题目内容：

关于x的方程a(x+m)²+b=0的解是x1=-2,x2=1,求方程a(x+m+2)平方+b=0的值

优质解答

a不为0.0=ax^2 + 2amx + am^2+b.由韦达定理,(x1)(x2)=-2=[am^2+b]/a,x1+x2=-1=-2am/a=-2m.m=1/2.-2a=[am^2+b]=a/4+b,b = -9a/4.0=a(x+m+2)^2+b=a(x+5/2)^2-9a/4,0=(x+5/2)^2 - 9/4 = (x+5/2 + 3/2)(x+5/2 - 3/2) = (...