x2-(p+q)x+pq=0因式分解为(x-p)(x-q)=0,根为:X1=__,X2=__
问题描述:
x2-(p+q)x+pq=0因式分解为(x-p)(x-q)=0,根为:X1=__,X2=__
答
根为:X1=p ,X2=q
x2-(p+q)x+pq=0因式分解为(x-p)(x-q)=0,根为:X1=__,X2=__
根为:X1=p ,X2=q