设a.b.c是互不相等的实数,且方程(b-c)x^2+(c-a)x+(a-b)=0有两个实数根,证明2b=a+c

问题描述:

设a.b.c是互不相等的实数,且方程(b-c)x^2+(c-a)x+(a-b)=0有两个实数根,证明2b=a+c

是不是有两个相等的实数根?
判别式等于0
(c-a)^2-4(b-c)(a-b)=0
(a-c)^2-4(b-c)(a-b)=0
[(b-c)+(a-b)]^2-4(b-c)(a-b)=0
(b-c)^2+(a-b)^2+2(b-c)(a-b)-4(b-c)(a-b)=0
(b-c)^2+(a-b)^2-2(b-c)(a-b)=0
[(b-c)-(a-b)]^2=0
(2b-a-c)^2=0
2b-c-a=0
2b=a+c