4.2: Operations on Radicals and Rational Exponents In this section, we will move from operations on polynomials to operations on radical expressions, including adding, subtracting, multiplying and dividing radical expressions.

In the previous section, we added and subtracted polynomials by combining like terms. In this section, we extend that idea to radicals. Combining like terms can take on many forms. For example,

2𝑥𝑥 + 3𝑥𝑥 = 5𝑥𝑥

2𝑥𝑥2 + 3𝑥𝑥2 = 5𝑥𝑥2

2 apples + 3 apples = 5 apples

27

+ 37

= 57

(2 sevenths + 3 sevenths = 5 sevenths)

2√7 + 3√7 = 5√7

2𝑥𝑥12 + 3𝑥𝑥

12 = 5𝑥𝑥

12

In each of these examples, the object we are adding does not change after we add it. That is, when we add 2𝑥𝑥 to 3𝑥𝑥 we get 5𝑥𝑥, not 5𝑥𝑥2. (Recall that the exponents change when we are multiplying the base, but not when we are adding.) Similarly, when we add like terms with radicals, the radicand (the number inside the radical) stays the same, so we get 2√7 + 3√7 = 5√7, not 5×7 and not 5√14. This makes sense, because √7 can also be written as 7

12, and when we add, the exponents do

not change: 2 ∙ 712 + 3 ∙ 7

12 = 5 ∙ 7

12. We see this again in the final example, 2𝑥𝑥

12 + 3𝑥𝑥

12 ,

where the exponent of 12 is the same as the square root – so 2𝑥𝑥

12 + 3𝑥𝑥

12 is the same as

2√𝑥𝑥 + 3√𝑥𝑥 , giving us 5√𝑥𝑥 = 5𝑥𝑥12. Whether we write 𝑥𝑥

12 or √𝑥𝑥, this part is the “like

term” that stays the same when we add. Similar to like terms in polynomials, like terms in radicals have two things in common: the same index and the same radicand. This means that we can only add or subtract radical terms together if the root is the same (square roots to square roots, cube roots to cube roots, etc.), and if the number on the inside of the radical is the same as well. EXAMPLES – Add or subtract. Assume all variables are positive. a. 5√6 − 6√5 + √6 + 2√5

See if you can figure out the answer before turning the page!

STEP 1: bring together like terms.

�5√6 + √6� + (−6√5 + 2√5) We can combine the two numbers that have √6 as the radicand and we can combine the two numbers that have √5 as the radicand.

6√6 − 4√5 STEP 2: add the coefficients (the numbers in front of the roots. Remember that √𝟔𝟔 is the same as 𝟏𝟏√𝟔𝟔.

We are finished because √6 and √5 are not like terms.

b. 3𝑎𝑎√53 − 7𝑎𝑎√53 + 8𝑎𝑎√53

STEP 1: bring together like terms. Since (3𝑎𝑎 − 7𝑎𝑎 + 8𝑎𝑎)√53 all three have the same root and radicand,

we can combine all three. 4𝑎𝑎√53 STEP 2: add all three coefficients.

c. 3𝑥𝑥12 − 5√𝑥𝑥 − 2�𝑦𝑦 + 8𝑦𝑦

12 STEP 1: remember that √𝑥𝑥 is the same as 𝑥𝑥

12.

This means that 3𝑥𝑥12 and −5√𝑥𝑥 are like terms.

Similarly, −2�𝑦𝑦 and 8𝑦𝑦12 are like terms.

−2√𝑥𝑥 + 6�𝑦𝑦

or, −2𝑥𝑥12 + 6𝑦𝑦

12

d. √8 + √32 − √18 STEP 1: bring together like terms.

But it looks like we don’t have any like terms!

TRY SIMPLIFYING FIRST! That’s right! If we cannot add or subtract as is, first try simplifying each radical,

then look for like terms again. Let’s break this up into 3 parts: √8 = ? √32 = ? −√18 = ?

See if you remember how to simplify these before checking on the next page.

Then why would this problem even be here?

Example d., continued √8 = √4 • 2 = √4 • √2 √32 = √16 • 2 = √16 • √2

= 2√2 = 4√2 −√18 = −√9 • 2 = −√9 • √2 = −3√2

Now, notice that all of these are like terms! So we can combine them into a single expression, and now we have

√8 + √32 − √18 = 2√2 + 4√2 − 3√2 = 3√2

e. √48 − 3√27 + 2√75

Again, we begin by simplifying each term in the expression, as there are no like terms yet.

