将方程x平方+（根号2+1)x+根号2=0配方后,是怎么变成『x+ （根号2+1）/2』平方 -（3-2倍根号2）/4 =0?
问题描述：
将方程x平方+（根号2+1)x+根号2=0配方后,是怎么变成『x+ （根号2+1）/2』平方 -（3-2倍根号2）/4 =0?
答
配方:
x^2+(√2+1)x+√2=0,
x^2+(√2+1)x+[(√2+1)/2]^2+√2-[(√2+1)/2]^2=0,
[x+(√2+1)/2]^2+√2-(3+2√2)/4=0,
[x+(√2+1)/2]^2+(3-2√2)/4=0