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1. Operations With Radical Expressions
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Chapter 3
1. 

Operations With Radical Expressions

This lesson delves into the complexities of working with radical expressions, focusing on techniques for simplification and rationalization. It provides a variety of examples and exercises that help in understanding how to operate with square roots effectively. These techniques are not just theoretical; they have practical applications in fields like engineering, physics, and computer science. For instance, understanding how to simplify radicals can be crucial when solving equations in electrical engineering or optimizing algorithms in computer science. The lesson also discusses the properties of square roots and how they can be manipulated for different mathematical operations. Overall, it serves as a comprehensive guide for anyone looking to master the subject and apply it in various professional fields.
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Lesson Settings & Tools
15 Theory slides
11 Exercises - Grade E - A
Each lesson is meant to take 1-2 classroom sessions
Operations With Radical Expressions
Slide of 15
When solving geometry problems, radical expressions — like square roots or cube roots — are present in a lot of concepts and formulas. For example, the diagonal of a square is calculated by asqrt(2), where a is the side length of the square. This lesson will explore the operations on more general radical expressions and how radicals might apply to geometry and physics.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Here are a few practice exercises before getting started with this lesson.

a Write 8^(25) as a radical.
b Consider the following algebraic expression. Assume that the variable x is positive.

sqrt(8x^3)*(x^()12)^5/sqrt(2) * x^3

What is the simplest form of the expression?
c Write the following expression in its simplest form. 1+sqrt(16)/sqrt(9) - 2sqrt(2)
Challenge

Length of a Triangular Fencing

Ignacio is planning to build an astronomical observatory in his garden. The building will be enclosed by a fence with a triangular shape. The dimensions of Ignacio's garden are presented in the following diagram.

Ignacio's garden with a fencing

In this diagram, all dimensions are measured in meters. Although some side lengths are still not decided, help Ignacio calculate the length of the fence L(x) with respect to x. What is the value of L(5)?

Discussion

n^\text{th} Root

The n^(th) root of a real number a expresses another real number that, when multiplied by itself n times, will result in a. In addition to the radical symbol, the notation is made up of the radicand a and the index n.
The resulting number is commonly called a radical. For example, the radical expression sqrt(16) is the fourth root of 16. Notice that sqrt(16) simplifies to 2 because 2 multiplied by itself 4 times equals 16. sqrt(16) = sqrt(2^4) = 2 The general expression sqrt(a) represents a number which equals a when multiplied by itself n times.


sqrt(a) * sqrt(a) * ... * sqrt(a)_(ntimes)=a or ( sqrt(a) )^n=a

Some numbers have more than one real n^\text{th} root. For example, 16 has two fourth roots, 2 and -2, because both 2^4 and (-2)^4 are equal to 16. The number of real n^\text{th} roots depends on the sign of a radicand a and an integer n.

n is Even n is Odd
a>0 Two unique real n^\text{th} roots, sqrt(a) and -sqrt(a) One real n^\text{th} root, sqrt(a)
a=0 One real n^\text{th} root, sqrt(0) = 0 One real n^\text{th} root, sqrt(0) = 0
a<0 No real n^\text{th} roots One real n^\text{th} root, sqrt(a)

By the definition of an n^\text{th} root, calculating the n^\text{th} power of the n^\text{th} root of a number a results in the same number a. The following formula shows what happens if these two operations are swapped.

Rule

n^\text{th} Roots of n^\text{th} Powers

To simplify an n^(th) root, the radicand must first be expressed as a power. If the index of the radical and the power of the radicand are equal such that sqrt(a^n), the radical expression can be simplified as follows.


sqrt(a^n)= a, & if n is odd |a|, & if n is even

The absolute value of a number is always non-negative, so when n is even, the result will always be non-negative. Consider a few example n^\text{th} roots that can be simplified by using the formula.

a n sqrt(a^n) Is n even? Simplify
4 3 sqrt(4^3) * 4
-6 5 sqrt((-6)^5) * -6
2 6 sqrt(2^6) |2| = 2
-3 4 sqrt((-3)^4) |-3| = 3

Proof

Informal Justification

To write the expression for sqrt(a^n), there are two cases to consider.

