D=0
4(a+1)2 -4 ( a+1)(a-2) =0
a+1[ (a+1) -(a-2)]=0
(a+1)(3) =0
a+1=0
a=-1
not possible since
in a quadratic eqn
ax2 + bx +c=0
a≠0
hence the eqn cannot have two roots equal
for second one
4(a+1)2 - 4(a+1)(a-1) =0
a+1[ (a+1) - (a-1)]=0
(a+1) (2)=0
a+1=0
a=-1
for second equation
a=-1