SOLVING RADICAL EQUATIONS WITH TWO RADICALS

Solve the following radical equation

Problem 1 :

√(4x + 1) = √(x + 10)

Solution :

√(4x + 1) = √(x + 10)

Raise both sides to the power 2.

[√(4x + 1)]² = (√x + 10)²

4x + 1 = x + 10

Subtracting x on both sides.

4x - x + 1 = 10

Subtracting 1 on both sides.

3x = 0

3x = 9

Dividing by 3 on both sides.

x = 3

Problem 2 :

√(3x – 3) - √(x + 12) = 0

Solution :

√(3x – 3) - √(x + 12) = 0

Add √(x + 12)

√(3x – 3) = √(x + 12)

Take squares on both sides.

3x - 3 = x + 12

Subtract x on both sides and add 3 on both sides.

3x - x = 12 + 3

2x = 15

Dividing by 2, we get

x = 15/2

Problem 3 :

∛2x – 5 - ∛8x + 1 = 0

Solution :

∛(2x – 5) - ∛(8x + 1) = 0

Add ∛(8x + 1)

(∛2x – 5)³ = (∛8x + 1)³

2x – 5 = 8x + 1

Subtract 8x and add 5 on both sides.

2x - 8x = 1 + 5

-6x = 6

Divide by -6 on both sides.

x = -6/6

x = -1

Problem 4 :

∛(x + 5) = 2∛(2x + 6)

Solution :

Given, ∛x + 5 = 2∛2x + 6

Raise both sides to the power 3.

(∛x + 5)³ = (2)³(∛2x + 6)³

x + 5 = (8)(2x + 6)

x + 5 = 16x + 48

x + 5 -16x – 48 = 0

-15x – 43 = 0

-15x = 43

x = -43/15

Problem 5 :

√(3x – 8) + 1 = √(x + 5)

Solution :

√(3x – 8) + 1 = √(x + 5) ----(1)

[√(3x – 8) + 1]² = √(x + 5)

√(3x – 8)² + 2 √(3x – 8) + (1)² = √(x + 5)²

3x – 8 + 1 + 2√(3x – 8) = x + 5

3x - 7 + 2√(3x – 8) = x + 5

2x - 12 = -2√(3x – 8)

6 - x = √(3x – 8)

Take square on both sides.

(6 - x)² = 3x - 8

36 - 2(6)x + x² = 3x - 8

x² - 12x + 36 = 3x - 8

Subtract 3x and add 8 on both sides.

x² - 15x + 44 = 0

(x - 11)(x - 4) = 0

x = 11 and x = 4

By applying the value of x in (1), we get the solution as x = 4.

Problem 6 :

√(x + 2) = 2 - √x

Solution :

√(x + 2) = 2 - √x

√(x + 2) + √x = 2

Take squares on both sides.

(√(x + 2) + √x)² = 2²

x + 2 + x + 2√(x + 2)√x = 4

2x + 2 + 2√[x(x + 2)] = 4

Subtract 2x and 2 on both sides.

2√[x(x + 2)] = 4 - 2x - 2

2√[x² + 2x)] = 2 - 2x

Dividing by 2, we get

√[x² + 2x)] = 1 - x

Take square on both sides.

x² + 2x = (1 - x)²

x² + 2x = 1 + x² - 2x

4x = 1

x = 1/4

Problem 7 :

√(2x + 4) = √(x + 2)

Solution :

√2x + 4 = √x + 2

√(2x + 4)² = √(x + 2)²

2x + 4 = x + 2

Subtracting x and 4, we get

2x - x  = 2 - 4

x = -2

Problem 7 :

The lateral surface area of a right circular cone, s, is represented by the equation 

s = πr√(r² + h²)

where r is the radius of the circular base and h is the height of the cone. If the lateral surface area of a large funnel is 236.64 square centimeters and its radius is 4.75 centimeters, find its height, to the nearest hundredth of a centimeter.

Solution :

s = πr√(r² + h²)

236.64 = 3.14(4.75) √(4.75² + h²)

236.64/3.14(4.75) = √(4.75² + h²)

15.86 = √(4.75² + h²)

Squaring both sides

(15.86)² = (4.75² + h²)

251.53 = 22.56 + h²

h² = 251.53 - 22.56

h² = 228.96

h = √228.96

h = 15.13

Problem 8 :

Solve algebraically for x: 

√(x² + x - 1) + 11x = 7x + 3

Solution :

√(x² + x - 1) + 11x = 7x + 3

Subtracting 11x on both sides

√(x² + x - 1) = 7x - 11x + 3

√(x² + x - 1) = - 4x + 3

(x² + x - 1) = (- 4x + 3)²

x² + x - 1 = (-4x)² + 2(-4x)(3) + 3²

x² + x - 1 = 16x² - 24x + 9

16x² - x² - 24x - x + 9 + 1 = 0

15x² - 25x + 10 = 0

3x² - 5x + 2 = 0

3x² - 3x - 2x + 2 = 0

3x(x - 1) - 1(x - 1) = 0

(3x - 1)(x - 1) = 0

x = 1/3 and x = 1