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
