The Mathematics 11
Competency Test
Rationalizing the Denominator
This document introduces an important method of algebraic simplification used when radicals
involve fractions, or equivalently, when a fraction contains radicals in its denominator.
Generally speaking, fractions with radicals in their denominators are considered to be less simple
than equivalent fractions with no radicals in their denominator. The process of starting with a
fraction containing a radical in its denominator and determining the simplest equivalent fraction
with no radical in its denominator is called rationalizing the denominator. (This terminology
arises from the fact that generally, radical expressions evaluate to give so-called irrational
numbers, whose decimal part never terminates and never achieves a repeated pattern of digits.
By eliminating the radical expression from the denominator, we are turning something which
gives irrational number values to something which does not give irrational number values – hence
the term “rationalizing” the denominator.) There is a general strategy for attempting to rationalize
denominators containing radicals of any order. However, in this document, we will focus
particularly only on fractions with square roots in the denominator.
Whenever we attempt to manipulate a fraction to get a new fraction which is equivalent but
simpler or easier to work with, it is important to ensure that we really do end up with an equivalent
fraction. There are many ways to get rid of a square root. (We could simply square it, or we
could even just erase it!). We must be careful to use a method which simultaneously makes the
square root in the denominator disappear, but leaves us with a mathematically equivalent
fraction.
To start with one fraction and turn it into another equivalent fraction, we can simply multiply the
numerator and denominator by the same thing. Secondly, to rationalize the denominator of a
fraction, we could search for some expression that would eliminate all radicals when multiplied
onto the denominator. Putting these two observations together, we have a strategy for turning a
fraction that has radicals in its denominator into an equivalent fraction with no radicals in the
denominator:
(i) deduce a factor which can be multiplied onto the denominator to eliminate all radicals in it.
(We will illustrate patterns of such factors for the most commonly occurring simple expressions
containing square roots.)
(ii) multiply both the numerator and the denominator of the fraction by this factor. (Then, the
radicals will disappear from the denominator, and since we are multiplying the numerator and
denominator by the same thing, the result will be a fraction which is equivalent to the original
fraction.)
(iii) carry out any necessary or feasible simplifications of the fraction coming out of step (ii).
The tricky part here is step (i). It is easily possible to write down fractions with radicals in their
denominators for which step (i) cannot be achieved. However, there are effective standard
approaches for most situations commonly occurring in technical applications.
For the BCIT Mathematics 11 Competency Test, you need to be able to rationalize denominators
of fractions only when those denominators are monomials, and the radicals present are square
roots. In this situation, the factor required in step (i) is just the square root part of the simplest
form of the original square root appearing in the denominator.
David W. Sabo (2003)
Rationalizing the Denominator
Page 1 of 10
5y
Example 1: Simplify
.
x
solution:
x , in the denominator of this fraction obviously cannot be simplified further.
The square root,
However, it is in the denominator of this fraction, and so to simplify this expression, we do need to
rationalize the denominator. To achieve this, we just need to multiply the numerator and
denominator of the fraction by the square root occurring in the denominator (since this square
root is already in simplest form). This gives
(
since
(
( 5y ) (
x
)
x
x
)(
)(
x
x
=
)
) (
=
x
5y
x
x
)
2
= x . The expression on the right just above is a fraction in
simplest form. It is equivalent to the original fraction given at the beginning of this example
because all we did was multiply the numerator and denominator of that original fraction by the
same thing. Finally it has no square root in the denominator. Hence it is the desired final answer
here.
Remark:
People sometimes mistakenly attempt to rationalize the denominator in a fraction such as the one
we started with in Example 1 by simply squaring the fraction:
5y
x
 5y 


 x 
→
2
=
(
(5y )
x
2
)
2
=
25 y 2
x
Now, it is true that this final form does have a rational denominator – the square root part has
disappeared. However, this final from is in no way equivalent to the original fraction. You can
easily see this by noting that
 5y 


x


2
=
5y
x
5y
i
x
=
5y i 5y
x i
x
and so, squaring the original fraction amounts here to multiplying the numerator by 5y and the
denominator by
x . Since 5y and
x are different values except by rare coincidence, this
operation does not produce an equivalent fraction. Hence simply squaring a fraction is not an
acceptable approach for rationalizing its denominator.
David W. Sabo (2003)
Rationalizing the Denominator
Page 2 of 10
7t
Example 2: Simplify
.
3 s+t
solution:
There is no factorization of the expression in the square root possible here, so the only
simplification necessary is to rationalize the denominator of the fraction. Since the square root in
the denominator is already as simple as possible, we need to just multiply the numerator and
denominator by that square root to achieve the required result.
7t
3 s+t
=
3
( 7t ) (
(
s+t
s+t
)(
)
s+t
=
)
7t s + t
3 (s + t )
Although there is a t in both the numerator and the denominator, it is not a factor in the
denominator, and therefore we cannot cancel the t’s between the numerator and denominator of
this last form. In fact, there is obviously no common factors to cancel between the numerator and
denominator of this last form, so this last expression is as simple as possible and so it must be
the required final answer.
