Question

find p for which (p+1)xsquare-6 (p+1)x+3 (p+q) has equal roots hence find the roots of the equation.

Answer 1

equal roots of quadratic means
b^2-4ac=0

given equation,
(p+1)x^2-6(p+1)x+3(p+q)=0

b=-6(p+1)
c=3(p+q)
a=(p+1)

{6(p+1)}^2-4 x 3 (p+1)(p+q)=0

3(p+1)^2-(p+1)(p+q)=0

(p+1){3p+3-p-q}=0

(p+1)(2p+3-q)=0

so, p+1 =0 and 2p+3-q=0
p=-1 and p=(q-3)/2
if p=-1 then equation doesn't exist
so, p isn't -1

p=(q-3)/2