Adding (and Subtracting)
Square Roots
Just as with "regular"
numbers, square roots can be added together. But you might not be able
to simplify the addition all the way down to one number. Just as "you
can't add apples and oranges", so also you cannot combine "unlike"
radicals. To add radical terms together, they have to have the same radical
part.
- Simplify:

Since the radical is
the same in each term (namely, the square root of three), I can combine
the terms. I have two copies of the radical, added to another three
copies. This gives me five copies:
That middle step, with
the parentheses, shows the reasoning that justifies the final answer.
You probably won't ever need to "show" this step, but it's what
should be going through your mind.
- Simplify:

The radical part is the
same in each term, so I can do this addition. To help me keep track
that the first term means "one copy of the square root of three",
I'll insert the "understood" "1":
Don't assume that expressions
with unlike radicals cannot be simplified. It is possible that, after
simplifying the radicals,
the expression can indeed be simplified.
- Simplify:

To simplify a radical
addition, I must first see if I can simplify each radical term. In this
particular case, the square roots simplify "completely" (that
is, down to whole numbers):
- Simplify:
![3 sqrt[4] + 2 sqrt[4]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t4boHwLoQwggHzkuCG0bOu9nPT_T60eZhpZ9CzOlD-CowRsTMMWJrDqQCBt1b6wv_9688lQICuB_ZZb2uwRkhMK02dlDVN_xqkmXhrakre0SfCwUHvSjtn=s0-d)
I have three copies of
the radical, plus another two copies, giving me— Wait a minute! I can
simplify those radicals right down to whole numbers:
Don't worry if you don't
see a simplification right away. If I hadn't noticed until the end that
the radical simplified, my steps would have been different, but my final
answer would have been the same:
- Simplify:
![3 sqrt[3] + 2 sqrt[5] + sqrt[3]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t4fM2YyX39mGRdGEqVgVkAwcYHb54EiGBTKdLgLIzZfbpNQL4LrLK3L3l02uV1rQNIWVyPiZvTLLViis6sPQkv_d8GWY0h_UAfZzz1lTWYC8BNqtFO4YVp=s0-d)
I can only combine the
"like" radicals, so I'll end up with two terms in my answer:
There is not, to my knowledge,
any preferred ordering of terms in this sort of expression, so the expression
should
also be an acceptable answer.
- Simplify:
Copyright ©
Elizabeth Stapel 1999-2011 All Rights Reserved
I can simplify the radical
in the first term, and this will create "like" terms:
- Simplify:
![sqrt[18] - 2 sqrt[27] + 3 sqrt[3] - 6 sqrt[8]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s2PxW9-LKpXPhhhW9M8sld-jPdZ4mG4Bjk_VOVt1HEZZ0CfU7y5C7eyUQ_BsoMRSGfhugIKxnJVx_CCCMt1uoDcEqFVcEHfImCZWcFYIc9xVf817tW_1US=s0-d)
I can simplify most of
the radicals, and this will allow for at least a little simplification:
- Simplify:

These two terms have "unlike"
radical parts, and I can't take anything out of either radical. Then
I can't simplify
the expression
any further and my answer has to be:
(expression
is already fully simplified)
- Expand:
](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uxgwlVWeNzWbGVtQEwLR5kV9IUZfezH0JBtPMGAktr1Dov25EZ89QNDUwO5ET-rF_Kl8ezdtG9Zvtt-tQ39fp1LK0Tq2W2McfWzcu-YIuL09-MZkQBtsT0=s0-d)
To expand (that is, to
multiply out and simplify) this expression, I first need to take the
square root of two through the parentheses:
As you can see, the simplification
involved turning a product of radicals into one radical containing the
value of the product (being 2×3
= 6). You should expect
to need to manipulate radical products in both "directions".
- Expand:

- Expand:
![(1 + sqrt[2])(3 - sqrt[2])](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v3LwpDe7vbv3X83QtAGrN5yjx94WcV9WPixeCIE_CMhzVShZDHfE9sp3QVH0wDG7BpfZtZYCPjH15CaD514_PjZ4Q3lTVe86zjOSHIbx-L038IdFNeiQQ=s0-d)
It will probably be simpler
to do this multiplication "vertically".
Simplifying gives me:
![3 + 2sqrt[2] - 2 = 1 + 2 sqrt[2]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uV5fOGw0S4EkdwiNzWRDX0u5KllsZfwkvPh6QKQUNm2zfc8TT0wC-z190Dcpbj-piEO8I5F-3-LW8wP9LQMXZMnJ9nH2rCXnSLYgaHZjjesy8ozKQUVG-9=s0-d)
By doing the multiplication
vertically, I could better keep track of my steps. You should use whatever
multiplication method works best for you.
- Simplify

