For what value of k

(1+p)x^2 +2(1+p)x +(1+p)=0

can be written in the form of a square.

 

well, the exp is independent of k.

guessing that by k u meant p,the exp. is (1+p)(x+1)² which is a square whenever p+1 is a square.

p must be of the form a²-1,a is integer.please tell me if your Q is something else.

since,we want it as a perfect square,

discriminant is zero.

but suprisingly,we get discriminant zero for all the values of p

if you were asking for rational squares,then it is true for all n>-1

if you were asking for integral squares,your answer is that p should be of form a^2-1

For what value of k

(1+p)x^2 +2(1+p)x +(1+p)=0

can be written in the form of a square.

Do you mean to write (1+p)x^2 +2(1+p)x +(1+p)=0 this as a perfect square?

If yes then the value of p can be k²-1 or p=k²-1 where k>1

here

(1+p)x² + 2(1+p)x+(1+p)=0 is the given equation

reduce it to ..

(1+p) [x²+2x+1] =0

(1+p)(x+1)² =0

Thus this eqn is already a square for all values of p except -1

Thus p belongs to R - {-1}