Sometimes in algebra
it is desirable to find
an equivalent expression
for a radical expression
that doesn't have any
radicals in the denominator.
This process is called
rationalizing the denominator.
You can also do the same
thing with the numerator,
but the denominator is
more common. I will only
explain rationalizing
the denominator and if
you need to rationalize
a numerator you do the
same thing with the numerator.
There are 3 cases that
commonly come up in algebra
classes.
- single square root
- single higher root
- sums and differences
of square roots
1. When you have a single
square root in the denominator
you just multiply top and
bottom by it.
2. When you have a single
nth root in the denominator,
multiply by something
so that you will get a
perfect nth power. You
can always do this by
multiplying by the denominator
to the n-1 power, but
that isn't always necessary.
For example if the denominator
is the cube root of 3
you can multiply by the
cube root of 9, because
9 is 3 to the 2nd power
and 2 is 1 less than 3,
but if the denominator
is the cube root of 9,
you don't have to multiply
by the cube root of 81,
you can multiply by the
cube root of 3.
3. Multiply top and bottom
by the difference if the
original is a sum and
the sum if the original
is a difference. That
way you will get the difference
of squares and get rid
of the square roots that
way.
Examples:
For the following problems
the instruction is to rationalize
the denominator, which means
to write an equivalent expression
for it that doesn't have
any radicals in the denominator.
Example 1:
Solution:
Explanation:
There is just one square
root in the denominator,
so this is case 1. So to
get rid of it we just multiply
top and bottom by it. The
square root of 3 times the
square root of 3 is 3 simply
from the meaning of square
root, you don't have to
write it first as the square
root of 9. Remember that
the square root of 3 means
the number you can square,
that is multiply by itself,
to get 3, so if you squared
it and didn't get 3 it wouldn't
be the square root of 3.
So by multiplying the top
and the bottom by the square
root of 3 we get rid of
the square root in the bottom.
In punishment for this the
top got more complicated,
and in fact the whole thing
does in fact look more complicated
after this 'simplification',
but many times, like
for example when you are
adding and need to find
common denominators, this
sacrifice is worth it, because
simplicity is more important
in denominators than in
numerators.
Example 2:
Solution:
Explanation:
Here we first need to split
up the quotient into two
separate radicals. (See
Square
Roots.) Then just like
the last one we multiply
top and bottom by the square
root that is downstairs,
the square root of 6, to
turn the square root of
6 into 6. The numerator
again suffer from it. Here
since there was a square
root upstairs it made it
look nicer to pull the radicals
together under one radical
and multiply. (See
Square
Roots.)
Example 3:
Solution:
Explanation:
This one is similar except
that there is something
multiplied by the radical
in the denominator. Here
we don't have to multiply
by the whole denominator,
just the radical part of
it. The 5 just sits there
and comes along for the
ride. another interesting
thing that happened is that
after we multiplied the
square roots of two to get
2, we were able to simplify
further, because that 2
canceled with the 2 we already
had upstairs.
Example 4:
Solution:
Explanation:
This one is an example of
case 2 where we have a higher
order root in the denominator.
Don't let the variables
bother you, we deal with
them just like we deal with
numbers, of course we do,
since they are just place
holders for numbers that
we haven't yet decided on.
Since this is a 3rd root,
in order to get a total
of 3 of them multiplied
together in the bottom we
need to multiply by 2 more,
that is the square of the
radical. When you do this
it doesn't matter whether
you put the square on the
inside or the outside of
the radical, because they
are equivalent. Putting
it on the inside makes it
easier to multiply in the
top.
Example 5:
Solution:
Explanation:
Here we have the same sort
of thing except for a couple
of difference. First we
have to split the quotient
into two separate radicals.
Then this is a very good
example of the exception
that I spoke of in the description
of case 2. Going by the
rule of multiplying by 1
less power we would have
multiplied by the cube root
of 49b
4. We could
have done that, but the
way we did it was simpler.
Since there was already
a b
2, we only
needed to multiply that
part by one more b to get
a perfect cube, so we just
multiplied top and bottom
by the cube root of 49b.
You could make all kinds
of elaborate rules for this,
but the easiest thing to
do is just think in terms
of trying to multiply the
thing inside the cube root
by something so that it
will make a perfect cube.
In this case, since there
are already two b's multiplied
we only need one more to
make the required 3.
We didn't really have
to multiply out the 7
times 49. Since we chose
49 specifically so that
it would multiply with
7 to make the perfect
cube 73, it
would be easier to just
write it is 73
and not worry about what
number it was. We are
only interested in what
its cube root is anyway,
and we can find that easier
by leaving it as 73.
Then just as with the
square root problems,
we need to put things
under one radical upstairs
and multiply out, but
this time we can do some
further simplifying afterward.
