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Today,we shall deal with the quadratic formula popularly known as almighty formula.

This formula is used to factorize and solve for the solutions ( real roots ) of any quadratic equation whose discriminant is not negative.

For countless years now, almighty formula always appears in this form
-b + √[b^2 - 4ac] / 2a
or
-b - √[b^2 - 4ac] / 2a.

This is nothing, but the fact that we normally express a general quadratic equation in the form given below
aX^2 + bX + c = 0, so that the principal of completing the square would apply for the derivation of the formula.

Suppose a general quadratic equation is expressed as a function F(X) = 0 ,where F is the quadratic Polynomial in X variable ,then how can we derive the quadratic formula without changing its original form?

This poses serious challenges to the methods used in the derivation of almighty formula,since no method could find its way.

However, I ( Mr Makasi Chinedu ) have developed a way we could follow so as to be able to derive a quadratic formula when represented as a function.

Derivation Of Almighty Formula When A Quadratic Equation Is Expressed As F(X) = 0

As we have expressed the general quadratic equation in the form F(X) = 0 ,then it is possible that a formula

F'(X) = +or--√[F(X)^2 - 4F(X)]

which represents the quadratic formula could be derived.

Proof:

Let F(X) =0 be a monic quadratic equation.

In this way, it is evident that we have already made the coefficient of the first term of F ( the quadratic polynomial) unity. That is ,

F( X ) = 0.............…...1

Now ,we let Y represents the LHS of equation 1 ,I.e,

Y = F ( X ) ...................2

Take the first order ordinary derivative of Y . That is ,

dY / dX = F'( X )...............3

Multiply the result of this differentiation by 1/2 ,I.e,

( dY / dX ) / 2 = F'( X ) / 2

( dY / 2dX ) = F'( X )............4

Raise both sides of equation four to 2 to have

(dY / 2dX )^2 = ( F'(X) / 2 )^2

(dY / 2dX )^2 = F'( X )^2 / 4.....................5

Now add both equation one and five. That is ,

F( X ) + ( dY / 2dX )^2 = F'( X )^2 / 4

Now take F(X) to the RHS to have

(dY / 2dX )^2 = F'( X )^2 / 4 - F( X )

Find the common denominator from the RHS. That is

( dY / 2dX )^2 = [ F'( X )^2 - 4F( X ) ] / 4

Take the square root and of both sides ,I.e,

√( dY / 2dX )^2 = +/- √[ F'( X )^2 - 4F( X ) ] / √4

(dY / 2dX) = +/- √[ F'( X )^2 -4F( X ) ] / 2

Now ,we clear out the fraction by multiplying both sides by 2 . That is

2( dY / 2dX ) = +/- 2√[ F'( X )^2 - 4F( X ) ] / 2

Which reduces to

dY / dX = +/- √[ F'( X )^2 - 4F( X )

Recall that

dY / dX = F ' ( X )

So in place of dY / dX from the LHS, we replace it with F ' ( X ) to have

F ' ( X ) = +/- √[ F ' ( X )^2 - 4F( X ) ]

This then becomes our Formula ( a.k.a almighty formula ) .

Lol!!!! Are you perplexed ? Don't be for i am to show you the veracity of this magic by given the few examples below.

Example 1

12X^2 - 7X - 10 = 0

Solution

First, make the question given to you a monic quadratic by dividing by 12 ,I.e,

X^2 - (7/12)X - (10/12) = 0

Here , the monic quadratic represents F(X) .

So, let

Y = F(X) = X^2 - (7/12)X - (10/12)

We work with the right hand side .

F( X ) = X^2 - (7/12)X - (10/12)

Differentiate F( X ) to have

F ' ( X ) = 2X - (7/12)

Now put 2X - (7/12) into F'(X)
In the formula ,I.e,

2X-(7/12) =+/- √[ (2X-7/12)^2 - 4F( X ) ]also recall that

F( X ) = X^2 - (7/12)X - (10/12 )

So in place of F(X) in the formula,we put the monic quadratic expression,i. E,

2X-(7/12) = +/-√[(2X - 7/12)^2 -4(X^2 -7/12X - 10/12)]

Let us evaluate the expressions inside for clarity.

(2X - 7/12)^2 - 4( X^2 - 7/12X - 10/12 )

Now expansion d to have

4X^2 - 7/3X + 49/144 - 4X^2+7/3X + 40/12

Taking like terms reduces the expression to

49/144 - 40/12

By LCM we have

529/144

So,we put our result in place of the expression inside the square root ,I.e,

2X - 7/12 = +/- √[ 529 / 144 ]
2X - 7/12 = - √[529] / √[144 ]
2X - 7/12 = [+/- 23] / 12

Divide by 2 to have

X = (7÷24) + or - (23÷24)
X = ( 7 + or - 23 ) ÷ 24

So X = 5 ÷ 4 or -2 ÷ 3

Visit my math blog to learn more @ makasistutors..com

Ur first equation is wrong,Y = F(X) = X^2 - (7/12) - (10/12).
Were u put x,DAT shows DAT UR solution is wrong.
Dis is d correction
Y=F(X)=X^2-(7/12)X-(10/12)

moblix:
Ur first equation is wrong,Y = F(X) = X^2 - (7/12) - (10/12).
Were u put x,DAT shows DAT UR solution is wrong.
Dis is d correction
Y=F(X)=X^2-(7/12)X-(10/12)

Na typo error. U should know that

K

U should correct it n resend it because small mistake can fail some 1

Op, by comparing your deductions with the original quadratic formula, I think the original is still minimalistic and simple to use.

Finding the zeros of a 2nd-degree polynomial is very much trivial in the mathematical world. There is a proof that goes all the way to Niels Henrik Abel, who found the impossibility of solving a quintic polynomial (polynomials higher than 5th degree).

You should do more work on Abel's proof, maybe to the possibility of solving n-dimensional polynomials.


CHEERS

BigotMan:
Op, by comparing your deductions with the original quadratic formula, I think the original is still minimalistic and simple to use.

Finding the zeros of a 2nd-degree polynomial is very much trivial in the mathematical world. There is a proof that goes all the way to Niels Henrik Abel, who found the impossibility of solving a quintic polynomial (polynomials higher than 5th degree).

You should do more work on Abel's proof, maybe to the possibility of solving n-dimensional polynomials.


CHEERS

Thanks

BigotMan:
Op, by comparing your deductions with the original quadratic formula, I think the original is still minimalistic and simple to use.

Finding the zeros of a 2nd-degree polynomial is very much trivial in the mathematical world. There is a proof that goes all the way to Niels Henrik Abel, who found the impossibility of solving a quintic polynomial (polynomials higher than 5th degree).

You should do more work on Abel's proof, maybe to the possibility of solving n-dimensional polynomials.


CHEERS

xup bro....i sent you a pm

I just love to hear Nigerian Maths gurus talk...they are specialist in doing what has already been done!

LordIsaac:
I just love to hear Nigerian Maths gurus talk...they are specialist in doing what has already been done!

How is it ? Have u seen this b4?