Solving Quadratic Equations

Date: 9/14/96 at 15:5:59
From: Anonymous
Subject: Solving Quadratic Equation

How do you solve

     3x^2 + x - 14 = 0

And another one:

     x^2 - 6x + 8 = 0

Date: 9/16/96 at 15:18:13
From: Doctor Mike
Subject: Re: Solving Quadratic Equation

Hello,
  
Each equation is in the form Ax^2 + Bx + C = 0 , so the Quadratic 
Formula says the solutions are (-B+R)/(2A) and (-B-R)/(2A) where 
R = sqrt(B^2 - 4*A*C) .  Is that familiar?
  
Using this, the 2 solutions for the first equation are 2 and -7/3 .  
The 2 solutions for the second equation are 2 and 4 .  So, 2 is the 
only solution for both equations. 
  
If you can factor these 2 expressions, then that will be another way 
to show this.  In factored form the 2 equations are (3x+7)*(x-2)=0 and 
(x-4)*(x-2)=0 .  This shows that x-2 is the common factor of both 
expressions, so that the value of x making x-2 = 0 true will be the 
common solution for the 2 equations. 
  
I hope this helps. 

-Doctor Mike,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/