Solving polynomial and rational inequalities
is a bit harder than solving similar
equations. Even in a simple problem
where you can factor like
x2-5x+6 >
0
it is not clear what to do after
you factor, because the usual principle
of zero factors doesn't hold for inequalities.
So once you get
(x-2)(x-3) > 0
it is not clear what to do. With
the equation
(x-2)(x-3)=0
you can use the principle of zero
factors because for a product of two
numbers to be zero one or the other
has to be zero, so you can set each
factor to zero. There is a way to
do something similar with the inequality,
but it is a bit more complicated.
You see it is not true that if a
product of two numbers is greater
than or equal to zero that one or
the other of them must be greater
than or equal to zero, but there is
something else that is true, and to
think of it you have to think of greater
than or equal to zero as positive
and think about the way positive and
negative numbers multiply. When is
the product of two numbers positive?
Well, there are two ways for this
to happen. They must be either both
positive or both negative. You could
then work this problem out by using
that and take both cases.
Case 1: x-2 > 0, x-3 >
0
This is the case when both factors are
positive. It will give us a positive
product, because the product of two
positive numbers is positive. By adding
2 to both sides for the first inequality
and 3 to both sides for the second inequality,
this gives us
x > 2 and x >
3.
This is a funny kind of condition,
though, because one of the conditions
isn't necessary, because it is weaker
than the other one. Certainly if x
is greater than or equal to 3, then
it must be greater than or equal to
2, since 3 is greater than 2, so we
can get rid of the first condition
and just use the second one, and this
reduces to
x > 3.
Case 2: x-2 < 0, x-3 <
0
This is the case when both factors are
negative. It will give us a positive
product, because the product of two
negative numbers is positive. By adding
2 to both sides for the first inequality
and 3 to both sides for the second inequality,
this gives us
x < 2 and x <
3.
Again one of these isn't necessary,
because it is weaker than the other.
This time, since 2 is less than 3,
certainly if x is less than or equal
to 2, then it is less than or equal
to 3, so only the first condition
is necessary, so this reduces to
x < 2.
Putting the two cases together we
get that there are two possible kinds
of numbers that will satisfy the inequality,
numbers that are less than or equal
to 2 and numbers that are greater
than or equal to 3. We can write this
in interval notation by using a union.
The square brackets indicate that
the 2 and the 3 are included. This
is true because for x=2 and 3, the
left side of the inequality is zero
and zero is certainly greater than
or equal to zero. The plus and minus
infinity have round brackets, because
infinity is not a number, so it is
not included. Here is a graph on the
number line of the solutions to the
inequality.
Notice how the interval notation is
sort of miniature picture of the graph.
The intervals represent pieces cut
out of the number line.
You could also use this method to
solve other kinds of inequalities
as well. If it was a less than inequality,
you would use the fact that in order
to multiply two numbers to get a negative
number, one of them has to be positive,
and the other has to be negative,
and again you could divide into two
cases, one where the first factor
was positive and the second factor
was negative, and the other where
the first factor was negative and
the second factor was positive. If
the inequality was strict, that is
< or >, then the only difference
would be that you wouldn't include
the endpoints, so you would use round
brackets in interval notation.
But there is a better way to do this
whole thing that is quicker and allows
you to deal with more complicated
inequalities much easier. Suppose
you had a more complicated inequality
like
(x-1)(x-2)(x-3)(x-4)(x-5) <
0
To break it down into cases like
I did with the equation above would
get very time consuming, because there
are too many possibilities. In this
case, because the product is negative,
there must be an odd number of minus
signs. But it is really not that complicated,
because a lot of them are impossible,
like for example, it is not possible
for x-1 to be negative and x-5 to
be positive, because x-1 will always
be bigger than x-5. But here is something
interesting that will happen with
the signs of the expression on the
left side of the inequality as you
travel up and down the real line.
At any place where one of the factors
is zero, there will be a sign change,
because there is a sign change of
one and only one factor. For example
for x=5.1 all of the factors are positive,
but for x=4.9, the first four factors
are positive, but x-5 is negative.
