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Solving Rational, Radical, and Polynomial Inequalities

Date: 12/28/2003 at 20:47:57
From: Chapmann Wong
Subject: How do I solve inequalties algebraically

How do I solve rational and radical inequalities algebraically?  I
understand how to solve a rational inequality using a number line. 
For example;

(x - 2)
------- > 0
(x + 4)

The number line is split into 3 parts:  x < -4, -4 < x < 2, x > 2.  A
"test point" number from each of the sections is taken and plugged
into the original inequality, and if it makes the inequality correct,
then the whole section is a solution to the inequality.  The solutions
for this inequality are x < -4, x > 2.  But when I use algebra, all I
get is  x > 2.

Why is this?  Is there something about rational inequalities that make
them have to be solved with a number line?  I also have trouble with
radical inequalities.  Is there something with the square root that
makes it not possible to solve by algebra?

Date: 12/29/2003 at 07:21:52
From: Doctor Rick
Subject: Re: How do I solve inequalties algebraically

Hi, Chapmann.

Thanks for writing.  First, the short answer--you should be able to
solve any inequality you run into using algebra.  But there are a few
ideas to keep in mind.

When solving your sample problem, you need to consider separate cases
for whether the denominator is positive or negative.

If x + 4 > 0 (positive), then when you multiply both sides by x + 4,
you don't change the direction of the inequality.  There are now two
conditions on x:

  x + 4 > 0 AND x - 2 > 0

If x + 4 < 0 (negative), then when you multiply both sides by x + 4
you need to change the direction of the inequality.  The two
statements for this case are:

  x + 4 < 0 AND x - 2 < 0

The two cases (positive or negative) lead to

  x > -4 AND x > 2       OR       x < -4  AND x < 2

The first of these cases reduces to x > 2, since any number greater 
than 2 is also necessarily greater than -4.

The second case reduces to x < -4, since any number less than -4 is
also necessarily less than 2.

Thus the algebraic solution is:

  x > 2 OR x < -4

Do you see where you went wrong?  Algebra does work, but you need to 
consider cases in a manner similar to the number-line method.  Any
time you multiply through by something which could be positive or
negative, you need to create a separate case for each possibility.

If you can't see yet how to correct your algebraic approach to 
inequalities with radicals, show me an example of such a problem and 
I'll go through it with you.

- Doctor Rick, The Math Forum 

Date: 12/29/2003 at 11:39:07
From: Chapmann Wong
Subject: How do I solve inequalties algebraically

Thank you for the information.  When trying to solve the following 

(x - 4)
------- >= -2
(x + 2)

I again got the wrong answer.  I tried making cases but it didn't seem
to work.  The correct answer is x < -2 or x >= 0.  Where did the zero
come from?  Thank you!

Date: 12/29/2003 at 12:26:08
From: Doctor Rick
Subject: Re: How do I solve inequalties algebraically

Hi again, Chapmann.

With this new problem, you need to multiply both sides by (x + 2), 
and again you need to keep track of whether the multiplier is 
positive (x + 2 > 0) or negative (x + 2 < 0), so there are two cases:

x + 2 > 0 AND x - 4 >= -2(x + 2)  OR  x + 2 < 0 AND x - 4 <= -2(x + 2)
  x > -2  AND x - 4 >= -2x -4     OR    x < -2  AND x - 4 <= -2x - 4
  x > -2  AND  3x >= 0            OR    x < -2  AND  3x <= 0
  x > -2  AND   x >= 0            OR    x < -2  AND   x <= 0

The first case reduces to x >= 0; the second reduces to x < -2. 
Again, the key idea is to consider what happens if the multiplier is
positive and what happens if it's negative.  The two conditions in
each case are the multiplier and the result of the inequality after
multiplying through.  If you don't follow any of these steps, please
ask about it.

- Doctor Rick, The Math Forum 

Date: 12/29/2003 at 14:35:12
From: Chapmann Wong
Subject: How do I solve inequalties algebraically

Dr. Rick -

Thank your for all your help.  I think I'm getting the rational
inequalities now.  But I'm still a little confused about radical
inequalities.  Can you help me with those?  Here's an example, and
what I did on it:
\/x + 2 >= x

I squared both sides and got the following possibilities:

x + 2 >= x^2  so then  0 >= x^2 - x - 2  so then  0 >= (x - 2)(x + 1)


x + 2 <= x^2  so then  0 <= x^2 - x - 2  so then  0 <= (x - 2)(x + 1)

So I took each of the two cases and solved them, and overall I came up
with -1 <= x OR x >= 2.  However, the answer is -2 <= x <= 2.  Why is
this?  I didn't get a -2 anywhere.  Or did I do something wrong in the
process?  Thanks again for all your help.

