設(shè)拋物線y=x²+(k+4)x+k與x軸兩交點為(x1,0),(x2,0),
即x²+(k+4)x+k=0兩根為x1,x2,
由根與系數(shù)的關(guān)系,得,
x1+x2=-(k+4),x1*x2=k
所以(x1-x2)²
=x1²-2x1*x2+x2²
=(x1+x2)²-4x1*x2
=(k+4)²-4k
=k²+8k+16-4k
=k²+4k+16
=（k+2）²+12
當k=-2時,(x1-x2)²有最小值為12
即二次函數(shù)y=x²+(k+4)x+k的圖像與x軸兩個交點間的最短距離為|x1-x2|=2√3