This learner's module discusses about the topic Addition and Subtraction of Radical Expressions. It also teaches about how to add and subtract Radical Expressions. It also includes some problems to be solved involving radical operations.
1. (Effective Alternative Secondary Education)
MATHEMATICS II
MODULE 4
Radical Expressions
BUREAU OF SECONDARY EDUCATION
Department of Education
DepEd Complex, Meralco Avenue, Pasig City
Y
X
2. Module 4
Radical Expressions
What this module is about
This module is about addition and subtraction of radical expressions.
Adding and subtracting radical expressions is very much similar to adding and
subtracting similar terms of polynomials. Similar terms or like terms are those
with the same literal parts or literal factors. Radicals are similar if they have the
same index and the same radicands. For example: x3 and 5 x3 are similar
radicals whose index is 2 and whose radicand is 3x.
What you are expected to learn
This module is designed for you to:
1. add and subtract radical expressions
2. solve simple problems involving radical operations.
How much do you know
Simplify and perform the necessary operations:
1. 5 5 + 6 5
2. 5 x3 + x12
3.
2
1
3 +
3
1
3
4. 2 aa 1842 +
5. 7 bb 5β
6. 7 5 - 5 20
2
3. 7. 16 2 - 3 32
8. yxyx 791057310 β++
9. 5 20 - 8 5 + 3 45
10. x72 + 2 x8 - 5 x2
What will you do
Lesson 1
Addition of Radical Expressions
Radicals may be combined into simple radical expression by adding
similar terms. The sum of two radicals cannot be simplified if the radicals have
different indices or different radicands.
3 + 3
3 cannot be simplified (indices are different
5 + 2 cannot be simplified (radicands are different
It is important that the radicals are written in simplest form before adding
and subtracting.
3 + 27 can further be simplified. 27 = 3 3
= 3 + 3 3 add
= 4 3
Examples:
Find the sum of the following radical expressions:
1. 4 + 5 2 + 6 2
Solution: Add coefficients and annex their common radical factor
= (4 + 5 + 6) 2
3
4. = 15 2
2. 6 2 + 3 5 + 2 2 + 5 5
Solution:
= (6 2 + 2 2 ) + (3 5 + 6 5 ) Group similar radicals.
= 8 2 + 9 5 Add the coefficients of similar radicals.
You cannot add the coefficients of different radicand.
3. 6 27 + 2 75
Solution:
= 6 39 β’ + 2 325 β’ Factor each radicand
= 6(3 3 ) + 2(5 3 ) Simplify
= 18 3 + 10 3 Add the coefficients and annex
their common radical factor
= 28 3
4. 33
10832 +
Solution:
= 33
42748 β’+β’ Factor each radicand such that one
factor is a perfect cube.
= 2 33
434 +
= (2 + 3)3
4
= 53
4
5.
25
3
3
25
3
4 +
Solution:
=
25
3
3
25
3
4 + Split the numerator and denominator
4
5. = 3
5
3
3
5
4
+ Simplify and add the coefficients
= 3
5
7
6.
2
3
3
2
+
Solution:
2
3
3
2
+
=
2
2
2
3
3
3
3
2
β +β Split the numerator and denominator
and rationalize the denominator.
=
4
6
9
6
+ Simplify
=
2
6
3
6
+ Get the LCD. LCD = 6
=
6
6362 +
Add
=
6
65
7. 37
9 aa +
Solution:
= aaaa β’+β’ 26
9 Split a7
such that the exponent is exactly
divisible by the index, 2.
= 3a2
aaa + Simplify
= (3a2
+ a) a
5
6. Keep in mind that our basic approach to these problems has been to first
put each term into simplest radical form before adding and /or subtracting.
Try this out
Perform the indicated operations.
A. 1. 4 2 + 5 2
2. 5 3 + 2 3 + 10 3
3. 16 7 + 8 7 + 7
4. 15 11 + 3 11 + 5 11 + 6 11
5. 13
3
2
13
2
1
+
B. 6. 2 27 + 3 16
7. 180 + 125 + 10 45
8. 3 a36 + a100
9. 75 + 2 27 + 12
10. 5 b + b4
C. 11. 80 + 3 45 + 20
12. x12 + x27 + x48
13.
4
3
+ 2
4
3
14. 22
22
5
4
5
yx
yx
+
15. 2
3
2
+ 5
2
3
D. Whatβs the Message?
6
7. Simplify the expression and find the correct answer in the box, then fill in
the small box with the corresponding letter. Once you have filled in all the small
boxes with the appropriate letter, darken all the blank spaces to separate the
words in the secret message. Whatβs the message? Have fun!
