p-q)^2 is always > 0 for any real values of p and q , as a square of a number (p-q) is always >=0... this does not mean that p = q.

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The equation editor is not properly working so i write in the normal text.

b^2- 4 ac = (p+q)^2 - 4 pq = (p-q)^2

If the roots of the given equation** both are equal,** then the discriminant is 0.

then (p-q)^2 = 0.** Hence p = q.**

if p= q, then equation will be x^+2x+1=0, hence roots are equal and = -1.

**If p > q and q > p -- in these case real roots exist but they are not equal.**

b^2- 4 ac = (p+q)^2 - 4 pq = (p-q)^2

If the roots of the given equation

then (p-q)^2 = 0.

if p= q, then equation will be x^+2x+1=0, hence roots are equal and = -1.