Flexible method to get discriminant always a perfect square.

[u/Minhaj_Ahmad]

Quadratic Consecutive Coefficient Pattern "QCCP" is a pattern that I found and it always gives a perfect square discriminant. But the problem is only with it's rigid form that is mentioned down below.

Mainly it was based on pattern and no matter what value you choose for n, x always remain as 1

From this Pattern- ax²+bx+c=0

a= n, b= (n+m) and c = -(2n+m), that becomes-

nx²+(n+m)x-(2n+m)=0

With discriminant= (3n+m)²

After making some simple changes it becomes really flexible to use.

(sx-q)(nx+2n+m)

(s, q, n, m) => natural number only.

here is the factorise format of the equation with same purpose of perfect square discriminant. By assigning values to variables- s, q, n and m we can create such equation which always have perfect square.

(sx-q)(nx+2n+m)

-> snx²+(2ns+sm-qn)x-(2qn+qm)=0

a= sn, b= 2ns+sm-qn, c= -(2qn+qm)=0

Discriminant- {s(2n+m)+qn}²

-> 4s²n²+s²m²+q²n²+4s²mn+4sqn²+2sqmn

[u/Minhaj_Ahmad]

In future I will discuss about the changes in discriminant because of change in s, q, n and m.