Roots of an Equation

Date: 4/16/96 at 22:42:39
From: Jonathan Stone
Subject: Solving Quadratic Equations

I'm trying to help my daughter with these. Example 6x^2+x-2 = 0.  
We can factor it to (3x+2)(2X-1), I think. We're not sure what to 
do now to solve it to two numbers.  A good method of factoring and 
solving these with an example or two would be greatly appreciated.  
Thanks again. 

     Jonathan Stone

Date: 4/17/96 at 13:6:35
From: Doctor Patrick
Subject: Re: Solving Quadratic Equations

Hi!  You're right so far - the factoring seems to be correct, and 
you do have to solve it for two answers.  Whenever you have 
problems with exponents, there will be as many answers as the 
highest power of the variable you are solving for, so in a problem 
with x^2 there will be two seperate answers.  

The reason for needing these two answers comes from the fact that 
x can me either positive or negative, but when it is squared it 
will be positive.  For example both 2^2 and -2^2 equals 4.  Since 
the x^2 in our equation can be either, we need to solve for both 
cases.

We do this be first making sure that the equation is set to 0, 
then factoring it as you did.  Then, we solve for each have of the 
equation separately.  So your  (3x+2)(2X-1) = 0 becomes:
3x+2=0 and 2x-1=0.

The justification for doing this is that since the two terms, when 
multiplied together, equal 0, one of the terms must also be equal 
to 0, since you only get 0 as a product of multiplication when one 
of the terms is also equal to 0. So since (3x+2)(2X-1) = 0 either 
3x+2 = 0, or 2x-1 = 0.  We then solve for both possiblities.

I'll let you try to solve these two equations on your own.  Good 
luck! Write back to us if you need more help with this, or other 
problems.
 
-Doctor Patrick,  The Math Forum