NOTE: Sometimes as they focus on simplifying fractions such as the final result of Example 2
above, people note that the numerator and denominator contain a power of the same expression
– in this case of (s + t). Without thinking they do the following:
7t s + t
3 (s + t )
=
7t s + t
3 (s + t )
=
7t ( s + t )
1
2
=
3 (s + t )
7t ( s + t )
1 −1
2
3
=
7t ( s + t )
−1
2
3
or
7t
3
(
s+t
s+t
)
2
=
7t
3 s+t
Superficially, it may appear that one or both of these lines represents a useful simplification, but
in fact, both lines really just amount to reversing the simplification previously done. The second
line obviously just gives back the original problem, and the first line does as well, though in a
somewhat less recognizable form. So, while it is very useful to break the simplification of
complicated algebraic expressions down into a series of relatively simple independent steps, it is
also important to review your entire solution to make sure that you haven’t lost track of what the
problem actually requires you to accomplish, and make sure that that is indeed what your final
answer has accomplished.
Example 3: Rationalize the denominator in
7 xy 2
8x3
and simplify the result as much as
possible.
solution:
David W. Sabo (2003)
Rationalizing the Denominator
Page 3 of 10
If we simply follow the steps illustrated in the first and second examples above, we get
7 xy 2
8x3
( 7 xy ) (
2
=
(
)(
8x3
)
8x3
8x3
7 xy 2 8 x 3
=
8x3
)
which has a rational denominator, but clearly can be simplified further. First
8x3 = 22•2•x2•x
so
( 2 )( x ) ( 2x )
8x3 =
2
=
2
22
x2
2x = 2x
2x
Thus
7 xy 2 8 x 3
8x3
=
(
)
7 xy 2 2 x 2 x
8x
3
7
14 x 2 y 2
=
4
8 x3
2x
x
=
7y 2 2x
4x
as the final simplified result.
Alternatively, we could have noted that the square root in the denominator of the original fraction
is not in simplest form, and so we could have started by simplifying both the square root and the
fraction itself first:
7 xy 2
8x3
=
7 x y2
2x
7y 2
=
2x
2 2x
Now the denominator of this fraction can be rationalized by multiplying the numerator and
denominator by
2x :
7y 2
2 2x
=
7y 2
2
(
(
2x
2x
)(
)
2x
)
=
7y 2 2x
7y 2 2x
=
2 ( 2x )
4x
giving the same final simplified result as we obtained earlier.
Example 4: Simplify
David W. Sabo (2003)
3−5
2 x
x
.
Rationalizing the Denominator
Page 4 of 10
solution:
No common factors can be found to cancel between the numerator and denominator here. Also,
the square roots which appear here obviously cannot be simplified further. So, all that’s left to do
to simplify this expression is to rationalize the denominator. This is accomplished by multiplying
x , giving:
the numerator and denominator by
3−5
x
=
2 x
(3 − 5 x )( x )
( 2 x )( x )
=
3 x −5 x
2x
x
3 x − 5x
2x
=
as the final result (since there are clearly no common factors that can be cancelled between the
numerator and denominator of this final expression). Notice that we used brackets explicitly to
ensure that the entire numerator was multiplied by
Example 5: Simplify
x in the first step here.
7y 3
.
6x
solution:
Since it is a basic property of radicals that
a
=
b
a
b
we see that the given expression actually amounts to a fraction with a radical in the denominator.
Therefore, to accomplish simplification of this expression, we need to rationalize the denominator:
7y 3
=
6x
7y 3
6x
(
=
)(
)
( 6x )( 6x )
7y 3
6x
=
42 xy 3
6x
A quick check before moving on indicates that there is a perfect square factor in the square root
that remains, and so we remove this factor from the square root before declaring a final answer:
7y 3
=
6x
42 xy 3
6x
=
y
42 xy
6x
There really is nothing else we can do to this final expression to simplify it further, so it must be
the required final answer.
David W. Sabo (2003)
Rationalizing the Denominator
Page 5 of 10
Example 6: Write the result of the following division in simplified form:
5st ÷
7t
solution:
We deal with division involving radicals in detail in a later document in this series. However, you
are already familiar with the way fractions express division, so that here we can write
5st ÷
7t =
5st
7t
The only possible simplification at this stage is to rationalize the denominator, which can be
accomplished by multiplying the numerator and denominator by
5st
7t
=
( 5st ) (
)
( 7t )( 7t )
=
7t
=
7t :
5s t 7t
7t
5s 7t
7
as the required simplest final result.
Example 7: Write the result of the following division in simplest form:
5st ÷ 7 t
solution:
This example is very similar to the previous Example 6, and so you should use it as a practice
problem. Attempt to work out the final answer yourself before looking at the steps of our solution,
which follows.
Here
5st ÷ 7 t =
5st
7 t
Now, we rationalize the denominator by multiplying the numerator and denominator by
t to
get
5st
7 t
=
David W. Sabo (2003)
( 5st ) (
7
t
)
( t )( t )
=
5s t t
7t
Rationalizing the Denominator
Page 6 of 10
=
5s t
7
as the required final answer in simplified form.
