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Polynomial Inequality Problem

Date: 2/7/96 at 12:44:6
From: Anonymous
Subject: polynomial inequalities

     x>6  x<-1  OR  x<6  x>-1

     How can I tell which one I'm suppose to use?


     What do I need to do next?
Can you please explain to me what I need to be doing with 
these kinds of problems?    

Date: 2/7/96 at 15:40:42
From: Doctor Elise
Subject: Re: polynomial inequalities


Let's look at the first one.  You've done a great job factoring it.

   (x - 6)(x + 1) > 0

Okay.  So we're multiplying two numbers together, and their 
product has to be a positive number.  This means we're either 
multiplying a positive number by a positive number, or a negative 
number by a negative number, right?  So EITHER:

   (x - 6) > 0  and  (x + 1) > 0  (both numbers are positive)
   x > 6 and x > -1

which basically boils down to x > 6, since 6 > -1


   (x - 6) < 0  and  (x + 1) < 0  (both numbers are negative)
   x < 6 and x < -1

which boils down to x < -1

So your answer is either:

   x > 6, or x < -1

The second one, 7y^2 - 63 < 0 , needs to be factored.
(I always write "y squared" as y^2).

If possible, I try to factor out the constant in front
of the y^2. We can do it here, so I'll show you what I
mean. Notice that:

   7y^2 - 63 = 7(y^2 - 9)

That way, now we just have to factor the smaller term
y^2 - 9. Remember though, that factoring out a constant
isn't always possible.  So now we need to factor:

   y^2 - 9

Because the first term is y^2, we know we at least have:

   (y    )(y    )

Since we have a -9, we know our signs must be like this:

   (y -  )(y +  )

Finally, since there's no "y" term, we know that the two 
middle terms will have to cancel out. This means that the
constants we are looking for must be equal. In this case, they
are both 3. Thus, 7y^2 - 63 factors to:

   7(y - 3)(y + 3)

So now we have:

   7(y - 3)(y + 3) < 0

First, we see that we can divide by 7. It doesn't change the
problem, and it makes the solution easier:

   (y - 3)(y + 3) < 0

In this case, we are multiplying two numbers together and 
getting a negative number, so one has to be positive and the 
other negative.

   negative   *   positive
   y - 3 < 0 and  y + 3 > 0
   y < 3     and  y > -3

   positive   *  negative
   y - 3 > 0 and y + 3 < 0
   y > 3     and y < -3

This second solution is obviously impossible, because a number 
cannot simultaneously be greater than 3 and less than -3, so you 
just throw this answer out as "impossible" and go with y is between 
-3 and 3.

Hope this helps!

-Doctor Elise,  The Math Forum

Associated Topics:
High School Basic Algebra

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