1.1、x^2-(k+1)x+k=0,(x-k)(x-1)=0,x1=k>=0,x2=1
1.2、x=k代入2,k^3-k(k+2)+k=0,k1=0,k2=(1±√5)/2
x=1代入2,k-(k+2)+k=0,k=2
2.1、若1-2a=0,2√ac-1=0,則x∈R,有無(wú)數(shù)解
若1-2a≠0,2√ac-1=0,則x1=x2=0,有兩個(gè)相等實(shí)數(shù)根
若1-2a=0,2√ac-1≠0,此時(shí)無(wú)解
若1-2a≠0,2√ac-1≠0,x^2=(1-2√ac)/(1-2a)>0,有兩個(gè)不相等的實(shí)數(shù)根
1-2√ac>0且1-2a>0,或1-2√ac