设a,b,c是△ABC的三边长，求证：方程x^2+2ax+b^2=0与x^2+2cx-b^2=0有公共跟的充要条件是∠A=90度。
热心网友
若x^2+2ax+b^2=0 ...(1) 与 x^2+2cx-b^2=0 ...(2)有公共根:(1)+(2): x^2 + (a+c)x = 0, x = -(a+c)(1)-(2): (a-c)x + b^2 = 0, x = b^2/(c-a)== -(a+c) = b^2/(c-a) == b^2 = a^2 - c^2 == ∠A=90度若∠A=90度:则，b^2 = a^2 - c^2 = (a+c)(a-c)方程化为：x^2+2ax+b^2 = x^2+2ax+(a+c)(a-c) = [x+(a+c)][x+(a-c)]=0 x^2+2cx-b^2 = x^2+2ax-(a+c)(a-c) = [x+(a+c)][x-(a-c)]=0可见，两个方程有公共根 -(a+c).因此，证毕。