USING DISTRIBUTIVE PROPERTY WITH RADICAL EXPRESSION

Expand and simplify :

Problem 1 :

-√2(3 - √2)

Solution :

-√2(3 - √2)

After distribution, we get

= -√2 ⋅ 3 + √2 ⋅ √2

= -3√2 + √(2 ⋅ 2)

= -3√2 + √4

     = -3√2 + 2

Problem 2 :

-√2(4 - √2)

Solution :

-√2(4 - √2)

= -√2 ⋅ 4 + √2 ⋅ √2

= -4√2 + √(2 ⋅ 2)

 = -4√2 + √4

 = -4√2 + 2

Problem 3 :

-√3(1 + √3)

Solution :

-√3(1 + √3)

 = -√3 ⋅ 1 - √3 ⋅ √3

= -√3 - √(3 ⋅ 3)

 = -√3 - √9

 = -√3 - 3

Problem 4 :

-√3(√3 + 2)

Solution :

-√3(√3 + 2)

= -√3 ⋅ √3 -√3 ⋅ 2

= -√(3 ⋅ 3) - 2√3

= -√9 - 2√3

= -3 - 2√3

Problem 5 :

-√5(2 + √5)

Solution :

-√5(2 + √5)

= -√5 ⋅ 2 - √5 ⋅√5

= -2√5 - √(5 ⋅ 5)

= -2√5 - √25

= -2√5 - 5   

Problem 6 :

-(√2 + 3)

Solution :

-(√2 + 3)

= -√2 – 3

Problem 7 :

-√5(√5 - 4)

Solution :

-√5(√5 - 4)

= -√5 ⋅√5 + √5 ⋅ 4

= -√(5 ⋅ 5) + 4√5

= -√25 + 4√5

= -5 + 4√5

Problem 8 :

-(3 - √7)

Solution :

-(3 - √7)

= -3 + √7

Problem 9 :

-√11(2 - √11)

Solution :

-√11(2 - √11)

= -√11 ⋅ 2 + √11 ⋅ √11

= -2√11 + √(11 ⋅ 11)

= -2√11 + √121

= -2√11 + 11

Problem 10 :

-2√2(1 - √2)

Solution :

-2√2(1 - √2)

= -2√2 ⋅ 1 + 2√2 ⋅ √2

= -2√2 + 2√(2 ⋅ 2)

= -2√2 + 2√4

= -2√2 + 2 ⋅ 2

= -2√2 + 4

Problem 11 :

-3√3(5 - √3)

Solution :

-3√3(5 - √3)

= -3√3 ⋅ 5 + 3√3 ⋅√3

= -15√3 + 3√(3 ⋅ 3)

= -15√3 + 3√9

= -15√3 + 3 ⋅ 3

= -15√3 + 9

Problem 12 :

-7√2(√2 + 3)

Solution :

-7√2(√2 + 3)

= -7√2 ⋅ √2 - 7√2 ⋅ 3

= -7√(2 ⋅ 2) – 21 ⋅ √2

= -7√4 - 21√2

= -7 ⋅ 2 - 21√2

= -14 - 21√2

Problem 13 :

A particle (√7 + 3) ft/sec. How long will it take the object to travel 30 feet ?

Solution :

Distance traveled by the particle = (√7 + 3) ft/sec

Distance traveled in 30 feet = 30(√7 + 3) 

Distributing 30, we get

= 30√7 + 30(3)

= 30√7 + 90

Problem 14 :

The ratio of the length to the width of a golden rectangle is (1 + √5) : 2. The dimensions of the face of the Parthenon in Greece form a golden rectangle. What is the perimeter of the this golden rectangle?

Solution :

Think of the length and height of the Parthenon as the length and width of a golden rectangle. The height of the rectangular face is 19 meters. You know the ratio of the length to the height. Find the length ℓ and then fi nd the perimeter using the formula P = 2ℓ + 2w.

(1 + √5) : 2 = l : 19

(1 + √5) / 2 = l / 19

19(1 + √5) = 2l

l = (19/2)(1 + √5)

= 9.5(1 + 2.23)

= 9.5(3.23)

l = 30.685

Perimeter of the rectangle = 2(30.68 + 19)

= 2(49.685)

= 99.37

So, the required perimeter is approximately 100 m.

Problem 15 :

√3 (8 √2 + 7 √32)

Solution :

= √3 (8√2 + 7√32)

Using distributive property, we get

= 8 √3 √2 + 7 (√3√32)

= 8 √6 + 7 √96

= 8 √6 + 7 √(2⋅2⋅2⋅2⋅2⋅3)

= 8 √6 + 7 (2⋅2)√6

= 8 √6 + 28√6

= 36√6

Problem 16 :

Are the expressions 3√2x and √18x equivalent? Explain your reasoning

Solution :

Given are, 3√2x and √18x

Simplifying √18x = √(3 ⋅ 3 ⋅ 2 ⋅ x)

= 3√2x

So, both are equal.

Problem 17 :

The orbital period of a planet is the time it takes the planet to travel around the Sun. You can find the orbital period P (in Earth years) using the formula P = √d³ , where d is the average distance (in astronomical units, abbreviated AU) of the planet from the Sun

a. Simplify the formula.

b. What is Jupiter’s orbital period?

Solution :

a) P = √d³

P = √(d ⋅ d ⋅ d)

P = d√d

b) P = √d³

Applying d = 5.2

= √(5.2)³

= √140.608

= 11.85