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設二元一次方程ax^2+bx+c=0(a不為0) 當△≥0時 x1=(-b+根號下△)/2a x2=(-b-根號下△)/2a 所以x1+x2=[(-b+根號下△)/2a]+[(-b-根號下△)/2a] =-2b/2a =-b/a 同理,x1x2=[(-b+根號下△)/2a]*[(-b-根號下△)/2a] =[(-b)^2-(b^2-4ac)]/4a^2 =4ac/4a^2 =c/a
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