√48 = √16 • 3 = √16 • √3 = 4√3

−3√27 = −3•√9 • 3 = −3•√9 • √3 = −3 • 3 •√3 = −9√3

2√75 = 2 • √25 • 3 = 2 • √25 • √3 = 2 • 5 • √3 = 10√3

Now we have √48 − 3√27 + 2√75 = 4√3 − 9√3 + 10√3 = 5√3

f. 3𝑥𝑥√45 − 8√5𝑥𝑥2 + √20𝑥𝑥2 + 7𝑥𝑥

The last term does not need to be simplified. Simplifying each of the other terms gives us:

3𝑥𝑥√45 = 3𝑥𝑥 • √9 • 5 = 3𝑥𝑥 • √9 • √5=3𝑥𝑥 • 3 • √5 = 9𝑥𝑥√5

−8√5𝑥𝑥2 = −8 • √5 • x2 = −8 • √5 • √𝑥𝑥2=−8 • √5 • 𝑥𝑥 = −8𝑥𝑥√5

√20𝑥𝑥2 = √20 • 𝑥𝑥2=√4 • 5 • 𝑥𝑥2 = √4 • √5 • √𝑥𝑥2 = 2 • √5 • 𝑥𝑥 = 2𝑥𝑥√5

Now we have: 3𝑥𝑥√45 − 8√5𝑥𝑥2 + √20𝑥𝑥2 + 7𝑥𝑥 = 9𝑥𝑥√5 − 8𝑥𝑥√5 + 2𝑥𝑥√5 + 7𝑥𝑥

= 3𝑥𝑥√5 + 7𝑥𝑥

Notice that in the end, the 7x is not a like term, since it does not have everything that the other terms have – it does not have 𝑥𝑥√5, it only has x.

g. √813 − 4√243 Before you turn the page, see if you can remember how to simplify cube roots, which is very similar to simplifying square roots – break up each cube root into what can be cube-rooted and what cannot.

Simplifying  each  term  gives  us:  

81! =   27! 3! = 3 3!

4 24! = 4   8! 3! = 4×2 3! = 8 3!

So  that,   81! −  4 24! =  3 3! −  8 3! =  −5 3!

!!

!!

h. 7! 2!!!! +  6!"! 16!

7! 2!!!! = 7  ! 2  ! !! ! !!   = 7  ! 2  ! ! = 7! 2  !!!!!! = 7! 2  !   !!  ! = 7!" 2!!

6!" 16!! = 6!"   8!   2!  !!!   = 6!"  ×2× 2!   !! = 12!" 2!!

So  that,  7 2!!!!! −  6!" 16!! =  7!" 2!! +  12!" 2!! =  19!" 2!!

Property:   ! ! (! !)  =  !" !"

In  other  words,  when  we  multiply  two  radicals  together  with  the  same  index,  we  multiply  numbers  on  the  outside  of  the  radical  together,  and  then  numbers  on  the  inside  of  the  

radical    together.  

EXAMPLES  

a.   3 5 (2 7  )  

         3  ×2  ×   5×7 = 6 35  

b.   3 2! 6 4!

       18 8!  

Multiply  numbers  on  the  outside  of  the  radical, (3  ×  2)  together,  and  numbers  on   the  inside  (5  ×  7)  together.  Since we cannot simplify the square root of 35, we stop.

Multiplying numbers on the outside of the radical gives us 18 (3 × 6), and numbers on inside  together  gives  us  8  (  2  ×  4)

Notice  that  we  CAN  simplify  the  radical  in  this  case,   8!  =  2.    So,

18 8! = 18  ×  2 = 36

c. 2 5 3 + 4 2

2  ×5 3 + ( 2  ×4 2)  =    5 6 +  4 4

In   this  case,  we  are  multiplying  a  monomial    by  a  binomial,  so  we  distribute   2  to  each    of  the  other  terms.      

5 6 +  4 4    =    5 6  + 4 2 =  5 6  + 8   Note  that  you  can  simplify  4 4 = 4 2 =  8  

Multiplying Radicals

d.   ! + 5 ! − 3 In  this  example,  we  are  multiplying  a  binomial  by  a  binomial, so  we  can use"FOIL" or the distributive property.  

FIRST:     !  × !    =   !!  

OUTER:     !  ×  −3 =  −3 !  

INNER:      5  × !  =  5 !  

LAST:    5  ×  −3 =  −15  

So,  initially,  we  get:     !! −  3 ! +  5 !  − 15.    We  can  simplify  because   !! = !,  then  we  have:                        x    −  3 ! +  5 !  − 15,  and  we  have  like  terms  (−  3 ! +  5 ! = 2 !  ),  

so  the  answer  is:                    x  +  2 !   − 15.  

e.   7 − 3 3 2 7 − 4 3 Again,  in  this  example,  because  we  have  a    binomial  times  a  binomial,  we  can  use "FOIL"  or the distributive property.  

FIRST:   7  ×2 7 = 2 49  

OUTER:   7  ×4 3 = 4 21  

INNER:    −3 3  ×2 7 =  −6 21  

LAST:    −3 3  ×−4 3 = 12 9    

Initially,  we  get  2 49  +  4 21    −6 21  +  12 9.      