1. E : 7437 ++ 9. U: 48122 +
2. N : 5032218 ++ 10. H: 2 33
6328 yy +
3. T: 3 1659281 ++ 11. O: 3
16
5
5
16
5
+
4. S :6 1010 + 12. L: 4 33
1802320 aa +
5. W: 3 xxx 934225 ++ 13. R: 55
9944 xx +
6. I: 25144100 ++ 14. C: 2 16327 +
7. Q: 8 3212 + 15. F:
49
64
8. Y: 122 + 16. V: 3 3381472200 ++
7
9. Solution:
3235 β
= (5 β 2) 3 Group their coefficients and subtract.
= 33
2. 96354 β
Solution:
96354 β
= 616369 β ββ Factor each radical in such a way that
one factor is a perfect square.
= ( )64363 β Simplify
= 61263 β Subtract
= 69β
3. 22
925 xx β
Solution:
22
925 xx β
= 5x β 3x
= 2x
4. 2323
1227 yxyx β
Solution:
2323
1227 yxyx β
= 2222
3439 yxxyxx β β β β ββ β β β Factor the radicals in such a way that
one factor is a perfect square.
= xxyxxy 3233 β 9, 4, x2
and y2
are perfect
squares.
= xxyxy 3)23( β Group the coefficients and subtract
9
10. = 1xy x3 or xy x3
5. 2 33
54516 β
Solution:
2 33
54516 β Factor the radicand in such a way
that
one factor is a perfect cube.
= 2 33
227528 β ββ 8 and 27 are perfect cubes hence, their
cube root is 2 and 3 respectively.
= 2 )23(5)22( 33
β Simplify
= 33
21524 β
= -113
2
6. 3 453 2
82 yxyx β
Rename the expression such that the radicand can be expressed as a
perfect cube and the exponents are exactly divisible by the index if
possible then simplify.
Solution:
3 453 2
82 yxyx β
= 3 3233 2
82 yyxxyx β 8, x3
and y3
are perfect cubes
= 3 23 2
22 yxxyyx β
= (2 - 2xy) 3 2
yx
Subtraction of radical expressions with fraction.
7.
9
3
32
2
a
a β
Solution:
10
11. 9
3
32
2
a
a β
=
9
3
32
2
a
a β Split the numerator and denominator
and simplify.
=
3
3
32
a
a β or
= 3
3
32
a
a β
= 3
3
2 ο£·
ο£Έ
ο£Ά

ο£

β
a
a Group the coefficients and simplify.
= 3
3
6
ο£·
ο£Έ
ο£Ά

ο£
 β aa
= 3
3
5a
Addition and Subtraction of Radical Expressions
With the above knowledge, you can now perform a combination of operations.
1. 25322635 +β+
Solution:
25322635 +β+
= 25263235 ++β Group similar terms and perform the
necessary operations.
= 21133 +
Addition of dissimilar radicands is not allowed.
2. yxyxyx 222
50818 +β
Solution:
yxyxyx 222
50818 +β
= yxyxyx 222
2252429 β +β ββ Factor in such a way that one factor is a
11
12. perfect square.
= yxyxyx 252223 +β Simplify
= (3x β 2x + 5x) y2 Perform the necessary operations.
= 6x y2
3. 7
9a - 3
a
Solution:
7
9a - 3
a
= aa6
9 - aa2
= 3a3
a - a a
= (3a3
β a) a
4. 4 223 44333
488 babaabba ββ+
In this problem, no two radicands are identical and the radical expressions
have indices 2, 3 and 4. The radical expression 4 22
4 ba can be converted into a
radical expression with index 2.
Solution:
4 22
4 ba = 4 2
)2( ab
Note that 4 2
)2( ab is the same as ( )4
2
2ab or ( )2
1
2ab = ab2 . You can
now complete the solution.
4 223 44333
488 babaabba ββ+
Since 4 22
4 ba = ab2 then,
abbaabba 288 3 44333
ββ+
= abbbaaabbbaa 2824 3 33322
ββ β β β+β β β β β Factor and simplify.
= abababababab 2222 33
ββ+
= 33
2222 abababababab β+β Group similar radicands.
12
13. = 3
)21(2)12( abababab β+β By addition
Try This Out
Perform the indicated operations.