This means that if we can find out
the sign of the product for just one
number, then we can figure out the
sign everywhere else and solve our
inequality, because after that it
will just flip flop. The easiest way
to keep track of this is to plot the
places where the product is zero on
the number line, and write in the
signs. Here are the places where the
product is zero plotted in order on
the number line.
Think of these numbers as cutting
the real line into intervals, with
alternating ones making up the solution
set of the inequality. The only problem
is which ones do we take. One way
we could determine this is to simply
pick one number on one of the intervals
and test out the sign of that interval
with it. So, for example, what happens
when x=2.5? Then the product is (2.5-1)(2.5-2)(2.5-3)(2.5-4)(2.5-5)=(1.5)(.5)(-.5)(-1.5)(-2.5),
which is negative, since there are
3 negatives. Then we could put a minus
sign in the interval between 2 and
3 and then alternate to fill in the
rest. But we could also do it even
easier by just reasoning about it,
and the easiest interval for that
is the right most one, the one that
has the big numbers. For any number
bigger than 5 we know that all of
the factors must be positive, because
x-5 is, and it is the smallest. So
we know that the right most interval
must be plus and can alternate from
there. We can even be a little bit
lazier about this and realize that
as long as all of the x coefficients
are positive, this will always happen,
and if we had a problem where one
of the x coefficients was negative,
we could always multiply by -1 on
both sides to get it to be positive,
remembering to reverse the direction
of the inequality of course. Anyway,
however we decide it, we get for this
problem the following map of plus
and minus signs for the product.
The meaning of this picture is that
the left side of the inequality is
positive, that is greater than zero,
wherever the pluses are and negative,
that is less than zero, where the
minuses are. So since we are trying
to find the numbers where it is less
than zero, we get
for the solution of the inequality.
An important thing to remember throughout
all of this is that positive numbers
are greater than zero and negative
numbers are less than zero.
This method can also be applied to
rational inequalities as well using
all the places where either the numerator
or the denominator is zero for the
sign changing places, because for
quotients as well as products the
total number of minus signs determines
the sign of the answer. The only difference
is that you never include any places
where the denominator is zero in your
answer even if the inequality is greater
than or equal or less than or equal,
because a zero in the denominator
doesn't make the quotient zero, it
makes it undefined.
Example:
Here the zeros of the numerator and
the denominator are -2 and 3. Here
they are on the number line.
If we alternate the pluses and minus
we get this.
This time since the inequality says
greater than or equal to zero we use
the intervals with the pluses, but
unlike with products we don't include
the 3, because it make the denominator
0, but we do include the -2, because
it is perfectly all right for numerators
to be zero and when the numerator
is zero the inequality will be satisfied,
because the quotient will be zero
and zero is greater than or equal
to zero. So the final answer is
If this all sounds too simple, though,
you are right, because it isn't. There
is an exception to this alternating
signs game, but it isn't really that
bad to deal with. The problem occurs
when a factor is raised to an even
power, like for example
(x-1)2(x-3) >
0.
The problem is that here there won't
be a change of signs at 1, because
(x-1)2 is always
greater than or equal to zero. This
is easy to deal with, though, you
just throw (x-1)2 out of
the game, and reduce the inequality
to
x-3 > 0.
Except not quite, because you still
have to take into account the fact
that the expression is going to be
equal to zero at x=1 and 0 >
0. So you throw it out and just use
that consideration to tweak your answer
a bit at the end. So here at first
we get that the answer should be
but 1 needs to be included as a solution
as well, so we union it as a singleton
and get
You also need to sometimes watch out
with strict inequalities. In that
case the problem is that there is
a number that you don't want to include
because it makes the product 0, because
0 is neither strictly less than 0
nor strictly equal to 0.
More Examples
For each of these the instruction is
to solve for x and write the solution
set in interval notation.