  • a≥ 0
  • a<0

Both cases will be considered one at a time.

a≥ 0

Start by noting that since a is non-negative, a^n is also non-negative. This means that the n^\text{th} root can be rewritten using a rational exponent. sqrt(a^n) = ( a^n )^(1n) Since both n and 1n are rational numbers, the Power of a Power Property for Rational Exponents can be applied to simplify the obtained expression.
( a^n )^(1n)
Simplify
a^(n* 1n)

a * 1/a=1

a^1
a
Therefore, sqrt(a^n) is equal to a if a is non-negative.

a<0

In case of a negative value of a, there are also two cases two consider.

  • n is even
  • n is odd

Even n

Recall that a root with an even index n_e is defined only for non-negative numbers. Although a is negative, a^(n_e) is positive. Also, a power with a negative base and an even exponent can be rewritten as a power with a positive base. a^(n_e) = (- a)^(n_e) Now, since - a is positive, the Power of a Power Property for Rational Exponents can be applied again to simplify sqrt(a^(n_e)).
sqrt(a^(n_e))
Simplify
sqrt((- a)^(n_e))
( (- a)^(n_e) )^()1n_e
(- a)^(n_e* 1n_e)

a * 1/a=1

(- a)^1
- a

Odd n

A root with an odd index n_o is defined for all real numbers. By the definition of the n^\text{th} root, the expression sqrt(a^(n_o)) is the number y that, when multiplied by itself n_o times, will result in a^(n_o). y * y * ... * y_(n_otimes)=a^(n_o) Because n_o is odd and a is negative, a^(n_o) is also negative. This means that the best candidate for y is simply a. sqrt(a^n) = a

Summary

If a is non-negative, sqrt(a^n) is always equal to a. However, in case of negative a, the value of sqrt(a^n) depends on the parity of n.

a≥ 0 a<0
Even n sqrt(a^n) = a sqrt(a^n) = - a
Odd n sqrt(a^n) = a sqrt(a^n) = a

To conclude, for odd values of n, the expression sqrt(a^n) is equal to a. On the other hand, if n is even, sqrt(a^n) can be written as |a|.


sqrt(a^n)= a, & if n is odd |a|, & if n is even

Extra

Simplifying sqrt(a^m)

Depending on the index of the root and the power in the radicand, simplifying sqrt(a^m) may be problematic. Because real n^\text{th} roots with an even index are defined only for non-negative numbers, the absolute value is sometimes needed.

If n is even, sqrt(a^m) is defined only for non-negative a^m.

The following applet presents a decision tree to simplify sqrt(a^m). In this applet, k is the greatest common factor of m and n.
Simplifying nth root of a to the mth power
Consider example n^\text{th} roots sqrt(a^m) that can be simplified by using the decision tree.
a m n sqrt(a^m) Simplify
2 8 4 sqrt(2^8) 2^()84 = 2^2 = 4
-3 6 2 sqrt((-3)^6) | (-3)^()62 | = | (-3)^3 | = |-27| = 27
x 6 3 sqrt(x^6) x^()63 = x^2
-3 2 8 sqrt((-3)^2) sqrt(|-3|) = sqrt(3)
2 3 6 sqrt(2^3) sqrt(2)
-2 3 9 sqrt((-2)^3) sqrt(-2)
2 6 8 sqrt(2^6) sqrt(2^3) = sqrt(8)
-2 6 8 sqrt((-2)^6) sqrt(|-2|^3) = sqrt(2^3) = sqrt(8)
Example

Homework on Radical Expressions

In addition to physics and astronomy, Ignacio is also interested in algebra. Unfortunately, he missed one class and he cannot solve his homework on radical expressions. Help Ignacio pair each expression to its simplified form.