I do the multiplication:
Then I complete the calculations
by simplifying:
- Simplify:

I do the multiplication:
Then I simplify:
Note
in the last example above how I ended up with all whole numbers. (Okay,
technically they're integers, but the point is that the terms do not
include any radicals.) I
multiplied two radical "binomials" together and got an answer
that contained no radicals. You may also have noticed that the two "binomials"
were the same except for the sign in the middle: one had a "plus"
and the other had a "minus". This pair of factors, with the
second factor differing only in the one sign in the middle, is very important;
in fact, this "same except for the sign in the middle" second
factor has its own name:
Given the
radical expression
,
the "conjugate" is the expression
.
The conjugate (KAHN-juh-ghitt)
has the same numbers but the opposite sign in the middle. So not only
is
the conjugate of
,
but
is the conjugate of
.
When you multiply conjugates,
you are doing something similar to what happens with a difference
of squares:
a2
– b2 = (a + b)(a – b)
Copyright ©
Elizabeth Stapel 1999-2011 All Rights Reserved
When you multiply the factors
a
+ b and a
– b, the middle
"ab"
terms cancel out:
The same thing happens
when you multiply conjugates:
We will see shortly
why this matters. To get to that point, let's take a look at fractions
containing radicals in their denominators.
Dividing by Square
Roots
Just as you can swap between
the multiplication of radicals and a radical containing a multiplication,
so also you can swap between the division of roots and one root containing
a division.
- Simplify:
![sqrt[ 8 / 2 ]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v6C37NjoyeZtjlsj8PwwNrYRpRI_RMbVlmaquWH0jRljd94rV7ZuTz2xEewJQmzEqIFD63QXJrHCuf7cIMM9c5TRvgUVrmCanyki-ceZRuQOUQ8Xfx3YYJ=s0-d)
I can simplify this by
working inside, and then taking the square root:
...or else by splitting
the division into two radicals, simplifying, and cancelling:
- Simplify:
![sqrt[ 25 / 3 ]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tTLmhcclO_Y-fmaz2BsHkRELDa_cMoSkY4pKS-YAYqo0qPMDP2uOtfh5iN9Cmx1q5HXneSTuXBVMbFpQLEJ1i0e2Pohz0hxUCV3VIoIYHnYck54szKhIH8=s0-d)
This looks very similar
to the previous
exercise, but this is the "wrong" answer. Why? Because the denominator
contains a radical. The denominator must contain no radicals, or else
it's "wrong". (Why "wrong" in quotes? Because this
issue may matter to your instructor right now, but it probably won't later
on. It's like when you were in elementary school and improper fractions
were "wrong" and you had to convert everything to mixed numbers
instead. But now that you're in algebra, improper fractions are fine,
even preferred. Once you get to calculus or beyond, they won't be so uptight
about where the radicals are.)
To get the "right"
answer, I must "rationalize" the denominator. That is, I must
find some way to convert the fraction into a form where the denominator
has only "rational" (fractional or whole number) values. But
what can I do with that radical-three? I can't take the 3
out, because I don't have a pair of threes.
Thinking back to those
elementary-school fractions, you couldn't add them unless they had the
same denominators. To create these "common" denominators, you
would multiply, top and bottom, by whatever the denominator needed. Anything
divided by itself is just 1,
and multiplying by 1 doesn't
change the value of whatever you're multiplying by the 1.
But multiplying that "whatever" by a strategic form of 1 could
make the necessary computations possible, such as:
We can use the same technique
to rationalize radical denominators.
I could take a 3 out
of the denominator if I had two factors of 3
inside the radical. I can create this pair of 3's
by multiplying by another copy of root-three. If I multiply top and
bottom by root-three, then I will have multiplied the fraction by a
strategic form of 1.
I won't have changed the value, but simplification will now be possible:
This last form, "five,
root-three, divided by three", is the "right" answer they're
looking for.
- Simplify:
![(6 sqrt[2]) / sqrt[3]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tAORD4j0VqLP37rLYH-XxYqhD2tHFfiv81VseWNZfaBiYKdEG44kI4JXTWYQnSAzAM3LlnC_LlvQaIXe8qGnez-cGLJwIctT4Ba6_2qW-OCy5AH1-Vn6SF=s0-d)
Don't stop once you've
rationalized the denominator. As the above demonstrates, you should always
check to see if something remains to be simplified.
- Simplify:
![3 / (2 + sqrt[2])](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uCs1wLUQo8HDhf1-B9P9DoXCKbAFoMNRLAt4kUiHTQwcEfEiLbRRbcV-nZvGcWxdZAuX8rBkqq0WpGGobiXuuL9ecueIdmIyZXGz6xPRucaC0bcAduiWjU=s0-d)
This expression is in
the "wrong" form, due to the radical in the denominator. But
if I try to multiply through by root-two, I won't get anything useful:
Multiplying through by
another copy of the whole denominator won't help, either:
But look what happens
when I multiply by the same numbers, but with the opposite sign in the
middle: Copyright
© Elizabeth Stapel 1999-2011 All Rights Reserved
This multiplication made
the radical terms cancel out, which is exactly what I want. This "same
numbers but the opposite sign in the middle" thing is the "conjugate"
of the original expression. By using the conjugate, I can do the necessary
rationalization.
Do not try to reach inside
the numerator and rip out the 6
for "cancellation". The only thing that factors out of the numerator
is a 3,
but that won't cancel with the 2
in the denominator. Nothing cancels!
- Simplify:
![(1 + sqrt[7]) / (2 - sqrt[7])](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t2EV2HtuXqRHAz6MhAgGjwxezkq7qJfUStHiNZlPRCcC_2hTaBaJSX-a1QRr-XIBiiA8K7gS4yJtwaAtGUft-s7bwn3kLAPmpj4kdVm4M1nVjvZCs6WT77=s0-d)
I'll multiply by the
conjugate in order to "simplify" this expression. The denominator's
multiplication results in a whole number (okay, a negative, but the
point is that there aren't any radicals):
The numerator's multiplication
looks like this:
Then the simplified (rationalized)
form is:
It can be helpful to do
the multiplications separately, as shown above. Don't try to do too much
at once, and make sure to check for any simplifications when you're done
with the rationalization.
Operations with cube roots,
fourth roots, and other higher-index roots work similarly to square roots.
Simplifying Higher-Index
Terms
- Simplify

