Associated Topics || Dr. Math Home || Search Dr. Math

### Other Ways to Get the Quadratic Formula

```Date: 02/19/2010 at 11:51:23
From: Davlon

Is there any way to derive the quadratic formula that does not
involve completing the square?

As a teacher, I like to find many different ways to approach
problems, so that my lessons can be more diversified for students.

I find it hard to believe there is only one derivation.

```

```
Date: 02/21/2010 at 03:41:42
From: Doctor Jacques

Hi Davlon,

Here are two methods I can think of.  The first one is actually
completing the square in disguise; the second one is interesting,
because it has connections with some deep concepts, but it may use
some ideas unfamiliar to the students.

In either case, we want to solve the equation:

f(x) = ax^2 + bx + c = 0                  [1]

Method 1 : Substitution
-----------------------

We notice that [1] is easily solved if b = 0 (and ac < 0).  We try to
transform the equation to that form by using a change of variable.

Let us write:

x = y + h                                 [2]

where y is the new unknown and h is to be determined to make the
equation easier to solve.

After substitution in [1], and re-arranging the terms, the equation
(in y) becomes:

a*y^2 + (b + 2ah)y + (ah^2 + bh + c) = 0  [3]

by choosing h = -b/2a, we can reduce [3] to the easy form, with the
coefficient of y equal to 0.  Working out the substitution gives:

a*y^2 + (c - b^2/4a) = 0                  [4]

This equation is easily solved for y, and, using [2] with h = -b/2a,
gives the solutions in x.

Method 2 : Symmetry
-------------------

Let us write p and q for the unknown roots of the equation.  We have
the well-known relations:

p + q = -b/a
pq = c/a                                  [5]

Note that you don't need the quadratic formula to prove these
relations.  You can prove them by equating coefficients in the
identity:

f(x) = ax^2 + bx + c = a(x - p)(x - q)    [6]

and this last identity can be proved by polynomial division, using
the fact that f(p) = f(q) by definition.

Now, [5] gives us the sum p + q.  If we could have the difference
p - q, the resulting system would be easy to solve for p and q.  The
problem is that there is no simple (more exactly, rational)
expression for p - q.  To see why, assume that we have an expression:

p - q = g(a,b,c)                          [7]

where g(a,b,c) is a rational function of the coefficients.  The right-
hand side of [7] has a single, well-determined value.  However, the
choice of p and q for the roots is quite arbitrary -- we don't know
which one is p and which one is q.  If we interchange p and q, the
sign of the expression would have to change, and this is not possible
for a rational expression.

To get rid of that sign change problem, we consider the expression
(p - q)^2.  This expression does not change if you swap p and q, and
this suggests that we could maybe find a rational expression for
(p - q)^2.  Let us try it:

(p - q)^2 = p^2 + q^2 - 2pq
= (p + q)^2 - 4pq
= b^2/a^2 - 4c/a
= (b^2 - 4ac)/a^2              [8]

(We made use of the first relation [5]).  This gives:

p - q = (+/-) sqrt(b^2 - 4ac)/a          [9]

Notice that swapping p and q changes the sign of the square root; as
this expression is defined up to the sign, this is OK.

It is now an easy matter to solve the linear system:

p + q = -b/a
p - q = sqrt(b^2 - 4ac)/a                [10]

for p and q (we can choose any sign for the RHS of the second
equation, as choosing the other sign will merely interchange p and q).

This idea is, in fact, a particular case of a method devised by
Lagrange to solve the cubic and quartic equations; this kind of idea
is at the heart of Galois theory (an undergraduate-level topic).

I believe this proof is accessible to high school students.  There
are two possible difficulties:

* The proof of the relations [5].
* If you have not yet taught complex numbers, an astute student may
object that [6] makes no sense if the original equation has no roots.

Please feel free to write back if you require further assistance.

- Doctor Jacques, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 02/21/2010 at 11:54:39
From: Davlon

Thank you very much for the added insight.  The second derivation is
very interesting.  I do try to get my students to think more.

Thanks again.

Davlon
```
Associated Topics:
High School Basic Algebra

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search