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Mathematics LibreTexts

7.3: Adding and Subtracting Radical Expressions

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Learning Objectives

  • Add and subtract like radicals.
  • Simplify radical expressions involving like radicals.

Adding and Subtracting Like Radicals

Adding and subtracting radical expressions is similar to adding and subtracting like terms. Radicals are considered to be like radicals16, or similar radicals17, when they share the same index and radicand. For example, the terms 26 and 56 contain like radicals and can be added using the distributive property as follows:

26+56=(2+5)6=76

Typically, we do not show the step involving the distributive property and simply write,

26+56=76

When adding terms with like radicals, add only the coefficients; the radical part remains the same.

Example 7.3.1:

Add: 735+335.

Solution

The terms are like radicals; therefore, add the coefficients.

735+335=1035

Answer:

1035

Subtraction is performed in a similar manner.

Example 7.3.2:

Subtract: 410510.

Solution

410510=(45)10=110=10

Answer:

10

If the radicand and the index are not exactly the same, then the radicals are not similar and we cannot combine them.

Example 7.3.3:

Simplify: 105+629572.

Solution

105+629572=10595+6272=52

We cannot simplify any further because 5 and 2 are not like radicals; the radicands are not the same.

Answer:

52

Caution: It is important to point out that 5252. We can verify this by calculating the value of each side with a calculator.

520.8252=31.73

In general, note that na±nbna±b.

Example 7.3.4:

Simplify: 5310+310310210.

Solution

5310+310310210=5310310+310210=4310+10

We cannot simplify any further, because 310 and 10 are not like radicals; the indices are not the same.

Answer:

4310+10

Adding and Subtracting Radical Expressions

Often, we will have to simplify before we can identify the like radicals within the terms.

Example 7.3.5:

Subtract: 3218+50.

Solution

At first glance, the radicals do not appear to be similar. However, after simplifying completely, we will see that we can combine them.

3218+50=16292+252=4232+52=62

Answer:

62

Example 7.3.6:

Simplify: 3108+324332381.

Solution:

Begin by looking for perfect cube factors of each radicand.

3108+324332381=3274+3833843273Simplify.=334+233234333Combineliketerms.=3433

Answer:

3433

Exercise 7.3.1

Simplify: 20+2735212.

Answer

53

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Next, we work with radical expressions involving variables. In this section, assume all radicands containing variable expressions are nonnegative.

Example 7.3.7:

Simplify: 935x32x+1035x.

Solution

Combine like radicals.

935x32x+1035x=935x+1035x32x=35x32x

We cannot combine any further because the remaining radical expressions do not share the same radicand; they are not like radicals. Note: 35x32x35x2x.

Answer:

35x32x

We will often find the need to subtract a radical expression with multiple terms. If this is the case, remember to apply the distributive property before combining like terms.

Example 7.3.8:

Simplify: (5x4y)(4x7y).

Solution

(5x4y)(4x7y)=5x4y4x+7yDistribute.=5x4x4y+7y=x+3y

Answer:

x+3y

Until we simplify, it is often unclear which terms involving radicals are similar. The general steps for simplifying radical expressions are outlined in the following example.

Example 7.3.9:

Simplify: 533x4+324x3(x324x+433x3).

Solution

Step 1: Simplify the radical expression. In this case, distribute and then simplify each term that involves a radical.

533x4+324x3(x324x+433x3)=533x4+324x3x324x433x3=533xx3+383x3x383x433x3=5x33x+2x332x33x4x33

Step 2: Combine all like radicals. Remember to add only the coefficients; the variable parts remain the same.

=5x33x+2x332x33x4x33=3x33x2x33

Answer: 3x33x2x33

Example 7.3.10:

Simplify: 2a125a2ba280b+420a4b.

Solution

2a125a2ba280b+420a4b=2a255a2ba2165b+445(a2)2bFactor.=2a5a5ba245b+42a25bSimplify.=10a25b4a25b+8a25bCombineliketerms.=14a25b

Answer:

14a25b

Exercise 7.3.2

32x6y+3xy3(y327x2x32x3y)

Answer

3x232y2y3x

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note

Take careful note of the differences between products and sums within a radical. Assume both x and y are nonnegative.

ProductsSumsx2y2=xyx2+y2x+y3x3y3=xy3x3+y3x+y

The property nab=nanb says that we can simplify radicals when the operation in the radicand is multiplication. There is no corresponding property for addition.

Example 7.3.11:

Calculate the perimeter of the triangle formed by the points (2,1),(3,6), and (2,1).

Solution

The formula for the perimeter of a triangle is P=a+b+c where a,b, and c represent the lengths of each side. Plotting the points we have,

5e3fa7e6e60851ebfb6d4b968dd7c614.png
Figure 7.3.1

Use the distance formula to calculate the length of each side.

a=[3(2)]2+[6(1)]2b=[2(2)]2+[1(1)]2=(3+2)2+(6+1)2=(2+2)2+(1+1)2=(1)2+(7)2=(4)2+(2)2=1+49=16+4=50=20=52=25

Similarly we can calculate the distance between (3,6) and (2,1) and find that c=52 units. Therefore, we can calculate the perimeter as follows:

P=a+b+c=52+25+52=102+25

Answer:

102+25 units

Key Takeaways

  • Add and subtract terms that contain like radicals just as you do like terms. If the index and radicand are exactly the same, then the radicals are similar and can be combined. This involves adding or subtracting only the coefficients; the radical part remains the same.
  • Simplify each radical completely before combining like terms.

