數學知識交流 － 二次方程 (2)續1

Case (1) Assume that a+b+c=2abc.
Since sum of roots = -b/a
-b/a=abc+2abc
=3abc
a^2*c=-1/3
a=sqrt(-1/(3c))
c=-1/(3a^2).........(1)
Since product of roots = c/a
c/a = 2(abc)^2
1=2a^3*b^2*c
a^3*b^2*c=1/2........(2)
Sub(1)into(2),
a*b^2=-3/2
a=-3/(2b^2)..........(3)
b=sqrt(-3/2a)
Sub (3) into (1),
c=-4b^4/27
b=quarticroot(-27c/4)
Case (2) Assume 2(a+b+c)=abc,
-b/a=3/2*abc
-2/3=a^2*c
a=sqrt(-2/(3c))
c=-2/(3a^2).....(4)
c/a=1/2*(abc)^2
2=a^3 * b^2 *c.....(5)
Sub (4) into (5),
a*b^2=-3
a=-3/(b^2)............(6)
b=sqrt(-3/a)
Sub (6) into (4),
c=-2/27*b^4
b=quarticroot(-27c/2)

Pointless to label someone's question~ Troublesome? Or else, 有人不想動腦筋? For opinions, they should be constructive or fun.
From a passer-by; 口痕痕~

If you don't like to answer , you can just see and leave instead of saying ' These kind of Qs aren't very meaningful... '

Too troublesome...

These kind of Qs aren't very meaningful...

Right?

2012-01-02 15:28:57 補充：
I'm just giving my opinion to you.