We know that   a quadratic expression is a perfect square if its discriminant is zero .but n^2 -1 is perfect square at n=1 even its discriminant is not zero  why. Pl explain this concept I am little bit confused.        Hence prove that there is no +ve  integer  for which  (n+1)^.5+ (n-1)^.5  is a rational number

Hey If D is zero then the quadratic is necessarily a perfect square (I am not sure I understand what you mean by a perfect square) but the inverse may not be true.

Example. If triangle ABC covers triangle PQR then it is not necssary that Triangle PQR covers triangle ABC.

One clarification: If the discriminant of a quadratic P(x) = ax²+bx+c equals zero, then P(x) = a (x-r)²

Now, if a and r are both integers and a is a perfect square P(x) is a square integer for all integer x. Note that this should happen for all integer x. P(n) = n²-1 is a perfect square only for n =1,-1. (as n² is a perfect square and the only consecutive squares are 0,1)

Another example is consider P(x) = x²+16. It becomes the perfect square of an integer when x =3,-3. and for no other value of x.

Now to your second problem:

Let x =  be a rational number.

For n =1, x =

Then  is a rational number. However, we have proved above that if  then n²-1 is not a perfect square and the square root of an integer that is not a perfect square is irrational. Hence we obtain a contradiction.

Another way:

Again Let x =  be a rational number

Then so must its reciprocal i.e. 

This means  and  are both rational

i.e.  are both rational. This happens if both n+1 and n-1 are both perfect squares. However, we know that there are no two squares whose difference is 2. Again, we obtain a contradiction.