kx^2+(k+2)x+k/4=0
1.(1)k=0
x=0
(2)k≠0
Δ=(k+2)^2-4*k*k/4
=k^2+4k+4-k^2
=4k+4>=0
k>=-1
所以 k>=-1有实根.
2.设2个实数根分别为x1,x2
1/x1+1/x2=(x1+x2)/x1x2
=-(k+2)/k/(1/4)=-4(k+2)/k=0
k=-2.