Just as I can pull from
a square (or second) root anything that I have two copies of, so also
I can pull from a fourth root anything I've got four of:
If you have a cube root,
you can take out any factor that occurs in threes; in a fourth root, take
out any factor that occurs in fours; in a fifth root, take out any factor
that occurs in fives; etc.
- Simplify the cube
root:

- Simplify the cube
root:

- Simplify:
Copyright ©
Elizabeth Stapel 1999-2011 All Rights Reserved
- Simplify:
![4 cbrt[27]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sKMXxar7KKZVP-E6QyjBJXyG02E0A-F26kqkYizFdbIMmGRCEQ4i-byfSJgUgYkLoxz20YUgK0E87w12hYZgGAxbwbrGOIW_Jz_2uoSK7obyyZaL3Fqb4=s0-d)
- Simplify:
![5th-rt[32 x^10 y^6 z^7]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uXISQDUbPY3NKJL6F-cWZR_G1u9OrXLadBrQxVZtWyceSNFVh_K5HRT0uXZ4Jc5RW50wQQTe6MQV2l3lLxEI4qabZ0Vbt_3PbAhKUSUvGTumyF7H2lNVj0=s0-d)
Multiplying Higher-Index
Roots
- Simplify:
![cbrt[9] cbrt[24]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vcGUaPDYZ7MRqg0f1AVua8OtgJG0tJKbwWS6UxmNALETmDHOD4YYeg7SEyVn6wNsBjpfPVDHa1bufsY7zdNf-vXNnHsatF7AQN96ix-I55hIyoH-o6HQgZ=s0-d)
- Simplify:
![4th-rt[75] (2 * 4th-rt[100])](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tNUA_2G1vhVQiMYcFHZveeuMzDPFKXcJPjumsJxpao7AXtBGjzU3vuBObHZGX6fQOxGV65TZUEo2gB74oCIf_0YTAvRDJuRALA8EPAXtgrUh608uX2UO8=s0-d)
Adding Higher-Index
Roots
- Simplify:
![cbrt[8] + cbrt[64]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v9Oz5IjXN2Cv8lqpZ18i217RFAXMaVnDRzMkodOD-szz1a4-dmfRk6AIbXv8HmI8sNB3i9JzKOvkvJno3adGeI_4hR95dlnfm6pYAJJrvzGoT2TK7lSOrZ=s0-d)
- Simplify:
![cbrt[81] + 5*cbrt[3]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tQmDrPwVEWdTtoUcjwZF-wHgDTgb__Sq5au1P--vuDK552BL4icRxvqUCVEbQB3qHcosFz6vdhTbxKbBW_v2ecY0hCKr25ZmN7kCA4hluCCwOWNEfLFLpK=s0-d)
Dividing Higher-Index
Roots
- Simplify:
![cbrt[ 5 / 27 ]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sfUHDs0GDygjJy3Y5zBkud19gnGofTnt6n_YDTzP2n0vynmTdtc2j2mTaUWbH-X0RsMlPRRZutiBfTr4xFfpR48zY5G8Go2wN-6X6r8KQDtVCvsJ0L9CMQ=s0-d)
- Simplify:
![cbrt[ 27 / 5 ]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_skJjAfHUo0c7TggYeyTgT8qP_M_qR5FETbvEociDs3u7pRczNwrbEay6nyA4LeJGD6nnXBq3f441lLVl6AhB4cxfy2qMDDNrW0j5ybuq5dovqn6FegWkx3=s0-d)
I can't simplify this
expression properly, because I can't simplify the radical in the denominator
down to whole numbers:
To rationalize a denominator
containing a square root, I needed two copies of whatever factors were
inside the radical. For a cube root, I'll need three copies. So that's
what I'll multiply onto this fraction:
- Simplify:
![4th-rt[ 5 / 72 ]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sdWPHT505xQQwD0Azmtlqtw2I970sGvFxGfkgkoCqCuA_BkBKMUtI-39EWH5AHOtme35rRotXCRlomyIjX6XsB_GsaxsRh1OcBpboQWcO7DsDLPzZAtmY=s0-d)
Since 72
= 8 × 9 = 2 × 2 × 2 × 3 ×3,
I won't have enough of any of the denominator's factors to get rid of
the radical. To simplify a fourth root, I would need four copies
of each factor. For this denominator's radical, I'll need two more 3s
and one more 2:
A Special Case of
