1：方程ax^2+bx+c=0
則其兩個(gè)實(shí)數(shù)根：x1或x2={-b±√（b^2－4ac）}/2a
另ax^2+bx+c=a(x-x1)(x-x2)
則ax^2+bx+c定能在實(shí)數(shù)范圍能分解因式..
2：當(dāng)a>0并且b^2-4ac＞0時(shí) x有兩個(gè)不相同的實(shí)數(shù)根
a+b+c=-1
b=-a-c-1=-(a+c+1)
b^2=(a+c)^2+2(a+c)+1=a^2+2ac+c^2+2a+2c+1
則
b^2-4ac
=a^2+2ac+c^2+2a+2c+1-4ac
=a^2-2ac+c^2+2a+2c+1
=(c-a)^2+2(c-a)+1+4a
=(c-a+1)^2+4a>0