EXPONENTS AND SQUARE ROOTS WORKSHEET

About "Exponents and Square Roots Worksheet"

Exponents and Square Roots Worksheet :

Worksheet given in this section is much useful to the students who would like to practice problems on exponents and square roots.

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Exponents and Square Roots Worksheet - Problems

Problem 1 :

Simplify the following :

(-4)5 ÷ (-4)8

Problem 2 :

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

Problem 3 :

If x2  =  y3 and x3z  =  y9, then find the value of z.

Problem 4 :

Solve for k : 

3x2 ⋅ 2x3  =  6xk

Problem 5 : 

Rationalize the denominator :  

/ (3 + √2)

Problem 6 : 

Rationalize the denominator :  

(1 - √5) / (3 + √5)

Problem 7 :

Find the value of :

√2.56

Problem 8 : 

Simplify the following square root expression :

(√17)(√51)

Problem 9 : 

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

Problem 10 :

If (√9)-7 ⋅ (√3)-4  =  3k, then solve for k. 

Exponents and Square Roots Worksheet - Solutions

Problem 1 :

Simplify the following :

(-4)5 ÷ (-4)8

Solution :

(-4)5 ÷ (-4) =  (-4)5-8

(-4)5 ÷ (-4) =  (-4)-3

(-4)5 ÷ (-4) =  1 / (-4)3

(-4)5 ÷ (-4) =  1 / (-64)

(-4)5 ÷ (-4) =  - 1 / 64

Problem 2 :

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 3 :

If x2  =  y3 and x3z  =  y9, then find the value of z.

Solution : 

x3z  =  y9

x3z  =  y⋅ 3

x3z  =  (y3)3

Substitute xfor y3.

x3z  =  (x2)3

x3z  =  x6

3z  =  6

Divide each side by 3.

z  =  2

So, the value of z is 2.

Problem 4 :

Solve for k : 

3x2 ⋅ 2x3  =  6xk

Solution : 

3x2 ⋅ 2x3  =  6xk

(3 ⋅ 2)(x⋅ x3)  =  6xk

(6)(x2+3)  =  6xk

6x5  =  6xk

Divide each side by 6. 

x5  =  xk

Using laws of exponents, we have

5  =  k

So, the value of k is 5.

Problem 5 : 

Rationalize the denominator :  

/ (3 + √2)

Solution :

To get rid of the radical in denominator, multiply both numerator and denominator by the conjugate of (3 + √2), that is by (3 - √2). 

1 / (3 + √2)  =  [1 ⋅ (3-√2)] / [(3+√2)  (3-√2)]

1 / (3 + √2)  =  (3-√2) / [(3+√2)  (3-√2)]

Using the algebraic identity a2 - b2  =  (a + b)(a - b), simplify the denominator on the right side.

1 / (3 + √2)  =  (3-√2) / [32 - (√2)2]

1 / (3 + √2)  =  (3-√2) / (9 - 2)

1 / (3 + √2)  =  (3 - √2) / 7

Problem 6 : 

Rationalize the denominator :  

(1 - √5) / (3 + √5)

Solution :

To get rid of the radical in denominator, multiply both numerator and denominator by the conjugate of (3 + √5), that is by (3 - √5). 

(1 - √5) / (3 + √5)  =  [(1-√5) ⋅ (3-√5)] / [(3+√5)  (3-√5)]

Simplify.

(1 - √5) / (3 + √5)  =  [3 √5 - 3√5 + 5] / [32 - (√5)2]

(1 - √5) / (3 + √5)  =  (8 - 4√5) / (9 - 5)

(1 - √5) / (3 + √5)  =  4(2 - √5) / 4

(1 - √5) / (3 + √5)  =  2 - √5

Problem 7 :

Find the value of :

√2.56

Solution :

To get rid of the decimal, multiply and divide 2.56 inside the radical by 100.

√2.56  =  √(2.56  100 / 100)

√2.56  =  √(256 / 100)

√2.56   √(16  16) / (10 ⋅ 10)

√2.56  =  16 / 10

√2.56  =  1.6

Problem 8 : 

Simplify the following square root expression :

(√17)(√51)

Solution : 

Decompose 17 and 51 into prime factors. 

Because 17 is a prime number, it can't be decomposed anymore. So, √17 has to be kept as it is. 

√51  =  √(3 ⋅ 17)  =  √3 ⋅ √17

So, we have

(√17)(√51)  =  (√17)(√3 ⋅ √17)

(√17)(√51)  =  (√17 ⋅ √17)√3

(√17)(√51)  =  17√3

Problem 9 : 

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 10 :

If (√9)-7 ⋅ (√3)-4  =  3k, then solve for k. 

Solution : 

(91/2)-7 ⋅ (31/2)-4  =  3k

(9)-7/2 ⋅ (3)-4/2  =  3k

(32)-7/2 ⋅ 3-2  =  3k

3⋅ (-7/2) ⋅ 3-2  =  3k

3-7 ⋅ 3-2  =  3k

3-7 - 2  =  3k

3-9  =  3k

k  =  -9

So, the value of k is -9.

After having gone through the stuff given above, we hope that the students would have understood "Exponents and Square Roots Worksheet". 

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