u/Minhaj_Ahmad

Finding square root of a number that I don't know was a pain for me.

Using that long division method 😭.

Like 841 it's the square root of 29, I didn't know that I tried many ways to over come but in the end I just used the calculator.

I was furious and curious can I find the square value of any number using (a+b)² expression then maybe I can also find the square root of perfect square numbers using it?

I came up with (10a+b)²= S [I know it's looks similar to that expression and actually it is I'm not claiming I found something new I just got rid of my problem]

100a²+20ab+b² = S (s is the targeted square value of which square root we are looking for.)

Now conditions; a= or < S, b can be only a single digit from 0 to 9.

Example: S=144.

100a²+20ab+b² =144

Then a=1 => 100+20b+b²=144

20b+b²=44

(For selecting b we can guess the number or just divide remaining number by 20a [a=1 in this case].)

44/20 ≈2

Now substituting as a test-

20(2)+(2)² =44

LHS = RHS then

a=1, b=2

(10×1+2)² =12².

I will do further case study related to this stuff.

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

After reading a lots of suggestions on Quadratic Consecutive Coefficient Pattern "QCCP" mainly about it's being rigid and with fixed value for x like x= 1 and x=-c/a.

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

Later on I will post about the discriminant and root changes due to change in variables.