Question

(1+p)x²+2(1+2p)x+(1+p)=0 has equal/coincident roots

Answer 1

If b^2 - 4ac (the discriminant) = 0, then there are 2 equal roots.

[2(1 + 2p)]^2 - 4(1 + p)(1 + p) = 0
4(1 + 4p + 4p^2) - 4(1 + 2p + p^2) = 0
4 + 16p + 16p^2 - 4 - 8p - 4p^2 = 0
12p^2 + 8p = 0
4p(3p + 2) = 0
4p = 0 or 3p + 2 = 0
p = 0............3p = -2
......................p = -2/3