system of equations

bearwing

Solve the following system of equations:

a^2 - b = 1
b^2 - 2b - c = 2
c^2 + 2c - a = 1

L.K.

咁難GE..
開估...

-終場ソ使者-

計計下計左個八次方程出黎[[tk_21]]

piyopiyo

計計下計左個八次方程出黎[[tk_21]]
-終場ソ使者- 發表於 18-6-2009 13:42

我都放棄左[[tk_21]]

Crossbone-X1

我都投降= =

[S]【YU】

a^2 - b = 1
b^2 - 2b - c = 2
c^2 + 2c - a = 1

b = a^2 - 1
(a^2-1)^2 - 2(a^2-1) = 2+c
c = a^4 - 4a^2 +1
(a^4 - 4a^2 +1)^2 + 2(a^4 - 4a^2 + 1) - a = 1
substitution and solve a  
咁開心,緊係唔做啦~

Crossbone-X1

估完啦
a=-1
b=0
c=-2  = =?
我好憎 trial&error

bearwing

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