Question

若x1 x2是函数fx=x2+mx-2（m属于R）的两个零点

Answer 1

△≥0 x1+x2=-m x1x2=-2

m^2+8≥0 满足

(x2-x1)^2=x1^2+x2^2-2x1x2=(x1+x2)^2-4x1x2

=m^2+8

当m=0时(x2-x1)^2有最小值8

x2-x1的最小值2√2