A. 1. 3 - 2 3
2. 5 x7 - 4 x7
3. 3 a6 - 7 a6
4. 4 x - 3 x - 2 x
5. 12 - 5 8 - 7 20
B. 6. 36 - 100
7. 6 18 - 32 + 2 50
8. -2 63 + 2 28 + 2 7
9. 5 8 + 3 72 - 3 50
10. - 27 + 2 48 - 75
C. 11. 3 2
72m - 5 2
32m + 3 2
18m
12. - 33
16254 β
13. 2 33
8227 xx β
14. 5 444
218162332 +β
15. 3 44 5
2 xyxyx β
D. Mental Math
13
14. Why
are
Oysters greedy?
Perform and Simplify the following radicals. Write the letter in the box above its
correct answer. Keep working and you will discover the answer the questions.
H. 5 - 18 5 + 4 5 - 5 5 E. 4 7 + 7
R. 35 - 6 5 I. 7 10 + 90
T.
3
1
27 - 3 3 F. 3 2 - 2 32
S. 2 27 - 3 48 + 75
5
1
Y. 4 27 + 5 7 - 2 48
L. 33
54316 + A.
4
5
2
20
+
Let βs Summarize
14
-2β3 -18β5 5β7 4β3+5β7
3β5
2 3β35 - 6β5 5β7
_
-5β3
_
-18β5
_
5β7 113
β2 113
β2 -5β2 10β10 -5β3 -18β5
15. When working with radicals, remember the following:
1. Put each radical into simplest form.
Sums and difference of radical expressions can be simplified by applying the
basic properties of real numbers.
The sum and difference of two radical expressions cannot be simplified if the
radicals have different indices and different radicands.
2. Perform any indicated operations, if possible.
3. Make sure the final answer is also in simplest radical
form.
What have you learned
A. Simplify each radical and perform the indicated operations.
1. 2 3 + 4 3
2. 5 5 - 5
3. 6 3 + 2 3 - 3 3
4. 2 7 - 5 7 -8 7
5. x 2 β 3x 2 + 5 2
6. 5 a3 + 5 a3 - 9 a3
7. 3 2 - 2 32
8. 12 + 5 8 - 7 20
9. 6 2 - 5 8 + 2 32
10.7 72 + 3 20 - 4 5
11. 3 +
3
1
2
15
19. The Message: SUCCESS IS ONLY FOR THOSE WHO NEVER QUIT
Lesson 2
Try this out
A.
1. - 3
2. x7
3. -4 a6
19
_
18β3
Q
_
8β3
U
27
I
53
T
__
7β10
S
_
8β3
U
_
12 + 6β3
C
_
12+6β3
+β3
C
_ _
5β7+β3
E
__
7β10
S
__
7β10
S
27
I
_
2β5
O
_
16β2
N
__
44aβ5a
L
_ _
β2 +2 β3
Y
8
7
F
_
2β5
O
___
5x2β
11x
R
53
T
__
7yβ7y
H
_
2β5
O
__
7β10
S
_ _
5β7 + β3
E
_
28βx
W
__
7yβ7y
H
_
2β5
O
_
16β2
N
_ _
5β7+β3
E
_ _
43β2+ 14β3
V
_ _
5β7+β3
E
___
5x2β
11x
R
__
7β10
S
20. 4. - x
5. 2 3 - 10 2 - 14 5
B.
6. -4
7. 24 2
8. 0
9. 13 2
10. 0
C.
11. 7m 2
11. -73
2
12. 2 3
x
13.19 4
2
15. x 4 xy
D. Mental Health
H. -18 5 E. 5 7
R. 35 - 6 5 I. 10 10
T. -2 3 F. -53
2
S. -5 3 Y. 4 3 + 5 7
L. 113
2 A.
2
53
20
21. Why are Oysters greedy?
What have you learned
1. 6 3
2. 4 5
3. 5 3
4. -11 7
5. -2x + 5 2
6. a3
7. -5 2
8. 2 3 + 10 2 - 14 5
9. 4 2
10. 42 2 + 2 5
11.
3
35
12.
3
211
13. 30 2
21
T
_
-2β3
H
_
-18β5
E
_
5β7
Y
_ _
4β3+5β7
A
3β5
2
R
__ _
3β35 - 6β5
E
_
5β7
S
_
-5β3
H
_
-18β5
E
_
5β7
L
_
113
β2
L
_
113
β2
F
_
-5β2
I
_
10β10
S
_
-5β3
H
_
-18β5