Rationalizing
If your class has covered
the formulas for factoring the sums
and differences of cubes,
then you might encounter a special case of rationalizing denominators.
The reasoning and methodology are similar to the "difference of squares"
conjugate process for square roots.
- Simplify:
![2 / (1 + cbrt[4])](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vWFX0adB53C5pldCnd6syRPNg9Oy_hH6Ly6nJROeqhGBPrH4t8ehdE0dB576d8O-ChvswVv1OSO18zCztX2_l85OYYtfn9G3oDuOGdK8RzWNQErD4qp2o=s0-d)
I would like to get rid
of the cube root, but multiplying by the conjugate won't help much:
But I can "create"
a sum of cubes, just as using the conjugate allowed me to create a difference
of squares earlier. Using the fact that a3
+ b3 = (a + b)(a2
– ab + b2),
and letting a
= 1 and b
equal the cube root of 4,
I get:
If I multiply, top and
bottom, by the second factor in the sum-of-cubes formula, then the denominator
will simplify with no radicals:
Naturally, if the sign in
the middle of the original denominator had been a "minus", I'd
have applied the "difference of cubes" formula to do the rationalization.
This sort of "rationalize the denominator" exercise almost never
comes up. But if you see this in your homework, expect one of these on
your next test.
Radicals Expressed
With Exponents
Radicals can be expressed
as fractional exponents. Whatever is the index of the radical becomes
the denominator of the fractional power. For instance:
The second root became a
one-half power. A cube root would be a one-third power, a fourth root
would be a one-fourth power, and so forth.This conversion process will
matter a lot more once you get to calculus. For now, it allows you to
simplify some expressions that you might otherwise not have been able
to. Copyright
© Elizabeth Stapel 1999-2011 All Rights Reserved
- Express
as a single radical term.
I will convert the radicals
to exponential expressions, and then apply exponent
rules to combine
the factors:
- Simplify:
![cbrt[5] / sqrt[5]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t3MwiTr0du6CNY81MTZNZH2JxTwHtHaxG-bUxAaWjelTicnQ4k8IRrSd9V5x1KbItP4FKvA3b6A8wL99hTS7H-sZtszGGzIJUwnbHqjkCPTQDUm57DeEg=s0-d)
A Few Other Considerations
Usually, we cannot have
a negative inside a square root. (The exception
is for "imaginary" numbers. If you haven't done the number "i"
yet, then you haven't done imaginaries.) So, for instance,
is not possible. Do not try to say something like "
", because it's not true:
. You must have a positive inside the square root. This can be important
for defining and graphing functions.
- Find the domain of
the following:
The fact that I have
the expression x
– 2 inside a square
root requires that x
– 2 be zero or greater,
so I must have x
– 2 > 0.
Solving, I get:
On the other hand, you
CAN have a negative inside a cube root (or any other odd root). For instance:
...because (–2)3
= –8.
- Find the domain of
the following:
For
,
there is NO RESTRICTION on the value of x,
because x
– 2 is welcome to
be negative inside a cube root. Then the domain is: