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is it possible to have a QUADRATIC equation of the form (a+1)x2+2(a+1)x+(a-2)=0 to have 2 equal roots?
is the same possible for the eqn (a-1)x2+2(a+1)x+(a+1)=0
if yes then find values for a in both cases and if no then why??
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4 Answers
1D=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
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