system of equations
Solve the following system of equations:a^2 - b = 1
b^2 - 2b - c = 2
c^2 + 2c - a = 1 咁難GE..
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-終場ソ使者- 發表於 18-6-2009 13:42 http://www4.nakuz.com/bbs/images/common/back.gif
我都放棄左[] 我都投降= = 本帖最後由 【YU】 於 18-6-2009 15:34 編輯
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:D
咁開心,緊係唔做啦~ 估完啦
a=-1
b=0
c=-2= =?
我好憎 trial&error 1234567890
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