應(yīng)該是相等實(shí)數(shù)根吧
2x^2+2(a-c)x+(a-b)^2+(b-c)^2=0的兩個相等的實(shí)數(shù)根,所以判別式=0,
[2(a-c)]^2-4*2*[(a-b)^2+(b-c)^2]=0
(a-c)^2-2*[(a-b)^2+(b-c)^2]=0
整理,a^2+2ac+c^2-4b(a+c)+4b^2=0
(a+c)^2-4a(a+c)+4b^2=0
(a+c-2b)^2=0
a+c-2b=0
b=(a+c)/2