Question

设函数f(x)=ax2+bx+c(a>0),且f(1)=-a/2,求证函数有两个零点

Answer 1

由f(1)=-a/2 => -a/2=a+b+c => -b = 3a/2+c

b2-4ac=9a2/4+c2+3ac-4ac=2a2+(a/2-c)2>0

所以该函数有2个解,即有两个零点.