Question

一道挺难的二次函数题 已知y=ax^2+bx+c.(a＞0)

Answer 1

ax1^2+bx1+c=a+b+c=0,c=-(a+b),x1=[-b+-(b^2-4ac)^0.5]/(2a)=[-b+-(b+2a)]/(2a)

a(x1^2-1)+b(x1-1)=0

(x1-1)*(ax1+a+b)=0,x1=1 or x1=-(a+b)/a

-a=am^2+bm+c,b=am^2/(m-1)

xi=1,or x1=-[a+am^2/(m-1)]/a=m^2/(1-m)-1