已知函数f(x)=ax^2+bx(ab不等于0),若x1不等于x2,且f(x1)=f(x2),则f(x1+x2)的值是?

Question

已知函数f(x)=ax^2+bx(ab不等于0),若x1不等于x2,且f(x1)=f(x2),则f(x1+x2)的值是?

Answer 1

f(x1)=f(x2)
所以ax1^2+bx1=ax2^2+bx2
a(x1^2-x2^2)+b(x1-x2)=0
a(x1+x2)(x1-x2)+b(x1-x2)=0
(x1-x2)[a(x1+x2)+b]=0
x1不等于x2
所以x1-x2不等于0
所以a(x1+x2)+b=0
x1+x2=-b/a
f(x1+x2)=a(x1+x2)^2+b(x1+x2)
=a*(-b/a)^2+b(-b/a)
=b^2/a-b^2/a
=0