Example 1:
(x-2)(x+5)>0
Solution:
Lining up the places where the left
side is equal to 0 in numerical order
on the number line and alternating the
pluses and minus we get
as a plot of the signs of the left side
of the inequality. Since the inequality
says greater than zero, we want the
pluses, so in interval notation that
is
Example 2:
x2+x < 12
Solution:
For this one first we need to get a
zero on one side of the inequality,
so we subtract 12 from both sides to
get
x2+x-12 <
0.
Then factoring this we get
(x+4)(x-3) < 0.
Plotting the zeros on the number
line and alternating pluses and minus
we get
Since then inequality is less than
or equal to zero, we want the minuses
and we also want the endpoints, because
they are where the left side is zero.
So in interval notation the solution
set is
[-4,3].
Example 3:
(x-1)3(x+2)4(x-3)
> 0
Solution:
Here we can partly ignore the second
factor, because it has an even exponent.
So I am just plotting the 1 and the
3 on the number line. Then alternating
pluses and minus we get
This would give
for the answer except that this would
include -2, which we don't want, because
the left side is zero there, so not
greater than zero, so we need to remove
just the single number -2 from the solution.
The way to write that in interval notation
is to break up the first interval into
two pieces using round brackets, so
that the -2 will not be included. Doing
that the solution can be written like
this.
Example 4:
(2x-3)(4x+3) > 0
Solution:
The only thing that is a little difficult
about this one is that it involves fractions.
To find the zeros we have to set 2x-3
and 4x+3 equal to zero.
2x-3=0
2x=3
x=3/2
4x+3=0
4x=-3
x=-3/4
Now line these two up on the number
line and alternate pluses and minuses.
Then again since the numbers that
are greater than zero are the positive
ones, the solution set is the union
of the two plus intervals
Example 5:
(-x+3)(2x-5) < 0
Solution:
To make it easier to tell where the
pluses and minuses are, multiply both
sides by -1 so that the coefficients
on x will all be positive. But when
we do that we have to make sure to use
the rule that when you multiply both
sides of an inequality by a negative
number the direction of the inequality
flips, so then the inequality becomes
(x-3)(2x-5) > 0
Then set each factor to zero to get
3 and 5/2 for the zeros and switching
places. Then plotting these on the
number line and alternating the pluses
and minuses we get
Since greater than or equal to zero
is plus, the solution is
Example 6:
Solution:
Since there are two negative coefficients
here we don't have to multiply both
sides to get rid of them. We can just
multiply top and bottom by -1 on the
left side to get
Now the zeros in the numerator and
denominator are 1,2, and 3. Plotting
these on the number line and alternating
pluses and minuses we get
and greater than or equal to zero
is plus we get the plus intervals.
Since the inequality is greater than
or equal, we should include the endpoints,
but this time with a quotient we don't
want to include the 3, because it
makes the denominator zero, and that
would be undefined, not zero, so the
solution set is
Example 7:
For this one since (x+1)2
cannot be negative we can temporarily
ignore it. -2 is the only place where
the sign will change. Plotting it
on the number line we get
Since less than zero is minus we get
for the solution set, but we have
to make sure that -1 is not in it,
because at -1 the quotient will be
zero, which is not less than zero,
but this time we don't have to worry,
because -1 is not in our set, because
it is greater than -2, so we leave
it as is.
Example 8:
Solution:
For this one we have to remember that
this whole method depends on having
0 on the right side of the inequality,
because we need to reason about signs
of numbers which have to do with whether
numbers are greater or less than 0,
not 5. So we have to do something to
get 0 on the right side. To do that
we have to subtract 5 from both sides
and then turn the left side into a single
fraction by finding a common denominator.
Then we need to multiply both sides
by -1 in order to have all positive
x coefficients and reverse the sense
of the inequality, so this becomes
Now the zeros in the numerator and
denominator are -13/4 and -3. -13/4
is less than -3 since -12/4=-3, so
plotting them on the number line and
alternating pluses and minuses it
looks like this.
We take the outer intervals because
greater than or equal to zero is plus,
but we must make sure to not include
the -3, because it makes the denominator
zero, so our solution set is
