Factoring to Solve a Problem

Date: 7/31/96 at 9:21:42
From: Keerthi Kumar
Subject: Factoring to Solve a Problem

I am a first year algebra student and I have a question involving 
factoring in problem solving. The instructions are:

Solve each problem, rejecting solutions that do not meet the 
conditions of the problem.

Let x = the smaller integer and x + 1 = the larger integer. 
Solve for x.

      x (x + 1) = 132

This is how I have been working the problem, but it comes out every 
time with one or the other of the factors being negative.

                                         2
x (x + 1) = 132 = x * x + x * 1 = 132 = x + x = 132

 2
x + x = 132

 2
x + x - 132 = 0
 
(x + 12)(x - 11) = 0

x + 12 = 0        x - 11 = 0
x = -12           x = 11                    
            

I know that the answer is 11 and 12, but how am I supposed to get both
integers to turn out positive?

D.S.K.

Date: 7/31/96 at 10:46:2
From: Doctor Jodi
Subject: Re: Factoring to Solve a Problem

Hi there! Thanks for your question.

Since you let x equal the smaller integer, and x + 1 equal the larger 
integer, the two solutions you get when you solve for x are NOT the 
smaller integer and the larger integer.  Rather they are two 
possibilities for the smaller integer.  Does that make sense?

So, if the smaller integer is 11, then the larger integer will be 12.  
There is still another possibility:  if the smaller integer is -12, 
then the larger integer is -11.  Do you see what I mean?

Since the question didn't specify positive or negative integers, both 
of these solutions-- 11 and 12; -11 and -12 -- are correct.

Hope this helps.  Let us know if you have any more questions.

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