(See Square
Roots.) We can re-split
it up so that the perfect
cube a3 is
under a separate radical,
and then we can take its
cube root and get a, which
goes on the outside of
the radical.
Example 6:
Solution:
Explanation:
Here since this is a 5th
root, we want to multiply
by something that gets it
raised to the 5th power.
Since there is already one
5th root of 2, we need 4
more to multiply, so that
would be the 5th root of
2 to the 4, the 5th root
of 16. One like this doesn't
come up quite so commonly
in algebra class, but I
did it so that you could
see that the idea will generalize.
In my classes we normally
only go up to cube roots.
Example 7:
Solution:
Explanation:
Even though there is only
one radical in the denominator
this is an example of case
3, because there is something
added to it. If we simply
multiplied top and bottom
by the square root of 10
it wouldn't work, because
we would have to multiply
it by both terms, so there
would still be a square
root of 10 multiplied by
the 9. If we multiplied
top and bottom by 9+sqrt(10)
that wouldn't work either
because there would be a
middle term from the inners
and outers. (See
How to Add, Subtract, Multiply, and Divide
Polynomials.) But there
is a special multiplication
formula that comes to rescue.
When you multiply a sum
and difference of the same
thing you can a difference
of squares, because the
outers and inners cancel
out. (See
How to Add, Subtract, Multiply, and Divide
Polynomials.) So the
thing to do is to multiply
by the difference instead
of the sum, so that all
together we get the product
of a sum and difference.
Then this gives us 9 squared
minus the square root of
10 squared. To square 9
we just do it, we multiply
it by itself to get 81.
To square the square root
of 10, just like in the
previous examples, we use
the meaning of the square
root. Since the square root
of 10 means the number you
can square to get 10, if
when you squared it you
got anything other than
10, it wouldn't be the square
root of 10, so when you
square the square root of
10, the only thing
you can get is 10. Then
to finish up the problem
we distribute the 7 in the
numerator and subtract in
the the denominator.
Example 8:
Solution:
Explanation:
This one is also case 3,
but now we have a difference,
so we need to multiply by
the sum. Then again we get
the difference of squares,
and when you square the
square root of 3 you get
3, and when you square the
square root of 5 you get
5. After we subtract there
are minuses both upstairs
and downstairs, so we can
simplify by getting rid
of both of them. Then we
just have to multiply out
in the top to get the final
answer.
Example 9:
Solution:
Explanation:
This one is similar to the
last two, but just a bit
more complicated because
of the difference in the
numerator. Since there is
a difference in the denominator,
to get a sum and difference,
we need to multiply by the
sum, and then that sum has
to get multiplied by the
difference in the numerator,
but in the numerator they
are not sum and differences
of the same thing, so there
is not shortcut to multiplying
them. We just have to use
FOIL , (see
How to Add, Subtract, Multiply, and Divide
Polynomials), and this
time there isn't even any
combining of like terms.
Also it is important to
realize that when we square
a product like 2sqrt(7),
we have to square both the
2 and the sqrt(7). To square
the 2 we just do it, multiplying
it by itself, and to square
the sqrt(7) we get 7 because
sqrt(7) is by definition
the number we can square
to get 7.
Example 10:
Solution:
Explanation:
This one is similar except
that it has variables instead
of numbers. Since we have
a difference in the denominator,
we multiply by the sum.
Then we something multiplied
by itself in the numerator,
so we can use the squaring
formula. (See
How to Add, Subtract, Multiply, and Divide
Polynomials.)
Extra for Experts
What if you have a sum or
difference of a higher root
in the denominator? Is there
a way to rationalize the
denominator then? Like,
for example, what if you
have a difference of cube
roots or 4th roots. Well,
I have never seen this taught
in algebra classes, but
you can rationalize the
denominator in such cases,
and the way to do it is
to use such things as the
sum and difference of cubes
factoring formula, and there
are such formula for higher
degrees as well. (See
Factoring Polynomials.)
Example 11:
Solution:
Explanation:
Here we can use the sum
of cubes factoring formula,
A3+B3=(A+B)(A2-AB+B2),
because we already have
the A+B part in the denominator.
Example 12:
Solution:
Explanation:
Here what we need is something
to multiply A-B by to get
at least A raised to the
4th power, so we can apply
it here to get rid of the
4th root. There isn't really
a specific formula for factoring
the difference of 4th powers,
but we can come up with
one by factoring it as a
difference of squares twice.
A4-B4=(A2-B2)(A2+B2)=(A-B)(A+B)(A2+B2)
Since our denominator
already has the difference
what we need to multiply
top and bottom by is the
other two factors here,
(A+B)(A2+B2),
where A is the 4th root
of 5 and B is 1.