Exercise 7.3.3

Simplify

  1. 10353
  2. 15686
  3. 93+53
  4. 126+36
  5. 457525
  6. 310810210
  7. 646+26
  8. 5101510210
  9. 1376257+52
  10. 10131215+5131815
  11. 65(4335)
  12. 122(66+2)
  13. (25310)(10+35)
  14. (83+615)(315)
  15. 436335+636
  16. 310+53104310
  17. (739433)(39333)
  18. (835+325)(235+6325)
Answer

1. 53

3. 143

5. 55

7. 6

9. 872

11. 9543

13. 5410

15. 1036335

17. 63933

Exercise 7.3.4

Simplify. (Assume all radicands containing variable expressions are positive.)

  1. 2x42x
  2. 53y63y
  3. 9x+7x
  4. 8y+4y
  5. 7xy3xy+xy
  6. 10y2x12y2x2y2x
  7. 2ab5a+6ab10a
  8. 3xy+6y4xy7y
  9. 5xy(3xy7xy)
  10. 8ab(2ab4ab)
  11. (32x3x)(2x73x)
  12. (y42y)(y52y)
  13. 53x123x
  14. 23y33y
  15. a53b+4a53ba53b
  16. 84ab+34ab24ab
  17. 62a432a+72a32a
  18. 453a+33a953a+33a
  19. (44xy3xy)(244xy3xy)
  20. (556y5y)(266y+3y)
  21. 2x233x(x233xx33x)
  22. 5y36y(6y4y36y)
Answer

1. 32x

3. 16x

5. 5xy

7. 8ab15a

9. 9xy

11. 22x+63x

13. 73x

15. 4a53b

17. 132a532a

19. 44xy

21. x233x+x33x

Exercise 7.3.5

Simplify.

  1. 7512
  2. 2454
  3. 32+278
  4. 20+4845
  5. 2827+6312
  6. 90+244054
  7. 4580+2455
  8. 108+48753
  9. 42(2772)
  10. 35(2050)
  11. 316354
  12. 381324
  13. 3135+34035
  14. 310833234
  15. 227212
  16. 350432
  17. 324321848
  18. 6216224296
  19. 218375298+448
  20. 24512+220108
  21. (2363396)(712254)
  22. (2288+3360)(272740)
  23. 3354+532504316
  24. 431622338433750
Answer

1. 33

3. 22+33

5. 5753

7. 55

9. 10233

11. 32

13. 435

15. 23

17. 23362

19. 82+3

21. 8366

23. 2632

Exercise 7.3.6

Simplify. (Assume all radicands containing variable expressions are positive.)

  1. 81b+4b
  2. 100a+a
  3. 9a2b36a2b
  4. 50a218a2
  5. 49x9y+x4y
  6. 9x+64y25xy
  7. 78x(316y218x)
  8. 264y(332y81y)
  9. 29m2n5m9n+m2n
  10. 418n2m2n8m+n2m
  11. 4x2y9xy216x2y+y2x
  12. 32x2y2+12x2y18x2y227x2y
  13. (9x2y16y)(49x2y4y)
  14. (72x2y218x2y)(50x2y2+x2y)
  15. 12m4nm75m2n+227m4n
  16. 5n27mn2+212mn4n3mn2
  17. 227a3ba48aba144a3b
  18. 298a4b2a162a2b+a200b
  19. 3125a327a
  20. 31000a2364a2
  21. 2x354x2316x4+532x4
  22. x354x33250x6+x232
  23. 416y2+481y2
  24. 532y45y4
  25. 432a34162a3+542a3
  26. 480a4b+45a4ba45b
  27. 327x3+38x3125x3
  28. 324x3128x381x
  29. 327x4y38xy3+x364xyy3x
  30. 3125xy3+38x3y3216xy3+10x3y
  31. (3162x4y3250x4y2)(32x4y23384x4y)
  32. (532x2y65243x6y2)(5x2y6x5xy2)
Answer

1. 11b

3. 3ab

5. 8x5y

7. 202x12y

9. 8mn

11. 2xy2yx

13. 4xy

15. 3m23n

17. 2a3ab12a2ab

19. 23a

21. 7x32x

23. 54y2

25. 442a3

27. 2x+23x

29. 7x3xy3y3x

31. 7x36xy6x32xy2

Exercise 7.3.7

Calculate the perimeters of the triangles formed by the following set of vertices.

  1. {(4,5),(4,3),(2,3)}
  2. {(1,1),(3,1),(3,2)}
  3. {(3,1),(3,5),(1,5)}
  4. {(3,1),(3,7),(1,1)}
  5. {(0,0),(2,4),(2,6)}
  6. {(5,2),(3,0),(1,6)}
  7. A square garden that is 10 feet on each side is to be fenced in. In addition, the space is to be partitioned in half using a fence along its diagonal. How much fencing is needed to do this? (Round to the nearest tenth of a foot.)
  8. A garden in the shape of a square has an area of 150 square feet. How much fencing is needed to fence it in? (Hint: The length of each side of a square is equal to the square root of the area. Round to the nearest tenth of a foot.)
Answer

1. 24 units

3. 8+42 units

5. 45+210 units

7. 54.1 feet

Exercise 7.3.8

  1. Choose values for x and y and use a calculator to show that x+yx+y.
  2. Choose values for x and y and use a calculator to show that x2+y2x+y.
Answer

1. Answer may vary

Footnotes

16Radicals that share the same index and radicand.

17Term used when referring to like radicals.


7.3: Adding and Subtracting Radical Expressions is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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