EVALUATE RADICAL EXPRESSIONS

Evaluate the following expressions using the given values of the variables.

Example 1 :

√x + √y for x = 4 and y = 9

Solution :

= √x + √y

Substitute x = 4 and y = 9.

= √4 + √9

= 2 + 3

= 5

Example 2 :

2√(185 - x) for x = 169

Solution :

= 2√(185 - x)

Substitute x = 169.

= 2√(185 - 169)

= 2√16

= 2√(4 ⋅ 4)

= 2(4)

= 8

Example 3 :

5√x³ for x = 3

Solution :

= 5√x³

Substitute x = 3.

= 5√3³

= 5√(3 ⋅ 3 ⋅ 3)

= 5(3√(3)

= 15√3

Example 4 :

5√x + 1 for x = 1

Solution :

= 5√x + 1

Substitute x = 1.

= 5√1 + 1

= 5√1 + 1

= 5(1) + 1

= 5 + 1

= 6

Example 5 :

√(x/49) for x = 25 

Solution :

= √(x/49)

Substitute x = 25.

= √(25/49)

= √25/√49

= √(5 ⋅ 5)/√(7 ⋅ 7)

= 5/7

Example 6 :

√(x² + y²) for x = 3 and y = 4

Solution :

= √(x² + y²)

Substitute x = 3 and y = 4.

= √(3² + 4²)

= √(9 + 16)

= √25

= √(5 ⋅ 5)

= 5

Example 7 :

³√x - ³√y for x = 27 and y = 8

Solution :

³√x - ³√y

Substitute x = 27 and y = 8.

= ³√27 - ³√8

= 3 - 2

= 1

Example 8 :

√12w + √27w for w = 3

Solution :

= √(12w) + √(27w)

= √(2 ⋅ 2 ⋅ 3 ⋅ w) + √(3 ⋅ 3 ⋅ 3 ⋅ w)

= 2√3w + 3√3w

= 5√3w

Substitute w = 3.

= 5√(3 ⋅ 3)

= 5(3)

= 15

Example 9 :

³√8x³y⁶ + √9x²y⁴ for x = 1 and y = 2

Solution :

= ³√8x³y⁶ + √9x²y⁴

= ³√(2 ⋅ 2 ⋅ 2 ⋅ x ⋅ x ⋅ x ⋅ y² ⋅ y² ⋅ y²) + √(3 ⋅ 3 ⋅ x ⋅ x ⋅ y² ⋅ y²)

= 2xy² + 3xy²

= 5xy²

Substitute x = 1 and y = 2.

= 5(1)(2²)

= 5(1)(4)

= 20

Example 10 :

√4p²q⁴ - ³√125p³q⁶ for p = -2 and q = -3

Solution :

= √4p²q⁴ - ³√125p³q⁶

= √(2 ⋅ 2 ⋅ p ⋅ p ⋅ q² ⋅ q²) - √(5 ⋅ 5 ⋅ 5 ⋅ p ⋅ p ⋅ p ⋅ q² ⋅ q² ⋅ q²)

= 2pq² - 5pq²

= -3pq²

Substitute p = -2 and q = -3.

= -3(-2)(-3)²

= -3(-2)(9)

= 54

Example 11 :

If the area A  of a square is known, then the length of its sides s, can be computed by the formula s = A^1/2

a) Compute the length of side of a square having an area of 100 square inches.

b)  Compute the length of the side of a square having an area of 72 square inches. Round your answer to the nearest tenth.

Solution :

Given that, side length of the square

s = A^1/2

a) Here area A = 100 square inches

s = 100^1/2

s = (10²)^1/2

= 10^{2 x (1/2)}

= 10

So, the side length of the square is 10 inches.

b)  Here area A = 72 square inches

s = 72^1/2

s = (72)^1/2

= 8.48

Example 12 :

The time T(in seconds) for a pendulum to make one complete swing back and forth is approximated by

T(x) = 2π√(x/32)

where x is the length of the pendulum in feet. Determine the exact time required for one swing for a pendulum that is 1 ft long. Then approximate the time to the nearest hundredth of a second.

Solution :

T(x) = 2π√(x/32)

x is the length of pendulum, that is x = 1 ft

T(1) = 2π√(1/32)

= 2π (√1/√32)

= 2π [√1/√(2 x 2 x 2 x 2 x 2)]

= 2π (1/4√2)

= (π/2√2)

Example 13 :

An object is dropped off a building x meters tall. The time T(in seconds) required for the object to hit the ground is given by

T(x) = √(10x/49)

Find the exact time required for the object to hit the ground if it is dropped off the First National Plaza Building in Chicago, a height of 230 m. Then round the time to the nearest hundredth of a second.

Solution :

x is the height and that is h = 230 m

T(230) = √(10(230)/49))

= √46.93

= 6.85

So, the required time is 7 minutes

Example 14 :

The length of the legs, s, of an isosceles right triangle is

s = √2A

where A is the area. If the lengths of the legs are 9 in., find the area.

Solution :

s = √2A

Length of legs = 9 inches

9 = √2A

Squaring on both sides, we get

9² = (√2A)²

81 = 2A

A = 81/2

A = 40.5 square inches.

So, the required area of the triangle is 40.5 square inches.

Example 15 :

Use the Pythagorean theorem to find a, b, or c.

a)  Find b, when a = 2 and c = y

b)  find b when a = h and c = 5

c)  find a, when b = x and c = 8

Solution :

a² + b² = c²

a)  when a = 2 and c = y

2² + b² = y²

b² = y² - 2²

b² = y² - 4

b = √(y² - 4)

b) when a = h and c = 5

h² + b² = 5²

b² = 5² - h²

b² = 25 - h²

b = √(25 - h²)

c)  when b = x and c = 8

a² + x² = 8²

a²  = 64 - x²

a = √(64 - x²)

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