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题目:若a、b、c是△ABC的三边,求证:b~2x~2+(b~2+c~2-a~2)x+c~2=0无实根.此题运用判别式△<0很容易证明,并无特别之处.值得注意的是:从这题出发可以引出一组题,而且都能用判别式△<0来证(或解),从而研究△<0应用的几种情况.现在将此题简证如下:证明∵△=(b~2+c~2-a~2)~2-4b~2c~2=(b~2+c~2-a~2+2bc)(b~2+c~2-a~2-2bc)=[(b+c)~2-a~2][(b-c)~2-a~2]=(b+c+a)(b+c-a)(b-c+a)(b-c-a).
Subject: If a, b, c are three sides of △ ABC, verify that: b ~ 2x ~ 2 + (b ~ 2 + c ~ 2-a ~ 2) x + c ~ 2 = 0 without roots. Discriminant △ <0 is easy to prove, there is no special place.It is noteworthy that: Starting from this question can lead to a group of questions, and can use the discriminant △ <0 to prove (or solution), to study △ < 0 apply several cases now briefly proof of this problem is as follows: Prove ∵ △ = (b ~ 2 + c ~ 2-a ~ 2) ~ 2-4b ~ 2c ~ 2 = (b ~ 2 + c ~ 2- a ~ 2 + 2bc = b + c ~ 2 - a ~ 2-2bc = b + c ~ 2 - a ~ 2 b + b - 2 b = c + a) (b + ca) (b-c + a) (bca).