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Proving/Deriving the Quadratic Equation


Date: 4/1/96 at 12:27:1
From: Anonymous
Subject: Quadratic equation solution

How can one derive the quadratic equation, i.e. prove it?  I 
memorized it in middle school along with every other seventh 
grader, but I don't remember ever seeing any sort of analysis or 
proof.  How did someone come up with it?


Date: 4/4/96 at 2:58:28
From: Doctor Aaron
Subject: Re: Quadratic equation solution

Hi,

Good question, I love to answer this one.

Do you remember how you solved quadratics before you memorized the 
formula?

Its a technique called completing the square.  That is 
(x + E)^2 = x^2 +2Ex +E^2.  The quadratic formula is just getting 
an arbitrary quadratic, Ax^2 + Bx + C = 0 into this format by 
adding and subtracting some stuff, then we get a big mess on the 
other side, but we have a perfect square so we can take its square 
root.

Lets do it.

Let A,B,C be real numbers and let Ax^2 + Bx + C = 0  

First we get rid of the A by dividing through

x^2 + (B/A)x + (C/A) = 0

Now (B/A) must equal 2*E if we are going to get a perfect square 
so we want

x^2 + (B/A)x + (B/(2A))^2 = something

but something is just what we get when we add (B/(2A))^2 and 
subtract (C/A) from both sides.

then x^2 + (B/A)x + (B/(2A))^2 = (B/(2A))^2 - (C/A)

we can multiply the right side by 4A/4A to get

B^2      4AC        B^2 - 4AC
---   -  ---    =  -----------  on the right.
4A^2     4A^2         4A^2

We have a perfect square on the left, so taking the square root of 
both sides, we get:

               +- sqrt (B^2 - 4AC)       +- sqrt (B^2 - 4AC)
 x + B/(2A) = ---------------------  =  ---------------------.
                    sqrt (4A^2)                   2A

Subtracting B/(2A) from both sides we get the familiar

x = -B (+ or -) sqrt (B^2 - 4AC)
    ------------------------------    
            2A.

-Doctor Aaron,  The Math Forum

    
Associated Topics:
High School Analysis

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