Polynomial and Rational Inequalities - DOC

Shared by:
Categories
Tags
-
Stats
views:
18
posted:
9/15/2012
language:
Unknown
pages:
5
Document Sample

```							   Polynomial and Rational Inequalities

Polynomial Inequalities

Solving polynomial inequalities, such as x 2  6x  8 , is like asking the question:
what are all the x-values that will make this inequality true? For x=2, for instance, this
inequality is true since 2 2  6  2  8 but for x=-3 it is not true since
 32  6  (3)  8 .

This type of inequalities, just as the Rational Inequalities that will be discussed later
in this study guide, can be solved by using a simple but elaborate routine. The ideas
presented in this routine are just one way of solving these inequalities; many other
similar ideas exist. Regardless of the chosen approach great care is required in
working these problems. A minor omission can have major consequences.

Let us use the inequality just presented and go through the routine.

Example 1: x 2  6x  8
Step 1: Put everything to one side of the inequality:
x 2  6x  8  0

Step 2: Factor the polynomial:
x  4x  2  0
Step 3: Make a number line and put all the so called critical values on this number
line. Critical values are those values that make each factor in the polynomial
equal to zero. Since the factors in this problem are x+4 and x+2, then the
critical values are x = -4 and x = -2 respectively.
The picture of the number line is as follows:

-4               -2
Step 4: Separate the number line into intervals: from -∞ to the first critical value,
from the first to the second critical value, from the second to the third etc. The
last interval goes up to +∞. In this case these intervals are x<-4, -4<x<-2 and
x>-2.

1
Step 5: Take a test value in the first interval, plug it into the original inequality and
see if it works. If it does, the claim is that all x-values in that interval will
work as well. If it doesn't, then no value in that interval will work. Do this for
every interval. The solution to the problem will be all those intervals for which
the test value works. In this example I have chosen test values of -5, -3 and 0
respectively.

Work should be summarized in the following table:

Interval   Test Value   Calculation        Result
x<-4       x=-5         (-5)2+6(-5)>-8 ?   Works
-4<x<-2    x=-3         (-3)2+6(-3)>-8 ?   No
x>-2       x=0          02+6(0)>-8 ?       Works

The answer to the problem will be all values where x<-4 or x>-2.

Example 2: x 3  4x .
Please ignore the = sign underneath the < symbol for now. It will return into the
discussion when it is time to present the answer.

Step 1: x 3  4x  0

Step 2: x(x  2)(x  2)  0

Step 3:

-2              0        2

Step 4: The intervals are x<-2, -2<x<0, 0<x<2 and x>2.

Step 5:     Interval    Test Value   Calculation   Result    :-
x<-2        x=-3         -27≤-12 ?     Works
-2<x<0      x=-1         -1≤-4 ?       No
0<x<2       x=1          1≤4 ?         Works
x>2         x=3          27≤12 ?       No
underneath the < symbol in the original problem. The inclusion of that =
means that also those x-values that will make x3=4x zero, must be listed in the
answer as well. These values are x=0 and x=±2. The complete answer to the
problem is then x  2 or 0  x  2 .

2
Rational Inequalities
Please recall the meaning of the word rational. A rational number is nothing more
than a number that can be written as a fraction. A rational inequality is an inequality
that can be written as a fraction. For instance, an example of a rational inequality is
3x  2
 3.
x5

Rational inequalities are solved in very much the same way as polynomial
inequalities. Only step 2 may be a little more elaborate in the case of a rational
inequality. Step 2 now says: combine the fractions on the left hand side into one
fraction only (use common denominators) and factor that fraction (the numerator)
completely, if possible.

3x  1
Example 3:                  0
5x  2

Steps 1 and 2 could be omitted here since everything is already on one side of the
inequality, it is written as 1 fraction, and there is no factoring to do. Thus, we continue
with step 3.

Step 3:

2               1

5               3

Step 4: The intervals are x<-2/5, -2/5<x<1/3 and x>1/3.

Step 5:     Interval       Test Value   Calculation   Result   :
x<-2/5         x=-1         -4/-3>0 ?     Works
-2/5<x<1/3     x=0          -1/2>0 ?      No
x>1/3          x=1          2/7>0 ?       Works

The answer is then is x<-2/5 or x>1/3.

3
4x  3
Example 4:                 2
x 1

4x  3
Step 1:            2  0
x 1

Step 2: Rewrite the nonzero side as a single fraction and factor both numerator and
denominator if this is possible. This usually means you have to add/subtract
the fractions by using a common denominator etc. Here it gives
4 x  3 2( x  1)
          0
x 1      x 1

2x  5
0
x 1

No factoring is needed.

For the rest, the routine follows the same steps as in the case of a polynomial
inequality.

Step 3:

-1             5
2
Step 4: The intervals are x<-1, -1<x<5/2 and x>5/2.

Step 5:     Interval    Test Value   Calculation   Result    :
x<-1        x=-2         -11/-1≥2 ?    Works
-1<x<5/2    x=0          -3/1≥2 ?      No
x>5/2       x=3          9/4≥2         Works

The preliminary answer is x<-1 or x>5/2. Looking at the = sign you note that
x=5/2 makes the numerator and hence the whole fraction equal to 2 and must
be added into the solution set. The complete answer is thus x  1 or x  5 .
2
x=1 cannot be put into the solution set since this value will make the
denominator equal to zero, and that is an illegal operation. Please recall that
you cannot divide any quantity by zero.

4
2   3
Example 5:           
x3 x3

Step 1: Putting everything on the left hand side gives:
2       3
       0
x 3 x 3

Step 2: Rewriting the left hand side as just one fraction we get
 2( x  3)  3( x  3)
0
( x  3)(x  3)

 5x  3
0
( x  3)(x  3)

Step 3:

-3            -3/5              3

Step 4: The intervals are x<-3, -3<x<-3/5, -3/5<x<3 and x>3

Step 5:
Interval        Test Value   Calculation   Result
x<-3            x=-4         -2/-1<3/-7?   No
-3<x<-3/5       x=-2         -2/1<3/-5?    Works
-3/5<x<3        x=0          -2/3<3/-2?    No
x>3             x=4          -2/7<3/1?     Works

3
Accounting for the equal sign will give us  3  x        or x  3 for an answer.
5

5

```
Related docs
Other docs by HC120915093451
david Harcombe
Variable Frequency Drives
Math Facts Frenzy