What is the minimum value of the expression x2+y2+2x−4y+8 for real x and y?
just complete the square in x and y
x2+y2+2x−4y+8(x2+2x+1−1)+(y2−4y+4−4)+8(x+1)2−1+(y−2)2−4+8(x+1)2+(y−2)2+3The minimum value is clearly seen to be 3
+1246 
