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New Way Of Representing Almighty Formula by makasimatics(m): 10:32am On Mar 22, 2017 |
Makasis Tutorials 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 |
Re: New Way Of Representing Almighty Formula by moblix: 11:21am On Mar 22, 2017 |
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) |
Re: New Way Of Representing Almighty Formula by makasimatics(m): 11:34am On Mar 22, 2017 |
moblix: Na typo error. U should know that |
Re: New Way Of Representing Almighty Formula by moblix: 11:53am On Mar 22, 2017 |
K |
Re: New Way Of Representing Almighty Formula by moblix: 11:57am On Mar 22, 2017 |
U should correct it n resend it because small mistake can fail some 1 1 Like |
Re: New Way Of Representing Almighty Formula by BigotMan(m): 11:59am On Mar 22, 2017 |
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 ![]() 1 Like 1 Share |
Re: New Way Of Representing Almighty Formula by makasimatics(m): 12:35pm On Mar 22, 2017 |
BigotMan: Thanks |
Re: New Way Of Representing Almighty Formula by uppa(m): 3:00pm On Mar 24, 2017 |
BigotMan:xup bro....i sent you a pm |
Re: New Way Of Representing Almighty Formula by LordIsaac(m): 6:07pm On Mar 24, 2017 |
I just love to hear Nigerian Maths gurus talk...they are specialist in doing what has already been done! |
Re: New Way Of Representing Almighty Formula by makasimatics(m): 12:38pm On Mar 29, 2017 |
LordIsaac: How is it ? Have u seen this b4? |
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