Equation Roots

Date: 01/22/2001 at 19:41:18
From: Sara Robichaux
Subject: Re: Algebra II

Dr. Math,

We are working on quadratic formulas, where we find the discriminant. 
Here's what I've done, and where I get stuck:

3xsquared + 5x - 2 = 0
a = 3, b = 5, c = -2
bsquared - 4ac = (5)squared - 4(3)(-2)
               = 25 + 24
               = 49

Now I get stuck. How do you get the real, imaginary, rational, and 
irrational(ROOTS) of the equation? 

-Sara

Date: 01/23/2001 at 21:19:00
From: Doctor Ian
Subject: Re: Algebra II

Hi Sarah,

You're right on track, but I'll walk you through the whole problem.  
The original equation is

  3x^2 + 5x - 2 = 0

This is a quadratic equation, so it has two roots.  These are:

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

and 

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

You've correctly noted that a = 3, b = 5, and c = -2.  So let's 
substitute those into the equation:

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


      -(5) +/- sqrt((5)^2 - 4(3)(-2))
  x = -------------------------------
                2(3)


      -5 +/- sqrt(25 - (-24))
  x = -----------------------
                 6


      -5 +/- sqrt(49)
  x = ---------------
            6

    
      -5 +  7    -5 - 7 
  x = ------- or ------
         6          6


  x = 2/6 or -12/6


  x = 1/3 or -2

And these are the roots.  You can check them by substituting them back 
into the original equation:

  3(1/3)^2 + 5(1/3) - 2 = 0

       3(1/9) + 5/3 - 2 = 0

          1/3 + 5/3 - 2 = 0           Looks good.


    3(-2)^2 + 5(-2) - 2 = 0

        3(4) - 5(2) - 2 = 0    

            12 - 10 - 2 = 0           Looks good. 

Note that these are both real roots, so we _could_ have factored the 
original equation this way: 

  3x^2 + 5x - 2 = (3x - 1)(x + 2)

But factoring is trickier when the first coefficient (a) is something 
other than 1...

As I said, you were already on track. Are you able to follow what I've 
done here?  Is there any part of it that you don't understand?  Would 
you like to move on to another problem? 

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