Example 8: The formula for the area, A, of a circle of radius r is given by
A = πr 2
Derive a formula for the radius, r, in terms of the area, A, and make sure that your result satisfies
the usual criteria of simplicity.
solution:
We deal systematically with the topic of rearranging formulas in another document on this
website. However, we can handle this problem without having to invoke a general strategy.
Since
A = πr 2
we can write
r2 =
A
π
by just dividing both sides of the original formula by π. Now we have a formula for r2. To get a
formula for r, we need to take the square root of both sides:
r =
A
r2 =
π
This formula looks quite simple already, but it does consist of the square root of a fraction, and so
the usual rules of simplification say we should rationalize the denominator. This is easily done:
A
π
=
A
π
=
(
(
A
π
)(
)(
π
π
)
)
=
Aπ
π
Thus, the final simplified formula for r is
r =
Aπ
π
David W. Sabo (2003)
.
Rationalizing the Denominator
Page 7 of 10
Example 9: One of the important discoveries in physics in the early twentieth century was that
electrons, tiny particles which carry negative electrical charge, also behave as if they were
“waves” with a wavelength λ given by
λ=
h
2mqV
where h is Planck’s constant, m is the mass of the electron, q is the electrical charge on the
electron, and V is the voltage through which the electron has been accelerated. Simplify this
formula for λ.
solution:
You may be thinking that something has gone wrong with your computer here – you thought you
were studying for a Math 11 test and suddenly here’s an example involving electrons and charges
and voltages and wavelengths and weird squiggles, and so on! (λ is a character in the Greek
alphabet. It is called “lambda,” and in fact amounts to the lower case “L” in the Greek alphabet.
Physicists conventionally use the symbol λ to represent the distance between successive crests
of wave motion – the “wavelength.”)
However, this example is not asking you to do any physics. Rather, it is asking you to simplify a
mathematical expression consisting of a fraction with a square root in its denominator. This we
can do in a flash without having to understand the physics behind the formula. Since
2mqV
cannot be simplified itself, it is the factor by which we need to multiply the numerator and
denominator of the formula for λ in order to rationalize the denominator and so achieve simplest
form.
λ=
h
2mqV
=
h
(
(
2mqV
2mqV
)(
)
2mqV
)
=
h 2mqV
2mqV
Thus, the simplified formula for λ is
λ=
h 2mqV
2mqV
This result may not appear to be simpler than the original expression. However, if this expression
for λ was to become part of a sum of several fractions, this “simplified” formula is more conducive
to finding the necessary common denominator to carry out such a sum.
Example 10: In stress analysis, the bending moment,
M, (a measure of its tendency to twist where it is
attached to its supports) of a beam of length L (in
metres) which is fixed at both ends is given by
M=
WL2
12
David W. Sabo (2003)
Beam
L
Rationalizing the Denominator
Page 8 of 10
where W is the mass in kilograms of a one metre length of the beam. Rearrange this formula to
give a formula for L, expressed in simplest form.
solution:
The sketch above to the right might help you understand a little of what this formula is
representing. The beam is supported at both ends. Its weight will tend to make it sag in between
the supports, which will be observed as a tendency to twist downwards (the dotted arrow in the
sketch) at its two supports. The value of M is a measure of this tendency to twist at the supports.
(You can see that the formula makes sense in general. The value of M will be bigger for a bigger
value of W, and we can well imaging that a heavier beam will have a greater tendency to twist in
this way. Similarly, a longer beam will tend to twist more than a shorter beam, and the formula
indicates that for larger values of L, we will get a larger value of M. Where the precise details of
the formula for M come from must be left to be explained by textbook dealing with the subject of
“statics” – something people in various civil and construction technologies must study.)
However, the problem here is about rearranging a formula, and we need not go into the physics
of the situation in order to be able to do that. In fact, this mathematical problem is very similar to
the one in Example 8 earlier. So, before looking at our solution which follows, try to solve this
problem on your own.
The first step in rearranging the given formula to obtain a formula for L is to isolate all references
to L on one side of the formula. This is easily done by multiplying both sides of the original
formula by
12
W
:
WL2
M=
12
⇒
12
WL2
12
Mi
=
i
=
W
12
W
( W L ) ( 12 )
( 12 ) ( W )
2
= L2
or
L2 =
12M
.
W
Now, since L2 is equal to 12M
L2 = L =
W
, their square roots must also be equal:
12M
W
Although this is a formula for L, it is not in simplest algebraic form because it amounts to a
fraction with a square root in the denominator. Thus, to finish, we need to rationalize the
denominator on the right-hand side.
L=
12M
W
=
12M
W
=
12M
W
W
W
=
12MW
W
In fact, since
12MW = 22•3•MW
we can do one more minor simplification:
David W. Sabo (2003)
Rationalizing the Denominator
Page 9 of 10
L=
12MW
=
W
22 i 3 i MW
=
W
22 3MW
W
=
2 3MW
W
This last expression cannot be simplified any further, so the required final result in simplest form
is
L=
2 3MW
W
David W. Sabo (2003)
Rationalizing the Denominator
Page 10 of 10