Hint

Write the radicand as a power. Then, use the n^\text{th} Roots of n^\text{th} Powers Property to simplify the radical expression.

Solution

To simplify the given expressions, the n^\text{th} Roots of n^\text{th} Powers Property will be used. sqrt(a^n)= a, & if n is odd |a|, & if n is even Unfortunately, only the second expression is written in the form that allows to use the property right away. Because the power and the index of the root are both 5 and 5 is an odd number, the second expression simplifies to x-2.

Radical Expression Simplified Form
sqrt(x^2-4x+4) ?
sqrt((x-2)^5) x-2
sqrt(16x^8) ?
sqrt(4x^2) ?
Each of the remaining expressions should be rewritten as a power with the exponent equal to the index of the corresponding root. Notice that the radicand in the first expression is a perfect square trinomial. x^2- 4x+ 4 = x^2- 2(2)x+ 2^2 Now, the radical expression can be simplified by writing the radicand as the square of a binomial. Because the index of the root is an even number, the result will require the absolute value.
sqrt(x^2-2(2)x+2^2)
sqrt((x-2)^2)
|x-2|
Therefore, the first expression simplifies to |x-2|.
Radical Expression Simplified Form
sqrt(x^2-4x+4) |x-2|
sqrt((x-2)^5) x-2
sqrt(16x^8) ?
sqrt(4x^2) ?
Next, properties of exponents will be used to write the radicand 16x^8 as a power with the exponent of 4.
sqrt(16x^8)
Rewrite
sqrt(2^4 x^8)
sqrt(2^4 x^(2* 4))
sqrt(2^4 (x^2)^4)
sqrt((2x^2)^4)

sqrt(a^4)=|a|

| 2x^2 |
Since x^2 is non-negative, 2x^2 is also non-negative. This means that | 2x^2 | is equal to 2x^2.
Radical Expression Simplified Form
sqrt(x^2-4x+4) |x-2|
sqrt((x-2)^5) x-2
sqrt(16x^8) 2x^2
sqrt(4x^2) ?
Now the final expression sqrt(4x^2) will be examined. Similar to the previous expression, properties of exponents will be used to rewrite the radicand as a square.
sqrt(4x^2)
Simplify
sqrt(2^2 x^2)
sqrt((2x)^2)
|2x|
This time, it is not known whether 2x is non-negative. Therefore, the absolute value is needed in this expression. Finally, all simplified forms may be written in the table.
Radical Expression Simplified Form
sqrt(x^2-4x+4) |x-2|
sqrt((x-2)^5) x-2
sqrt(16x^8) 2x^2
sqrt(4x^2) |2x|
Discussion

Product and Quotient Properties of Radicals

Usually, the n^\text{th} Roots of n^\text{th} Powers Property is not enough to simplify radical expressions. Therefore, more properties will be presented and proven in this lesson. The first one refers to the root of a product.

Rule

Product Property of Radicals

Given two non-negative numbers a and b, the n^\text{th} root of their product equals the product of the n^\text{th} root of each number.


sqrt(ab) = sqrt(a)* sqrt(b), for a≥ 0 and b≥ 0

If n is an odd number, the n^\text{th} root of a negative number is defined. In this case, the Product Property of Radicals for negative a and b is also true.

Proof

First, a special case of the property will be proven. Assume that a or b is equal to 0. By the Zero Property of Multiplication, the radicand in left-hand side of the formula is equal to 0. \begin{array}{ccc} \underline\textbf{Left-Hand Side} & & \underline\textbf{Right-Hand Side} \\[0.5em] \sqrt[n]{0} & \stackrel{?}{=} & \sqrt[n]{a}\cdot \sqrt[n]{b} \end{array} Because 0^n = 0, n^\text{th} root of 0 is equal to 0. This means that both the left-hand side and one of the factors on the right-hand side equal 0. Again, by the Zero Property of Multiplication, the right-hand side is also 0. \begin{array}{ccc} \underline\textbf{Left-Hand Side} & & \underline\textbf{Right-Hand Side} \\[0.5em] 0 & = & 0 \end{array} Therefore, the property is true when a or b is equal to 0. Next, assume that both a and b are nonzero. Let x, y, and z be real numbers such that x = sqrt(a), y = sqrt(b), and z = sqrt(ab). By the definition of an n^\text{th} root, each of these numbers raised to the n^\text{th} power is equal to its corresponding radicand. x^n = a & (I) y^n = b & (II) z^n = ab & (III) Next, because b is not equal to 0, Equation (I) can be multiplied by y^n, which is equal to b. x^n = a ⇓ x^n y^n = a y^n Now, substitute Equations (II) and (III) into the obtained equation.
x^n y^n = a y^n
Substitute values and simplify
x^n y^n = a b
x^n y^n = z^n
(xy)^n = z^n
z^n = (xy)^n
Since xy and z are of the same sign, the final equation implies that z=xy. z^n=(xy)^n ⇒ z=xy The last step is substituting z=sqrt(ab), x=sqrt(a), and y=sqrt(b) into this equation. z=xy ⇔ sqrt(ab)=sqrt(a)* sqrt(b)

The following property indicates how to work with roots of a quotient.

Rule

Quotient Property of Radicals

Let a be a non-negative number and b be a positive number. The n^\text{th} root of the quotient ab equals the quotient of the n^\text{th} roots of a and b.


sqrt(a/b) = sqrt(a)/sqrt(b), for a≥ 0, b > 0

If n is an odd number, the n^\text{th} root of a negative number is defined. In this case, the Quotient Property of Radicals for negative a and b is also true.

Proof

First, a special case of the property will be proven. Assume that a is equal to 0. Because 0b=0, the radicand in left-hand side of the formula is equal to 0. \begin{array}{ccc} \underline\textbf{Left-Hand Side} & & \underline\textbf{Right-Hand Side} \\[0.5em] \sqrt[n]{0} & \stackrel{?}{=} & \dfrac{\sqrt[n]{0}}{\sqrt[n]{b}} \end{array} Because 0^n = 0, n^\text{th} root of 0 is equal to 0. This means that both the left-hand side and the numerator on the right-hand side equal 0. Again, since dividing 0 by a nonzero number results in 0, the right-hand side is also 0. \begin{array}{ccc} \underline\textbf{Left-Hand Side} & & \underline\textbf{Right-Hand Side} \\[0.5em] 0 & \stackrel{{\color{#009600}{\checkmark}}}{=} & 0 \end{array} Therefore, the property is true when a is equal to 0. Next, assume that both a and b are nonzero. Let x, y, and z be real numbers such that x = sqrt(a), y = sqrt(b), and z = sqrt(ab). By the definition of an n^\text{th} root, each of these numbers raised to the n^\text{th} power is equal to its corresponding radicand. x^n = a & (I) y^n = b & (II) z^n = ab & (III) Next, because b is not equal to 0, Equation (I) can be divided by y^n, which is equal to b. x^n = a ⇓ x^n/y^n = a/y^n Now, substitute Equations (II) and (III) into the obtained equation.
x^n/y^n = a/y^n
Substitute values and simplify
x^n/y^n = a/b
x^n/y^n = z^n

a^m/b^m=(a/b)^m

(x/y)^n = z^n
z^n = (x/y)^n
Since xy and z are of the same sign, the final equation implies that z= xy. z^n=(x/y)^n ⇒ z=x/y The last step is substituting z=sqrt(ab), x=sqrt(a), and y=sqrt(b) into this equation. z=x/y ⇔ sqrt(a/b)=sqrt(a)/sqrt(b)
Example

The Voltage in an Electric Circuit

To work on physics experiments in his astronomical observatory, Ignacio needs the right lighting for the new workstation. He has already designed a simple electric circuit for a 48-watt light bulb.

A simple electric circuit representing the situation

The voltage V required for a circuit is given by V = sqrt(P)*sqrt(R). In this formula, P is the power in watts and R is the resistance in ohms.

a What voltage of battery is needed to light the light bulb if its resistance is 12 ohms?
b Does the battery need to have a higher, lower, or the same voltage in order to light a 20-watt light bulb with a 32-ohm resistance?

Hint

a Substitute the given values into the formula for the voltage V. Then, evaluate the expression using the Product Property of Radicals.
b Similar to Part A, evaluate the voltage needed to light the other light bulb. How is the lower bound of the obtained root calculated?

Solution

a The formula for the voltage V required for a circuit is given. In this equation, P represents the power in watts and R represents the resistance in ohms.
V=sqrt(P)*sqrt(R) The circuit connects the battery to a 48-watt light bulb with a resistance of 12 ohms. The voltage V_1 needed to light the bulb is calculated by substituting P= 48 and R= 12 into the given formula.
V_1=sqrt(P)*sqrt(R)
V_1=sqrt(48)*sqrt(12)
Evaluate right-hand side
V_1 = sqrt(48*12)
V_1=sqrt(576)
V_1 = 24
Therefore, a 24-volt battery is needed to power the light bulb.
b In Part A, it was found that the voltage needed to light a 48-watt light bulb is 24 volts. Now, the number of volts V_2 needed to light a 20-watt light bulb with a resistance of 32 ohms will be calculated. To do so, substitute P= 20 and R= 32 into the given formula and simplify.
V_2=sqrt(P) * sqrt(R)
V_2=sqrt(20)* sqrt(32)
Evaluate right-hand side
V_2 = sqrt(20* 32)
V_2=sqrt(640)
V_2=25.298221...
V_2≈ 25.3
This means that a battery with a higher voltage is needed to light the 20-watt light bulb.
Example

Calculating the Length of the Diagonal in TV Screen

Ignacio wants to organize a movie night to celebrate the grand opening of his astronomical observatory. He plans to buy a brand new TV for the occasion, but he does not know what size of TV screen will fit on his wall.

TV screen with a 16:9 aspect ratio

The shape of a TV screen is represented by its aspect ratio, which is the ratio of the width of a screen to its height. The most common aspect ratio for TV screens is 16:9, which means that the width of the screen is 169 times its height.

a Write an expression for the width of a TV screen with this aspect ratio in terms of the length of the diagonal d.
b Find the area of the screen if the length of the diagonal is sqrt(3033) inches.

Hint

a The length of the diagonal of a rectangle can by found by using the Pythagorean Theorem.
b Use the expression found in Part A. The area of a rectangle is calculated by multiplying its length times its width.

Solution

a A TV screen can be modeled by a rectangle. Let w and h denote the width and height of the rectangle, respectively. Since it is given that the aspect ratio of the screen is 16:9, the height can be written in terms of the width.

w = 16/9h ⇕ h = 9/16w Next, the height and the width of the rectangle will be labeled on a diagram. The length of the diagonal d will be marked in the rectangle as well.

A rectangle modeling a TV screen with a marked diagonal
Because two sides and the diagonal of the rectangle form a right triangle, the length of the diagonal can be related to the width by applying the Pythagorean Theorem. d^2 = w^2 + (9/16w )^2 Now, the equation will be solved for w. Keep in mind that the absolute value of a positive number is equal to itself.
d^2 = w^2 + (9/16w )^2
Solve for w
d^2 = w^2 + (9/16 )^2 w^2
d^2 = w^2 + 81/256 w^2
d^2 = w^2 + 81w^2/256
d^2 = 256w^2/256 + 81w^2/256
d^2 = 256w^2+81w^2/256
d^2 = 337w^2/256
d^2 = 337/256w^2
256/337d^2 = w^2
w^2 = 256/337d^2
sqrt(w^2) = sqrt(256/337d^2)
|w| = sqrt(256/337d^2)
|w| = sqrt(256/337)sqrt(d^2)
|w| = sqrt(256)/sqrt(337)sqrt(d^2)
|w| = 16/sqrt(337)sqrt(d^2)
|w| = 16/sqrt(337)|d|
w = 16/sqrt(337)d
b It is given that the length of the diagonal is sqrt(3033) inches. Therefore, d=sqrt(3033) will be substituted into the equation written in Part A to calculate the width of the rectangle.
w = 16/sqrt(337)d
w = 16/sqrt(337) sqrt(3033)
Evaluate right-hand side

