设f(x)=ax^2+bx+c,依题设得f(0)=c=1.
设方程ax^2+bx+c=0的两个根分别是x1和x2,则有(x1-x2)^2=(b/a)^2-4c/a=(2√2)^2=8
另由f(x-2)=f(-x-2)得a(x-2)^2+b(x-2)+c=a(x+2)^2-b(x+2)+c,对比系数得b=4a.
联立方程得a=(1+√5)/4,b=1+√5,c=1或a=a=(1-√5)/4,b=1-√5,c=1
故f(x)=(1+√5)/4*x^2+(1+√5)x+1或f(x)=(1-√5)/4*x^2+(1-√5)x+1