已知实数x,y 满足x2+y2+2x=0则x+y得最小值如题

已知实数x,y 满足x2+y2+2x=0则x+y得最小值
如题

设x+y=k,代入x2+y2+2x=0
x2+(k-x)2+2x=0
x2+k2-2kx+x2+2x=0
2x2-(2k-2)x+k2=0
判别式=(2k-2)2-4*2k2>=0
4k2-8k+4-8k2>=0
4k2+8k-4