能
對(duì)于方程ax^2+bx+c=0
(x1-x2)^2=x1^2+x2^2-2x1x2
=(x1+x2)^2-4x1x2
而x1+x2=-b/a x1x2=c/a
代入得
(x1-x2)^2=(b^2-4ac)/a^2
所以|x1-x2|=根號(hào)下(x1-x2)^2=根號(hào)下[(b^2-4ac)/a^2]=根號(hào)下[(b^2-4ac)]/a
所以x1-x2=根號(hào)下[(b^2-4ac)]/a 或-根號(hào)下[(b^2-4ac)]/a