Question

已知:（b-c)x的平方+（c-a)x+(a-b）=0有俩个相等的实数根,求证:2b=a+c

Answer 1

两个等跟则判别式等于0

所以(c-a)²-4(b-c)(a-b)=0

(a-c)²-4(b-c)(a-b)=0

[(a-b)+(b-c)]²-4(b-c)(a-b)=0

(a-b)²+2(b-c)(a-b)+(b-c)²-4(b-c)(a-b)=0

(a-b)²-2(b-c)(a-b)+(b-c)²=0

[(a-b)-(b-c)]²=0

(a+c-2b)²

a+c-2b=0

a+c=2