已知:x1,x2是方程ax2+bx+c=0(a＞0,b2-4ac≥0）的两个根.求证：x1-x2的绝对值=根号b2-4a

证明
|x1-x2|
=√(x1-x2)²
=√(x1²+x2²-2x1x2)
=√[(x1+x2)²-4x1x2]
=√[(-b/a)²-4c/a]
=√[(b²-4ac)/a²]
=√(b²-4ac)/a
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