韦达定理x1-x2的绝对值

x1+x2=-b/a
x1x2=c/a

(x1+x2)^2=b^2/a^2
(x1-x2)^2=b^2/a^2-4c/a=b^2-4ac/a^2
绝对值x1-x2=√b^2-4ac/a的绝对值
即绝对值a分之√△

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