# EXPONENTS AND SQUARE ROOTS WORKSHEET

Problem 1 :

Simplify :

(a7 ⋅ a⋅ a-4) / (a⋅ a-3 ⋅ a4)

Problem 2 :

Simplify :

(a6 ⋅ b3) / (a⋅ b-3)2

Problem 3 :

If a-1/2  =  5, then find the value of a.

Problem 4 :

If 3x+3 - 3x+2  =  k(3x), then solve for k.

Problem 5 :

If x2y3  =  10 and x3y2  =  8, then find the value of x5y5.

Problem 6 :

Simplify the following square root expression :

√40 + √160

Problem 7 :

Simplify the following square root expression :

√27 ⋅ √3

Problem 8 :

Simplify the following square root expression :

(14√117) ÷ (7√52)

Problem 9 :

Simplify the following square root expression :

(√3)3 + √27

Problem 10 :

Rationalize the denominator :

(3 - √3) / √3 ## Solutions

Problem 1 :

Simplify :

(a7 ⋅ a⋅ a-4) / (a⋅ a-3 ⋅ a4)

Solution :

(a7 ⋅ a⋅ a-4/ (a⋅ a-3 ⋅ a4)  =  a7+2-4 / a2-3+4

(a7 ⋅ a⋅ a-4) / (a⋅ a-3 ⋅ a4)  =  a5 / a3

(a7 ⋅ a⋅ a-4) / (a⋅ a-3 ⋅ a4)  =  a5-3

(a7 ⋅ a⋅ a-4) / (a⋅ a-3 ⋅ a4)  =  a2

Problem 2 :

Simplify :

(a6 ⋅ b3) / (a⋅ b-3)2

Solution :

(a6 ⋅ b3) / (a⋅ b-3) =  (a6 ⋅ b3) / [(a2)⋅ (b-3)2]

(a6 ⋅ b3) / (a⋅ b-3) =  (a6 ⋅ b3) / (a4 ⋅ b-6)

(a6 ⋅ b3) / (a⋅ b-3) =  a6-4 ⋅ b3+6

(a6 ⋅ b3) / (a⋅ b-3) =  a2b9

Problem 3 :

If a-1/2  =  5, then find the value of a.

Solution :

a-1/2  =  5

a  =  5-2/1

a  =  5-2

a  =  1/52

a  =  1/25

Problem 4 :

If 3x+3 - 3x+2  =  k(3x), then solve for k.

Solution :

3x+3 - 3x+2  =  k(3x)

Using laws of exponents, we have

3x ⋅ 33 - 3x ⋅ 32  =  k(3x)

3x ⋅ 27 - 3x ⋅ 9  =  k(3x)

3x(27 - 9)  =  k(3x)

3x(18)  =  k(3x)

Divide each side by 3x.

18  =  k

So, the value of k is 18.

Problem 5 :

If x2y3  =  10 and x3y2  =  8, then find the value of x5y5.

Solution :

x2y3  =  10 -----(1)

x3y2  =  8 -----(2)

Multiply (1) and (2) :

(1) ⋅ (2) -----> (x2y3) ⋅ (x3y2)  =  10 ⋅ 8

x5y5  =  80

So, the value x5yis 80.

Problem 6 :

Simplify the following square root expression :

√40 + √160

Solution :

Decompose 40 and 160 into prime factors using synthetic division. √40  =  √(2 ⋅ 2 ⋅ 2 ⋅ 5)  =  2√10

√160  =  √(2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 5)  =  4√10

So, we have

√40 + √160  =  2√10 + 4√10

√40 + √160  =  6√10

Problem 7 :

Simplify the following square root expression :

√27 ⋅ √3

Solution :

√27 ⋅ √3  =  √(27 ⋅ 3)

√27 ⋅ √3  =  √(3 ⋅ 3 ⋅ 3 ⋅ 3)

√27 ⋅ √3  =  3 ⋅ 3

√27 ⋅ √3  =  9

Problem 8 :

Simplify the following square root expression :

(14√117) ÷ (7√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

(14√117) ÷ (7√52)  =  14(3√13) ÷ 7(2√13)

(14√117) ÷ (7√52)  =  42√13 ÷ 14√13

(14√117) ÷ (7√52)  =  42√13 / 14√13

(14√117) ÷ (7√52)  =  3

Problem 9 :

Simplify the following square root expression :

(√3)3 + √27

Solution :

(√3)3 + √27  =  (√3 ⋅ √3  √3) + √(3 ⋅ 3 ⋅ 3)

(√3)3 + √27  =  (3  √3) + 3√3

(√3)3 + √27  =  3√3 + 3√3

(√3)3 + √27  =  6√3

Problem 10 :

Rationalize the denominator :

(3 - √3) / √3

Solution :

To get rid of the radical in denominator, multiply both numerator and denominator by √3.

(3 - √3) / √3  =  [(3-√3) ⋅ √3] / (√3 ⋅ √3)

(3 - √3) / √3  =  (3√3 - 3) / 3

(3 - √3) / √3  =  3(√3 - 1) / 3

(3 - √3) / √3  =  √3 - 1 Apart from the stuff given in this section if you need any other stuff in math, please use our google custom search here.

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