Aa+Bb+Cab+D=0 and Ea+Fb+Gab+H=0

I have used your approach, and got about the same thing, except that I have the complete details for your reference.
Since a has two solutions, so does b. So there are two sets of solutions, (a1,b1) and (a2,b2). Here's what I've got, and let me know if they agree with your solutions:

a1=(-(B*(-sqrt(C^2*H^2 ((4*A*B-2*C*D)*G-2*A*C*F - 2*B*C*E)*H D^2*G^2 (-2*A*D*F-2*B*D*E)*G A^2*F^2 (4*C*D-2*A*B)*E*F B^2*E^2) C*H-D*G A*F-B*E)) / (2*B*G-2*C*F)-D)/((C*(-sqrt(C^2*H^2 ((4*A*B-2*C*D)*G-2*A*C*F-2*B*C*E)*H D^2*G^2 (-2*A*D*F-2*B*D*E)*G A^2*F^2 (4*C*D-2*A*B)*E*F B^2*E^2) C*H-D*G A*F-B*E))/(2*B*G-2*C*F) A)
b1=(-sqrt(C^2*H^2 ((4*A*B-2*C*D)*G-2*A*C*F-2*B*C*E)*H D^2*G^2 (-2*A*D*F-2*B*D*E)*G A^2*F^2 (4*C*D-2*A*B)*E*F B^2*E^2) C*H-D*G A*F-B*E)/(2*B*G-2*C*F)

a2=(-(B*(sqrt(C^2*H^2 ((4*A*B-2*C*D)*G-2*A*C*F-2*B*C*E)*H D^2*G^2 (-2*A*D*F-2*B*D*E)*G A^2*F^2 (4*C*D-2*A*B)*E*F B^2*E^2) C*H-D*G A*F-B*E)) / (2*B*G-2*C*F)-D)/((C*(sqrt(C^2*H^2 ((4*A*B-2*C*D)*G-2*A*C*F-2*B*C*E)*H D^2*G^2 (-2*A*D*F-2*B*D*E)*G A^2*F^2 (4*C*D-2*A*B)*E*F B^2*E^2) C*H-D*G A*F-B*E))/(2*B*G-2*C*F) A)
b2=(sqrt(C^2*H^2 ((4*A*B-2*C*D)*G-2*A*C*F-2*B*C*E)*H D^2*G^2 (-2*A*D*F-2*B*D*E)*G A^2*F^2 (4*C*D-2*A*B)*E*F B^2*E^2) C*H-D*G A*F-B*E)/(2*B*G-2*C*F)

其實都差唔多, 雖然有少少錯.

用電腦計的答案係:

a=(-Y(+-)√Y^2-4WZ)/(2X)

X = C E - A G
Y = B E - A F + D G - C H
W = C F- B G
Z = D E - A H

你自己對一對啦

2008-05-16 06:24:24 補充：
係 a=(-Y(+-)√Y^2+4WZ)/(2X)

ABCDEFGH都係constant呀, 你所謂的unknown 只有a同b咋
舉個例:

Ax^2+Bx+C=0

x is:

x=-(B(+-)√(B^2-4AC))/2A

我expect一個類似咁的答案...
我問果條問題Year 1時好似教過, 但我數學太差唔太記得點拆解

朋友 你講緊笑啊?!
呢條都叫數?!
你唔好再多D unknown @@