We  can  simplify  2 49 = 2×7 = 14,  and  12 9 = 12×3 = 36.  

So,  we  have    14 +  4 21    −6 21 +  36  .    However,  note  that  we  do  have  like  terms,  because  4 21    −6 21 =  −2 21.    14  +  36  are  like  terms  as  well,  and  gives  us  50.    So  the  answer  is:  50 − 2 21  

f.   5 ! − 2 !!

Recall  when  we  are  squaring  something  first  write  it  in  expanded  form,  so  that:  

5 ! − 2 !!=   5 ! − 2 ! 5 ! − 2 ! Once  again,  we  have  a  binomial  times  a    

binomial,  so  we  can  FOIL  this  as  in  the  other  examples,  and  we  get:  

FIRST:    5 !×  5 ! = 25 !!  

OUTER:  5 !  ×−2 ! = −10 !"  

INNER:    −2 !  ×5 ! =  −10 !"  

LAST:    −2 !×  −2 ! = 4 !!  

Initially,  we  have  25 !! −  10 !" −  10 !" +  4 !!,  

!"#$%"&'"()  !"  !"#:                25! − 20 !" +  4!  

Dividing  Radicals:  

Dividing Radicals

A radical is not considered to be in standard form if there are any radicals in the denominator. Therefore, the main point in dividing radical expressions is going to be to get rid of the radical in the denominator. This process of getting rid of radicals in the denominator is also called RATIONALIZING. We call it rationalizing because we want to turn the denominator into a rational number.

For example, the expression 20√8

has a radical in the denominator, so the denominator is

irrational. On the calculator, √8 = 2.82842712 …. This is a decimal that goes on forever. It can be difficult to divide by a decimal that goes on forever without repeats, so it may be difficult to do 20

√8= 20 ÷ √8. In the days before calculators, it was much

worse!

😱😱 So instead of dividing right away, we try to first see if we can get rid of the radical in the denominator: What can we multiply the fraction 20

√8 by, that will get rid of the radical in the

denominator? There are several possibilities.

Option 1: multiply the numerator and the denominator of the fraction by √2. Why does that work? Because then we have a number that can be square-rooted in the denominator. And multiplying by √2

√2 is okay because √2

√2= 1.

20√8

∙√2√2

=20√2√8√2

=20√2√16

=20√2

4

Now we have a rational number in the denominator, 2. We also now have a fraction that can be simplified! 20√2

4= 5√2. Thus, we were able to accomplish our division

without using a calculator at all!

Option 2: multiply the numerator and the denominator of the fraction by √8. This also works to get a number that can be square-rooted in the denominator.

Nooooooooo!

20√8

∙√8√8

=20√8√8√8

=20√8√64

=20 √4•2

8 =20 • 2√2

8 =40√2

8 = 5√2

Of course, we get the same result using either option!

How do you know what will work? Tip: your goal is to get a number that can be square rooted in the denominator. Since the square root is the same as the ½ power, you want an even exponent in the denominator, one that can be cut in half.

√8 = √23 = (23)12

You cannot cut 3 in half evenly.

√16 = �24 = (24)12 = 22 = 4

But you can cut 4 in half evenly.

√64 = �26 = (26)12 = 23 = 8

And you can cut 3 in half evenly.

Division (Rationalizing) Examples a. Divide (Rationalize): 14

√32.

Again, there is more than one option. Option 1: Since 32 = 25, it is enough to multiply numerator and denominator by one more radical 2, to get 26, an even power:

14√32

•√2√2

=14 • √2√32 • √2

=14√2√64

=14√2

8 =7√2

4

This works because √64 = √26 = (26)12 = 23 = 8.

Even though we were not able to completely get rid of the denominator, we do have a “nice,” rational number in the denominator.

Option 2: multiply numerator and denominator by √32. Doubling the square root will always get us an even power!

14√32

•√32√32

=14 • √32√32 • √32

=14√16 • 2√1024

=14 • 4√2

32 =7√2

4

This works because √1024 = √210 = (210)12 = 25 = 32.

You may have learned in the past that you can always rationalize by multiplying the numerator and denominator by the denominator. This is true for square roots, but

sometimes results in very large numbers, as we saw in option 2. And it does not work for cube roots!

b. Divide (Rationalize): 10√23

Now we have a cube root, or 1/3 power. That means our new cube root in the denominator needs to have an exponent that is divisible by 3. For example, if we had 23 in the root in the denominator, that would work. We currently only have 21 in the root in the denominator. So we need two more 2’s! Since 22 = 4, we have:

10 • √43

√23 • √43 =10 • √43

√83 =10 • √43

2 = 5√43

😧😧 c. Divide (Rationalize): 30

√253

Again, we have a cube root, or 1/3 power. That means our new root in the denominator needs to have an exponent that is divisible by 3. But 25 is 52. We would love to have 53 in the root in the denominator instead! So, multiply by one more 5.

