(9*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3))-5/6)/2-sqrt((36*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(2/3)-15*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3)+1552)/((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3))/12-11/4

     
解 3
x = (-sqrt((-225/(2*sqrt((36*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(2/3)-15*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3)+1552)/((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3))))-((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3)-388/(9*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3))-5/6)/2)+sqrt((36*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(2/3)-15*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3)+1552)/((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3))/12-11/4

     
解 4
x = sqrt((-225/(2*sqrt((36*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(2/3)-15*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3)+1552)/((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3))))-((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3)-388/(9*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3))-5/6)/2+sqrt((36*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(2/3)-15*((4*sqrt(14969)*%i)/sqrt(3)+460/27)^(1/3)+1552)/(…………

ysr

输入1:  a=1,  b=2,  c=3,  d=4，e=-5;  输出结果1:    x1,2=-0.687509062557641972761017201778+ -1.371633382010711683016399697636,  x3,4=-0.312490937442358027238982798221+ -1.857874439171989445346701282261i  

输入2:  a=1,  b=11,  c=46,  d=96，e=120;  输出结果2:    x1,2=-4.414731491148088975319044433426+ -1.294583421496928798415482164043i,  x3,4=-1.085268508851911024680955566574+ -2.119366801671099328593384848563i

ysr

输入1:  a=1,  b=-2,  c=3,  d=4，e=5;  输出结果1:    x1,2=-0.5795978599+ -0.7253449538i,  x3,4=1.5795978599+ -1.8179487826i

这种情况应该是对的，求A的时候，根号里面是正值

代码如下：
Private Sub Command1_Click()
Dim a, b, c, d, f, g
Dim ja, jb, jc
k = DeleteSpace(Text1.Text)
ja = DeleteSpace(Text2.Text)
jb = DeleteSpace(Text3.Text)
jc = DeleteSpace(Text4.Text)
jd = DeleteSpace(Text5.Text)
sd1 = DeleteSpace(Text15.Text)
If Abs(Val(sd1)) <= 10 Then
sd = 10 + 2
Else
If Len(sd1) > 3 And Abs(Val(sd1)) > 300 Then
sd = 300 + 2
Else
sd = Int(Abs(Val(sd1))) + 2
End If
End If

a2 = zhengliys2(Trim(ja), Val(sd))
b2 = zhengliys2(Trim(jb), Val(sd))

c2 = zhengliys2(Trim(jc), Val(sd))
D2 = zhengliys2(Trim(jd), Val(sd))
k2 = zhengliys2(Trim(k), Val(sd))
If MBJC(Trim(k2), 0) = 0 Then
Text4.Text = "a 不能为 0"
   Text5.Text = "a 为 0可能已不是1元4次方程"
  Else
   
a3 = mcc2(Trim(a2), Trim(k2), Val(sd))
b3 = mcc2(Trim(b2), Trim(k2), Val(sd))
c3 = mcc2(Trim(c2), Trim(k2), Val(sd))
d3 = mcc2(Trim(D2), Trim(k2), Val(sd))
jk = 8 & String(sd, "0")
ja1 = mbc2("-4" & String(sd, "0"), Trim(b3), Val(sd))
jb1 = mpc2(mbc2(2 & String(sd, "0"), mbc2(Trim(a3), Trim(c3), Val(sd)), Val(sd)), mbc2(8 & String(sd, "0"), Trim(d3), Val(sd)))
jc1 = mpc2(mbc2(Trim(d3), mpc2(mbc2(4 & String(sd, "0"), Trim(b3), Val(sd)), mbc2(Trim(a3), Trim(a3), Val(sd))), Val(sd)), mbc2(Trim(c3), Trim(c3), Val(sd)))
y = jie3cifc(Trim(ja1), Trim(jb1), Trim(jc1), Trim(jk), Val(sd))
Y1 = zhengliys2(Trim(y), Val(sd))
End If
Text6 = zhengliys2(Trim(y), Val(sd))
Text8 = y
za = 1 & String(sd, "0")
z2 = mpc2(mpc3(mbc2(8 & String(sd, "0"), Trim(Y1), Val(sd)), mbc2(Trim(a3), Trim(a3), Val(sd))), mbc2(4 & String(sd, "0"), Trim(b3), Val(sd)))
z2 = mbbc2(qdfh(Trim(z2)), Val(sd))
zb1 = mcc2(mpc3(Trim(a3), Trim(z2)), 2 & String(sd, "0"), Val(sd))
zb2 = mcc2(mpc2(Trim(a3), Trim(z2)), 2 & String(sd, "0"), Val(sd))
zc1 = mpc3(Trim(Y1), mcc2(mpc2(mbc2(Trim(a3), Trim(Y1), Val(sd)), Trim(c3)), Trim(z2), Val(sd)))
zc2 = mpc2(Trim(Y1), mcc2(mpc2(mbc2(Trim(a3), Trim(Y1), Val(sd)), Trim(c3)), Trim(z2), Val(sd)))
y3 = jie2cifc(Trim(za), Trim(zb1), Trim(zc1), Val(sd))
y4 = jie2cifc(Trim(za), Trim(zb2), Trim(zc2), Val(sd))
Text9 = y3
Text10 = y4
Label1.Caption = Val(Label1.Caption) + 1
Text14.Text = Text14.Text & "  输入" & Label1.Caption & ":  " & "a=" & k & ",  b=" & ja & ",  c=" & jb & ",  d=" & jc _
  & "，e=" & jd & ";  输出结果" & Label1.Caption & ":    " & "x1,2=" & Text9.Text & ",  x3,4=" & Text10.Text
End Sub

ysr

Public Function jie2cifc(a2 As String, b2 As String, c2 As String, sd As String) As String '2次方程
Dim d, y
d = mbc2(Trim(b2), Trim(b2), Val(sd))
D1 = mpc2(Trim(d), mbc2(4 & String(sd, "0"), mbc2(Trim(a2), Trim(c2), Val(sd)), Val(sd)))
d3 = qdfh(Trim(D1))
D2 = mbbc2(Trim(d3), Val(sd))
y = mcc2(Trim(b2), mbc2(2 & String(sd, "0"), Trim(a2), Val(sd)), Val(sd))
sf = fhys(Trim(y))
sf = Val(-1 * sf)
y = tjfh(qdfh(Trim(y)), Val(sf))
y = shuchujg(Trim(y), Val(sd))
If mbjc2(Trim(D1), 0) >= 0 Then
d3 = mcc2(Trim(D2), mbc2(2 & String(sd, "0"), Trim(a2), Val(sd)), Val(sd))
d3 = shuchujg(Trim(d3), Val(sd))
Else
d3 = mcc2(Trim(D2), mbc2(2 & String(sd, "0"), Trim(a2), Val(sd)), Val(sd))
d3 = shuchujg(Trim(d3), Val(sd)) & "i"
End If
jie2cifc = y & "+ -" & d3
End Function

ysr

一元4次方程的系数依次为：1,11,46,96,120，结果：
-4.4147314911+ -1.2945834215i

-1.0852685089+ -2.1193668017i
顶一下，研究解方程的程序

风花飘飘

这不是5次方程，

llshs好石

drc2000再来 发表于 2021-4-18 17:32
谢谢各位老师!辛苦了!
x(x+1)(x+2)(x+3)(x+4)=120的根除了x=1,其余的那么复杂,简直没天理!

zhuanlan.zhihu.com/p/345101278
一元四次方程解法全集
请多多指教，谢谢

drc2000再来

谢谢各位老师!辛苦了!
x(x+1)(x+2)(x+3)(x+4)=120的根除了x=1,其余的那么复杂,简直没天理!

ysr

两种方法理论上应该是一致的一样的，精确度好象不一样，下午上班，有空再计算一下，并代入方程验证一下，或者编程验证一下，那个结果精确度高？