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Derivation of Quadratic Formula

Date: 6/12/96 at 20:46:1
From: Milton F. Lopez
Subject: Derivation of quadratic formula

Confession: my K-12 days are long gone ...

Request: I would like to derive the solution to ax(2) + bx + c = 0. 
I did this years ago in a college level analysis class, but I've lost 
a lot of neurons since then. Can you help? Most places I've 
looked simply state the soluion without proof.

Thanks in advance.

Date: 6/13/96 at 20:54:32
From: Doctor Ceeks
Subject: Re: Derivation of quadratic formula

ax^2+bx+c = a(x+b/(2a))^2-b^2/(4a)+c = 0, so that

Now take square roots and subtract b/(2a) from both sides.

-Doctor Ceeks,  The Math Forum
 Check out our web site!   

Date: 6/13/96 at 20:52:30
From: Doctor Luis
Subject: Re: Derivation of quadratic formula

In order to find a solution to the general quadratic equation 
ax^2+bx+c=0 one needs to "complete the square" on the terms ax^2+bx. 
By doing this you are transforming the general quadratic into the form 
(x+m)^2-n=0, which we know has the solution x= -m +/- sqrt(n). 
("+/-" plus or minus, and sqrt() is the square root function)


Claim: The solutions to the general quadratic ax^2+bx+c=0 are
(-b+sqrt(b^2-4ac))/2a and (-b-sqrt(b^2-4ac))/2a

 Consider the quadratic equation:

  ax^2+bx+c=0                 (General form)

 Now divide by a


 Add and subtract (b/2a)^2 (a form of 0)


 Now, examine very closely the parenthesized expression. It looks
 like the perfect square of a binomial (and it is)

   [x^2+(b/a)x+(b/2a)^2] - (b/2a)^2+(c/a)=0
   [x^2+2(b/2a)x+(b/2a)^2] - (b/2a)^2+(c/a)=0

 If you let m = b/2a, then

   [x^2+2mx+m^2] - (b/2a)^2+(c/a)=0
   [x+m]^2 - (b/2a)^2+(c/a)=0

 Okay. So far we have:

   (x+m)^2 + (c/a)-(b/2a)^2 = 0
   (x+m)^2 + (c/a)-(b^2/4a^2) = 0
   (x+m)^2 + (4ac-b^2)/(4a^2) = 0
   (x+m)^2 = (b^2-4ac)/(4a^2)
   (x+m)   = (+/-)sqrt(b^2-4ac)/sqrt(4a^2)
   (x+m)   = (+/-)sqrt(b^2-4ac)/2a
    x      = -m +/- sqrt(b^2-4ac)/2a
    x      = -b/2a +/- sqrt(b^2-4ac)/2a
    x      = (-b +/- sqrt(b^2-4ac))/2a

                                  Quod erat demonstrandum
                             [Latin for "which was to be proven"]

Right now the proof might look complicated, but if you write it on 
paper it looks clearer than with ASCII.

-Doctor Luis,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Basic Algebra

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