已知二次函数y=ax2+bx(a≠0),当x取x1,x2(x1≠x2)时,函数值相等,求当x取x1+x2时的函数值

2025-06-23 22:20:50
推荐回答(1个)
回答1:

根据题意得ax12+bx1=ax22+bx2
ax12-ax22+bx1-bx2=0,
a(x1-x2)(x1+x2)+b(x1-x2)=0,
(x1-x2)(ax1+ax2+b)=0,
∵x1≠x2
∴ax1+ax2+b=0,即x1+x2=-

b
a

∴当x=x1+x2=-
b
a
时,y=a×(-
b
a
2+b×(-
b
a
)=0.