Understanding the quadratic formula


All quadratic equations have two solutions.

In special cases the two solutions might be the same,

or they might be complex numbers of the for A + B i where

A and B are real numbers, and i is square root of minus one.


X^2  means x squared.

x^2 - 5 x + 6  = 0 has the solutions x = 2 and x = 3.

we can know this by factoring x^2 - 5 x + 6 into

( x - 2 ) (x - 3) = 0

The product of two numbers is zero only if at least one of
the numbers is zero.

Thus  if x - 2 = 0, or x = 2, then (x - 2) (x-3) will = 0.

if x - 3 = 0, or x = 3,  then ( x-2) (x-3) will = 0.

And

since  ( x -2) (x - 3) = x^2 - 5 x + 6 ,

if x = 2, then x^2 - 5 x + 6 will be zero.

and 

if x= 3, then x^2 - 5 x + 6 will be zero.

Another way to see this is

2^2 - 5 * 2 + 6 = 4 - 10 + 6 = 0

3^2 - 5 * 3 + 6 = 9 - 15 + 6 = 0.

Now, what about the quadratic formula.  How does it relate.

The quadratic formula is the formula for the solution
to the quadratic equation  

A x^2 + B x + C = 0,  for any numbers, A,B,C.


x = (-B/[2A]) + sqrt( (B/[2A])^2 -  C/A ) = [  - B + sqrt( B^2 - 4 A C )  ] / [ 2 A ]

or

x = (-B/[2A]) - sqrt( (B/[2A])^2 -  C/A ) = [  - B - sqrt( B^2 - 4 A C )  ] / [ 2 A ]


It can be derived several different ways.


But first, let's illustrate the formula.


Let's look at  x^2 - 5 x + 6 = 0


A = 1;  B = -5 ;  C = 6


x = -(B/[2A]) + sqrt( (B/[2A])^2 -  C/A) = [  - B + sqrt( B^2 - 4 A C )  ] / [ 2 A ]

x = -(-5/2) + sqrt( (-5/2)^2 - 1 * 6)  = [  - B - sqrt( B^2 - 4 A C )  ] / [ 2 A ]

x = 2.5 + sqrt(6.25 - 6)  or  x = 2.5 - sqrt(6.25 - 6)

x = 2.5 + sqrt(.25)  or x = 2.5 - sqrt(.25)

x = 2.5 + .5  or x = 2.5 - .5

x = 3  or x = 2.

The only requirement on the quadratic formula, is that A, the coefficient
of x^2 must not be zero.

How might the quadratic formula be derived.

Consider the general quadratic equation,

A x^2 + B x + C = 0, where A is not zero.

We make the first simplication by dividing through by A.

This makes the coefficient of x^2 equal to 1.

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


Now recall when a trinomial is square.

( x + 1 )^2 = x^2 + 2 x + 1

( x + 2 )^2 = x^2 + 4 x + 4

( x + 3 )^2 = x^2 + 6 x + 9

( x + 4 )^2 = x^2 + 8 x + 16

Note the pattern.

The middle term of the trinomial is twice the constant of 
the binominal times x.

The constant term in the trinomial is the square of 
the constant term of the binomial

So, if we can transform the trinomial equation

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


into a square trinomial equal to a number, se can
solve it.

We take the next step.

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


Now (B/A) has to be twice something.

What would that be?

It would of course be twice (1/2)(B/A) = B/(2A).

so

(x + B/(2A) )^2  = x^2 + 2 B/(2A)  x + (B/[2 A])^2

How do we relate this square to



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


By adding the square term  ( B/[2A])^2 to both sides of the equation!



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




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

By taking square roots, we get

x + B/[2 A]  = + sqrt(   (B/[2A] )^ - (C/A)  )

or


x + B/[ 2 A ] = - sqrt( (B/[2 A] )^2 - (C/A)  )


One more step gives us the formula.


x = -B/[2A] + sqrt( (B/[2A])^2 - (C/A) )

or 

x = -B/[2A] - sqrt( (B/[2A])^2 - (C/A))


By multiplying and dividing by 2A
and remembering that

sqrt(4A^2) = 2 A,

we get the  traditional forms of the quadratic formula.

x = [ - B + sqrt( B^2 - 4 A C) ] / [2 A]

or 

x = [ - B - sqrt( B^2 - 4 A C) ] / [2 A]