# 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.

Before look at the worksheet, if you would like to learn the stuff exponents and square roots in detail,

## 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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