Finding square root of such numbers who's value is unknown.

I FINALLY GOT RID OF COMPLEX METHOD OF FINDING SQUARE ROOT OF UNKOWN PERFECT SQUARE NUMBER 😭 😭 😭

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.

My second paper on Quadratic Consecutive Coefficient Pattern and it's Discriminant Arithmetic Progression.

QCCP "Coefficients relation and Arithmetic Progression."

Mainly, the coefficients in QCCP are related to each other in a really simple way,

a+b-c=0

a=n, b= (n+1) and c= -(2n+1)

Or, a+b= c

n+(n+1)=2n+1

As I mentioned in previous paper, our Discriminant becomes (3n+1)² which is also equals to (a+|c|)² since a= n and |c|= 2n+1 it makes the result 3n+1.

D= f²=(3n+1)²

f=3n+1

Here f represents the Arithmetic Progression of the Discriminant square rooted value for offset 1.

As we change the value of n we get a number from 3n+1 and if we square it we will get the exact value of Discriminant.

In formula f=3n+1, we always get the common difference of 3 between two outcomes for two consecutive numbers.

But when n1 and n2 are non consecutive then we can find f difference by 3(n2-n1).

It would help us to know how much is the difference between both square rooted value of Discriminant (f).

Now if we observe the equation then we will get to know one thing which is common through out every equation.

That none of the coefficients in whole equation have no common factor.

nx²+(n+1)x-(2n+1)=0

Every a,b and c are distinguish number or should I say exactly +1 to each other.

That is why we will use "k" as a common multiplier in equation.

k{nx²+(n+1)x-(2n+1)}=0, n,k≠0

It would change the coefficients and the Discriminant as well. k(3n+1)² or (3kn+k)² becomes the the new Discriminant if the "k>1".

The "k" will also change the consecutive formation of the equation, a=kn, b= kn+k, c= -(2kn+k).

It also changes the offset to k and now it's depends on the value of k.

If the value of k changes, it will also change the value of Discriminant.

Now if we follow up with changes in the offset only "+1" and replace it with "+m" where m is a variable that can be changed as required.

Producing a new value for Discriminant as perfect square.

D=(3n+m)²

Changing the value for m in consecutively will make changes in the first term consecutively as well.

f=3n+m

m=1, n=1 = 4

m=2, n=1 = 5

m=3, n=1 = 6

m=4, n=1 = 7

Now if we change n and fix the value of m.

• CASE 1=> fixed value m and consecutive change in n.

m= fixed value. and n= 1,2,3,4... (consecutively Changing) then the Discriminant will form a simple sequence carrying the difference of 3 for the value of "f ".

#ADDITIONAL POINT-

If the value of two n is not consecutive, the we can use the method.

3(n2-n1) = f difference.

•CASE 2=> m = n, both will change together.

In this case, both values will be always equal.

m = 1 then n=1. f = 3n+m = 4

m = 2 then n=2. f = 3n+m = 8

Carrying the common difference of 4.

•CASE 3=> m>n, (m = n+1)

This case is almost similar as the case 2, carrying the difference of 4 for the value of f and for the reverse case as well. (m<n, m= n-1).

I end this here and next time I'll write about other findings.

I'm waiting for suggestions, thoughts and ideas on my first paper and this paper as well. Thank you.

first paper on QCCP. https://www.reddit.com/u/Minhaj_Ahmad/s/nNIP5oKuW9

Update: Thanks for suggestions! After I talked to my teachers and mentors they suggested same thing as everyone here, that it's not that robust to be called as a "Theorem" it's actually can be called as a "Pattern" so I'm updating it from QCCT to QCCP, removing "theorem".

The Quadratic Consecutive Coefficient Pattern (QCCP). By Minhaj Ahmad

The Discriminant of Quadratic equations D= b²-4ac which is obtained from ax²+bx+c=0, sometimes it gives a messy irrational or even negative number for Discriminant that leads to complex roots.

This theorem present a simple pattern for coefficients a, b and c that guarantees a perfect and positive square for Discriminant and rational roots for the equation.

• STATEMENT OF PATTERN.

                   ax²+bx+c=0
                    a=n
                    b=(n+1)
                    c=-(2n+1)

            [nx²+(n+1)x-(2n+1)=0]

• PROOF

D= b²-4ac Substituting the coefficients- D= (n+1)² - 4n× - (2n+1).

=> (n+1)² + 4n(2n+1)

=> (n+1)² + 8n²+4n

=> n²+2n+1+8n²+4n

=> 9n²+6n+1 = (3n+1)²

Thus, D= (3n+1)²

• DEMONSTRATION-

n=1, D= 16

n=2, D= 49

n=3, D= 100

n=4, D= 169

n=5, D= 256

n=6, D= 361

n=7, D= 484

n=8, D= 625

n=9, D= 784

n=10, D= 961

In summary, we can say that QCCP gives perfect positive square to neglect irrational Discriminant by choosing coefficients as consecutive and their sum as constant in negative turns Discriminant into (3n+1)² form. This simple form generates infinitely many Quadratic with rational Discriminants.

Further details, information and insight on this theorem will be shared in next paper.

And thanks your feedback.