A Quadratic Inequality

Date: 4/27/96 at 17:10:42
From: Anonymous
Subject: Quadratic Inequality

Solve the inequality, and express the solutions in terms of intervals 
whenever possible.

1. (3x+1)(d-10x)>0
    x > -(1/3)
    x < (1/2)

I am able to solve for an answer but I do not understand how to 
express it in intervals.

Date: 8/4/96 at 16:31:47
From: Doctor Mike
Subject: Re: Quadratic Inequality
  
The (d-10x) was probably intended to be (5-10x), right?  That would
match up with your partial answer x<(1/2). You are correct in the way 
that you have solved the problem so far.  What you need to understand 
is that BOTH x > -(1/3) AND x < (1/2) has to be true at the same time 
for the same value of x. If you have such an x value, that x value 
will make (3x+1) > 0 and (5-10x) > 0 so that the product will be > 0 
too.
  
Where the interval part of the question comes into the picture is that 
any x value between -(1/3) and +(1/2) will work.  The interval on the 
number line from -(1/3) to +(1/2) is usually represented as 

  ( -(1/3), 1/2 ). 

Sometimes you see intervals represented like this:

  [ -(1/3), 1/2 ]

using square brackets instead of parentheses.  That means that 
the end-points, in this case the numbers -(1/3) and 1/2, are included 
in the interval.  In your problem you DO NOT want to write the 
interval in this way because including these two x values will make 
the product equal to zero instead of greater than zero.  If your 
original problem had been written " > = 0 " meaning greater-than-or-
equal-to-zero, only then would the correct interval answer have been 
[ -(1/3) , 1/2 ]. 
  
Hope this helps.  Please write back if you have other math questions
and we will try to be faster in answering them.  

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