Question

X^2+2(M+1)X+3M^2+4MN+4N^2+2=0

Answer 1

X^2+2(M+1)X+3M^2+4MN+4N^2+2=(X+M+1)^2-M^2-2M-1+3M^2+4MN+4N^2+2=(X+M+1)^2+M^2-2M+1 +M^2+4MN+4N^2=(X+M+1)^2+(M-1)^2+(M+2N)^2=0 有实根所以(M-1)^2+(M+2N)^2