Date: 12/29/2003 at 15:21:52
From: Doctor Rick
Subject: Re: How do I solve inequalties algebraically

Hi Chapmann -

Problems with radicals take some thinking; you always need to check
your results to see if they make sense.  When you square both sides of
an equation or inequality, funny things can happen.  For example,
here's a true statement:

-2 < 2

But if you square both sides, you get a false statement:

4 < 4

Similarly, you can turn a false statement into a true one by squaring:

-3 = 3  (False)  
 9 = 9  (True after squaring both sides!)

So we tend to approach radical problems the same way we always do, but
keep it in your mind that you need to really think about and check the
answer you get.

There's one other important idea to keep in mind when working with
radicals--you can't take the square root of a negative number if you
are working with real numbers only.  So hold onto that thought, too.

I like your idea of squaring both sides, but I don't agree that you
would get two cases as you show.  In fact, if you look at your two cases

0 >= (x - 2)(x + 1)  OR  0 <= (x - 2)(x + 1)

closely, you'll see that you are saying that the product of (x - 2)
and (x + 1) can be negative (case 1), positive (case 2) or zero (both
cases).  In other words, it can be anything!  So the answer to your
approach would be that x can be any real number, which is clearly not
the correct solution to the problem!

Squaring both sides just produces one case, and it does not change the
inequality (though we need to check our answers, remember), so the
correct case is your first one

0 >= (x - 2)(x + 1)

In order for the product to be negative, one factor must be positive 
and the other negative. In other words, one of these is true:

  x - 2 <= 0 AND x + 1 >= 0   OR  x - 2 >= 0 AND x + 1 <= 0
      x <= 2 AND x >= -1      OR      x >= 2 AND x <= -1
        -1 <= x <= 2          OR          impossible

Thus the apparent solution is x between -1 and 2.  However, when we
look back at the original problem, there are two considerations.

1. The result of the square root can never be negative since the way
it is written refers to the principal or positive root.  Thus, if x is
negative, the inequality must be true (as long as the radical makes
sense) since a positive will be greater than any negative.

2. The square root only makes sense for positive numbers (if we're 
talking about real numbers, not complex numbers).  Thus the equation 
only makes sense if x + 2 >= 0, that is, if x >= -2.

For positive numbers x, we know x must be less than or equal to 2. 
For negative numbers, we have now observed that the equation will be 
true for any negative x that is allowed, namely x greater than or 
equal to -2.  The correct solution therefore is -2 <= x <= 2.

The thing to remember is that whenever you square both sides, you are 
likely to introduce spurious solutions, so you must check your answer 
and adjust the solution set as needed.

- Doctor Rick, The Math Forum 

Date: 12/29/2003 at 16:02:12
From: Chapmann Wong
Subject: How do I solve inequalties algebraically

Hi Doctor Rick, 

Thanks so much for the information that you've given me.  It has
helped so much.

By the way, do I follow the same format for solving polynomial 
inequalities such as (x - 2)(x + 1)^2(x - 4) <= 0?  Or is there a
different method?

Date: 12/29/2003 at 17:25:31
From: Doctor Rick
Subject: Re: How do I solve inequalties algebraically

Hi, Chapmann.

In general for polynomial inequalities you should set one side to 0
and factor the other side completely.  Your problem is already in that
condition.  There are several methods that could be used, or combined
in various ways.

One would be to do a number line analysis as you mentioned at the
beginning, with critical values at -1, 2 and 4.

Another would be to make a series of statements realizing that for the
result to be negative (< 0) you would need either one negative factor
and two positive factors or three negative factors.  This task is
reduced slightly if you realize that (x + 1)^2 will always be
positive, but it's still a lot of statements and is not recommended.

My first thought when I see your example is to think about the graph
of the polynomial function y = (x - 2)(x + 1)^2(x - 4).  The
polynomial has four roots: -1 (double root), 2, and 4.  I can sketch
the graph.  It is positive to the right of x = 4, since the
coefficient of the highest power is positive.  From right to left, it
crosses the x axis to the negative side (below the x-axis) at x = 4,
recrosses to the positive side at x = 2, then touches the x axis on
the positive side at x = -1 before rising again.  Thus I can see that
it is negative in 2 < x < 4, and zero at each of the roots. The
solution is

  2 <= x <= 4, x = -1

This method is quite similar to your number line analysis, and as you
learn more about polynomial functions and their graphs, you'll see why
they are related.

Good luck!

- Doctor Rick, The Math Forum 

Date: 01/04/2004 at 12:48:08
From: Chapmann Wong
Subject: Thank you (How do I solve inequalties algebraically)

Thank you for helping me learn more about math.  It has helped me
greatly.  Once again, thank you very much!
Associated Topics:
High School Basic Algebra

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