The Quadratic Formula

Date: 12/12/2005 at 00:04:25
From: Chris
Subject: The Quadratic formula

We've been studying the quadratic formula as a way to solve quadratic
equations, but I'm a little confused by it.  Can you give me a summary
of what the formula is, how and when you apply it, and why it's important?

Date: 12/12/2005 at 11:36:11
From: Doctor Ian
Subject: Re: The Quadratic formula

Hi Chris,

_When_ you apply it is:  When it's difficult or impossible to factor
the quadratic expression in the equation you are trying to solve. 
Usually, this means when the coefficient of x^2 is something other
than 1, or when the roots aren't integers.  

_How_ you apply it is, you write down the formula, 

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

and then you figure out what a, b, and c are.  For example:
  
  0 = 3x^2 - 4x - 9
  0 = 3x^2 + -4x + -9
      |       |    |____ c = -9
      |       |_________ b = -4
      |_________________ a =  3

Then you substitute into the formula, using ()'s around the values to
keep from messing up any signs:

       -(-4) +/- sqrt((-4)^2 - 4(3)(-9))
   x = ---------------------------------
                   2(3)

and then you simplify that.  You will usually get two answers, 
although sometimes you can get one answer twice.  You can tell how 
many and what type of answers you will get by the expression in the
square root, which is called the "discriminant". 

If the "discriminant" evaluates to zero, you just get one root:

      -b
  x = --
      2a

If it evaluates to a positive perfect square, you get a couple of
rational roots:

      -b + n  -b - n
  x = ------, ------
        2a      2a

If it evaluates to a positive non-square, you get a couple of 
irrational roots:
             _         _
      -b + \/n  -b - \/n
  x = --------, --------
         2a        2a

And if it evaluates to a negative number, you get a couple of 
imaginary roots:
             __         __
      -b + \/-n  -b - \/-n
  x = ---------, ---------
         2a         2a

               _           _
      -b + i*\/n  -b - i*\/n
    = ----------, ----------        where i^2 = -1
         2a          2a

(Can you see why the name "discriminant" was chosen?)

Note that sometimes you can _partially_ apply it when you're trying to
factor, just to see if you can expect factoring to work.  For example,
if I'm looking at something like 

  0 = 3x^2 + -4x + -9
      |       |    |____ c = -9
      |       |_________ b = -4
      |_________________ a =  3
  
I'll quickly evaluate the discriminant,

  b^2 - 4ac = (-4)^2 - 4(3)(-9)

            = 16 + 128

            = 124

Since this isn't a perfect square, I know I'm not going to get
factoring to work out, so I'll go ahead and apply the rest of the
formula.  On the other hand, if I'm looking at something like 

   2x^2 + -4x + -16

the discriminant is 

   b^2 - 4ac = (-4)^2 - 4(2)(-16)

             = 16 + 128

             = 144

             = 12^2

so that tells me that factoring will work.  Of course, at this point,
it's easier to apply the rest of the formula to get the roots, 

  -(-4) +/- 12
  ------------ = 1 +/- 3 = -2 or 4
        4 

so depending on what I'm trying to do, it might not be worth going 
back to complete the factoring.

Does this help? 

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