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Date: 10/15/98 at 21:03:19
From: Anne Smith
Subject: How the quadratic formula was formed from  ax^2+bx+c=0

I am wondering how the quadratic equation was derived from
ax^2 + bx + c = 0. I have tried numerous ways to figure it out and I am
very frustrated. I am working with it in a proof form and I first
divide it all by "a" but then I remember that "a" is also on top.
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Date: 10/22/98 at 00:21:27
From: Doctor Santu
Subject: Re: How the quadratic formula was formed from  ax^2+bx+c=0

Anne, this is a classic problem solved by Italian mathematicians of
the 12th century (if I remember right), and probably by even earlier
mathematicians, using painfully difficult methods, but we're lucky,
since we have algebra.

First, some background.

A. You know that if you multiply (A + B)(A - B) you get A^2 - B^2,
right? So if we ever see something of the form A^2 - B^2, we know
immediately that it can be factored into (A+B)(A-B).

B. What about things like A^2 - 5?  Of course 5 isn't obviously of
the form (something)^2, but we can still "factor" this in the form
[A + sqrt(5)][A - sqrt(5)], where Sqrt(5) is the square root of 5.
In particular, if I see something like:

(x+r)^2 - (p+q)

I can factor it as [(x+r) + SQRT(p+q)][(x+r) - SQRT(p+q)].

C. We need to complete squares. In other words, by inspecting things
like:

(x+1)^2  =  x^2  +  2x  +  1
(x+2)^2  =  x^2  +  4x  +  4
(x+3)^2  =  x^2  +  6x  +  9
...
(x+12)^2  =  x^2  +  24x  +  144

and so on, we get to be able to answer the puzzle: "if X^2 + 14X are
the first two terms of a perfect square, what should the third term
be?"

In each line, observe that the last term is the square of half of
the middle coefficient. For instance, take 6, halve it (you get 3),
and then square that, you get 9. So to get the proper last term for
X^2 + 14X, take the middle term, which must be 14X, take the
coefficient, 14, halve it, 7, and square it, 49. So the perfect
square whose first two terms is X^2 + 14X is X^2 + 14X + 49. That's
called completing the square.

ax^2 + bx + c = 0

Divide through with a. You get:

b        c
x^2  +  - x  +  ---  = 0
a        a

Now complete the square: take half of (b/a) to get b/(2a). Now
square that, and you get b^2/(4a^2). We add this, and then we
subtract it (which is the same as adding 0 and so doesn't change the
equation):

2      2
b        b      b    c
x^2  +  - x  +  ---  - --- + -  =  0
a         2      2   a
4a     4a

Now regroup as follows:

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

If that's hard to follow, look at the graphic file:

which was created using Maple V, which will give you the algebra in
almost-normal-looking form.

Now you basically have a rather simple equation, something like:

(x + something)^2  -  (something else) = 0

so you can say:

(x + something)^2 = (something else)

Then the (x + something) must be either the square root of the
something else on the right side, or the negative of the square root
of the something else. And that's exactly what you get in the

x + b/2a  = (+ or -)SQRT(b^2 - 4ac)/(2a)

so:

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

For another version of this derivation, see:

http://http://mathforum.org/library/drmath/view/52950.html

- Doctor Santu, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Equations, Graphs, Translations
Middle School Algebra
Middle School Graphing Equations

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