Problem 1 :

√169 + √121

Problem 2 :

√20 + √320

Problem 3 :

√117 - √52

Problem 4 :

√243 - 5√12 + √27

Problem 5 :

-√147 - √243

Problem 6 :

(√13)(√26)

Problem 7 :

(3√14)(√35)

Problem 8 :

(8√117) ÷ (2√52)

Problem 9 :

(8√3)2

Problem 10 :

(√2)3 + √8

Problem 11 :

4√(x4/16)

Problem 12 :

3√(125p6q3)

Problem 1 :

√169 + √121

Solution :

Decompose 169 and 121 into prime factors using synthetic division.

 √169  =  √(13 ⋅ 13)√169  =  13 √121  =  √(11 ⋅ 11)√121  =  11

So, we have

√169 + √121  =  13 + 11

√169 + √121  =  24

Problem 2 :

√20 + √320

Solution :

Decompose 20 and 320 into prime factors using synthetic division.

 √20  =  √(2 ⋅ 2 ⋅ 5)√20  =  2√5 √320  =  √(2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 5)√320  =  2 ⋅ 2 ⋅ 2 ⋅ √5√320  =  8√5

So, we have

√20 + √320  =  2√5 + 8√5

√20 + √320  =  10√5

Problem 3 :

√117 - √52

Solution :

Decompose 117 and 52 into prime factors using synthetic division.

 √117  =  √(3 ⋅ 3 ⋅ 13)√117  =  3√13 √52  =  √(2 ⋅ 2 ⋅ 13)√52  =  2√13

So, we have

√117 - √52  =  3√13 - 2√13

√117 + √52  =  √13

Problem 4 :

√243 - 5√12 + √27

Solution :

Decompose 243, 12 and 27 into prime factors using synthetic division.

√243  =  √(3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3)  =  9√3

√12  =  √(2 ⋅ 2 ⋅ 3)  =  2√3

√27  =  √(3 ⋅ 3 ⋅ 3)  =  3√3

So, we have

√243 - 5√12 + √27  =  9√3 - 5(2√3) + 3√3

Simplify.

√243 - 5√12 + √27  =  9√3 - 10√3 + 3√3

√243 - 5√12 + √27  =  2√3

Problem 5 :

-√147 - √243

Solution :

Decompose 147 and 243 into prime factors using synthetic division.

√147  =  √(7 ⋅ 7 ⋅ 3)  =  7√3

√243  =  √(3 ⋅ 3 ⋅ 3 ⋅ 3 ⋅ 3)  =  9√3

So, we have

-√147 - √243  =  -7√3 - 9√3

-√147 - √243  =  -16√3

Problem 6 :

(√13)(√26)

Solution :

Decompose 13 and 26 into prime factors.

13 is a prime number. So, it can't be decomposed anymore.

√26  =  √(2 ⋅ 13)  =  √2 ⋅ √13

So, we have

(√13)(√26)  =  (√13)(√2 ⋅ √13)

(√13)(√26)  =  (√13 ⋅ √13)√2

(√13)(√26)  =  13√2

Problem 7 :

(3√14)(√35)

Solution :

Decompose 14 and 35 into prime factors.

√14  =  √(2 ⋅ 7)  =  √2 ⋅ √7

√35  =  √(5 ⋅ 7)  =  √5 ⋅ √7

So, we have

(3√14)(√35)  =  3( √2 ⋅ √7)(√5 ⋅ √7)

(3√14)(√35)  =  3(√7 ⋅ √7)(√2 ⋅ √5)

(3√14)(√35)  =  3(7)√(2 ⋅ 5)

(3√14)(√35)  =  21√10

Problem 8 :

(8√117) ÷ (2√52)

Solution :

Decompose 117 and 52 into prime factors using synthetic division.

 √117  =  √(3 ⋅ 3 ⋅ 13)√117  =  3√13 √52  =  √(2 ⋅ 2 ⋅ 13)√52  =  2√13

(8√117) ÷ (2√52)  =  8(3√13) ÷ 2(2√13)

(8√117) ÷ (2√52)  =  24√13 ÷ 4√13

(8√117) ÷ (2√52)  =  24√13 / 4√13

(8√117) ÷ (2√52)  =  6

Problem 9 :

(8√3)2

Solution :

(8√3)=  8√3 ⋅ 8√3

(8√3)2  =  (⋅ 8)(√3 ⋅ √3)

(8√3)2  =  (64)(3)

(8√3)2  =  192

Problem 10 :

(√2)3 + √8

Solution :

(√2)3 + √8  =  (√2 ⋅ √2  √2) + √(2⋅ ⋅ 2)

(√2)3 + √8  =  ( √2) + 2√2

(√2)3 + √8  2√2 + 2√2

(√2)3 + √8  =  4√2

Problem 11 :

4√(x4/16)

Solution :

4√(x4/16)  =  4√(x4) / 416

4√(x4/16)  =  4√(x ⋅ x ⋅ x ⋅ x) / 4√(2 ⋅ 2 ⋅ 2 ⋅ 2)

4√(x4/16)  =  x / 2

Problem 12 :

3√(125p6q3)

Solution :

3√(125p6q3)  =  3√(5 ⋅ 5 ⋅ 5  p2 ⋅ p2 ⋅ p2 ⋅ q ⋅ q ⋅ q)

3√(125p6q3)  =  5p2q

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

## Recent Articles

1. ### Simplifying Algebraic Expressions with Fractional Coefficients

May 17, 24 08:12 AM

Simplifying Algebraic Expressions with Fractional Coefficients

2. ### The Mean Value Theorem Worksheet

May 14, 24 08:53 AM

The Mean Value Theorem Worksheet