c,求d=|x1-x2|的取值范围.各位大神帮帮忙啊

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∵原方程有两个实数根,∴判别时△=b^2-4ac≥0.
由韦达定理得：x1+x2=-b/a,x1*x2=c/a.
(x1+x2)^2=(-b/a)^2=b^2/a^2.
(x1-x2)^2=(x1+x2)^2-4x1x2.
=b^2/a^2-4c/a.
=(b^2-4ac)/a^2
|x1-x2|^2=(x1-x2)^2.
|x1-x2|=√(b^2-4ac)/a.
∵b^2-4ac≥0,a＞0.
∴0＜|x1-x2|≤√(b^2-4ac)/a.