30 • √53

√253 • √53 =30 • √53

√1253 =30 • √53

5 = 6√53

d. Divide (Rationalize): 7√274

Now we have a fourth root, or 1/4 power. That means our new root in the denominator needs to have an exponent that is divisible by 4. But 27 is 33. We would love to have 34 in the root in the denominator instead! So, multiply by one more 3.

7 • √34

√274 • √34 =7 • √34

√814 =7 • √34

√344 =7 • √34

3

e. Divide (Rationalize): 12√275

Now we have a fifth root, or 1/5 power. That means our new root in the denominator needs to have an exponent that is divisible by 5. But 27 is 33. We would love to have 35 in the root in the denominator instead! So, multiply by two more 3’s!

12 • √95

√275 • √95 =12 • √95

√2435 =12 • √95

√355 =12 • √95

3 = 4√95

There are two many twos, err, too many toos! Too many twos! Please do an example that doesn’t have twos!

What if there is a variable involved? The same principles apply!

f. Divide (Rationalize): 10√5𝑥𝑥3

Again, we want to have an even power of 5 and an even power of x. If we have one more 5 and one more x, we will have an even exponent for both:

10√5𝑥𝑥3

•√5𝑥𝑥√5𝑥𝑥

=10√5𝑥𝑥√25𝑥𝑥4

=10√5𝑥𝑥

5𝑥𝑥2

g. Divide (Rationalize): 11√16𝑥𝑥25

Now we have a fifth root, or 1/5 power. That means our new root in the denominator needs to have an exponent that is divisible by 5. But 16 is 24. We need one more 2. We also need three more x’s!

11 • √25

√16𝑥𝑥25 • √2𝑥𝑥35 =11 • √25

√32𝑥𝑥55 =11 • √25

√25𝑥𝑥55 =11 • √25

2𝑥𝑥

What  if  we  have  2  terms  in  the  denominator,  as  in  the  example  below?  

23 + 1

Well,  if  we  have  an  expression  with two terms in  the  denominator,  then  we  multiply  both  the  numerator  and  the  denominator  by  what  we  call  the  CONJUGATE.  

Definition:  

Conjugate:    The  conjugate  of  any  binomial  term  is  found  by  negating  the  second  term  in  the  binomial.    The  conjugate  of  a  +  b  is  a  –  b.    The  conjugate  of  –x  –  y    is  –x  +  y.  

Let’s  look  at  some  examples:  

 EXAMPLES:    Rationalize  (divide)  the  following  completely:  

h.  !!!!

We  first  have  to  find  the  conjugate  of  the  

denominator.    The  conjugate  of   3 + 1  is  3 − 1.    We  must  multiply  both  the  

numerator  and  the  denominator  of  the  fraction  by  the  conjugate  

!!!!

 × !!!!!!

Now,  we  distribute  the  conjugate  into  the  

the  numerator  and  the  denominator.  

! !!!!! !! !!!

We  can  now  simplify  the  denominator.    

Notice  that  we  can  rewrite   9  !"  3,  and  that  − 3  !"# 3  cancel  out!    So,  we  are  left  with:

! !!!!!!

 =  ! !!!!

Note  that  there  is  NO  radical  left  in  the  

denominator!    This  is  what  we  wanted!    However,  note  as  well  that  we  can  simplify  these  fractions  a  little  further,  each  term  in  the  numerator  and  denominator  share  a  factor  of  2,  so  we  can  simplify  the  fractions  as  below:  

! !!!!

=   ! !!−  !

!=   3 − 1

i.  !

!!   !Again,  we  first  have  to  determine  what  the  

conjugate  of   ! − !  is,  as  before,  the  conjugate  is  just  the  change  in  the  middle  sign,  so  the  conjugate  of   ! − !  !"   ! + !.    Therefore,  we  multiply  the  numerator  and  the  denominator  by   ! + !.  

!!!   !

× !! !!! !

Multiplying  these  together  gives  us:  

!!!   !

× !! !!! !

= !!! !"!!! !"! !"! !!

=    !+   !"!−!  

We  distribute  the  2  into  each  term  in  the  numerator  and  we  FOIL  the  

denominator  

j.  !!!!!!

We  begin  by  finding  the  conjugate  of  the  

denominator.    The  conjugate  of   ! − 3  !"  

! + 3.    Multiplying  the  numerator  anddenominator  by  the  conjugate  give  us:  

!+3!−3  ×  

!!!!!!

=   !!!  ! !!! !!!!!!! !!! !!!

=   !!! !!!!!!