a/c* b =a * b/c

w = 16*sqrt(3033)/sqrt(337)
w = 16*sqrt(3033/337)
w = 16*sqrt(9)
w = 16* 3
w = 48
The width of the rectangle is 48 inches. The area A of the rectangle is calculated by multiplying the rectangle's width by its height. The height will now be found by using the aspect ratio of the TV screen. Substitute w = 48 into the formula h= 916w.
h = 9/16w
h = 9/16( 48)
Evaluate right-hand side

a/c* b =a * b/c

h = 9*48/16
h = 9* 3
h = 27
Finally, the area of the TV screen can be found.
A = w* h
A = 48* 27
A = 1296
Therefore, the area of the TV screen is 1296 square inches.
Example

The Surface Area of the Earth

Ignacio wants to decorate his observatory by hanging a model of the solar system on the ceiling. He has already bought some of the planets, which are modeled by gleaming spheres. The volume of the miniature Earth is 4π3 cubic inches.

Image of the Earth

The volume of a sphere is given by the formula V= 43π r^3. In this formula, r is the radius of the sphere. Ignacio wants to find the surface area of the model to approximate the surface area of the Earth by using the model scale.

a Using the formula for the surface area of a sphere, S=4π r^2, represent S in terms of the volume V. Choose the correct answer written as a product of radicals.
b Calculate the surface area of the model of the Earth.

Hint

a Solve the formula for the volume of a sphere for r. Then, substitute the expression for r into the formula for the surface area of a sphere.
b Substitute the given volume into the formula written in Part A. Use the properties of radicals to calculate the surface area.

Solution

a The formulas for both the volume and the surface area of a sphere are given. Note that both quantities depend on the radius of the sphere.
Volume:& V= 43π r^3 Surface Area:& S=4π r^2 Therefore, the formula for the volume can be solved for r, which can then be substituted into the formula for the surface area. This will give the formula for the surface area in terms of the volume.
V=4/3π r^3
Solve for r
V=4π/3 r^3
3/4πV = r^3
3V/4π = r^3
r^3 = 3V/4π
sqrt(r^3) = sqrt(3V/4π)
r = sqrt(3V/4π)
Now, the rewritten expression in terms of r will be substituted into the formula for the surface area S=4π r^2. Use the properties of exponents and radicals to simplify the formula.
S = 4π r^2
S = 4π( sqrt(3V/4π) )^2
Simplify right-hand side
S = 4π( (3V/4π)^()13 )^2
S = 4π (3V/4π)^(13* 2)
S = 4π (3V/4π)^(23)
S = 4π (3V)^(23)/(4π)^(23)

a * b/c= a/c* b

S = 4π/(4π)^(23) (3V)^(23)

a=a^1

S = (4π)^1/(4π)^(23) (3V)^(23)
S = (4π)^(1- 23) (3V)^(23)
S = (4π)^(33- 23) (3V)^(23)
S = (4π)^(13) (3V)^(23)
S = (4π)^(13) (3V)^(2* 13)
S = (4π)^(13) ((3V)^2)^()13
S = sqrt(4π) * sqrt((3V)^2)
This corresponds to option C.
b The volume of the model of the Earth is given as 4π3 cubic inches. The surface area of the model can be calculated by substituting V = 4π3 into the formula found in Part A.
S = sqrt(4π) * sqrt((3V)^2)
S = sqrt(4π) * sqrt((3( 4π/3 ))^2)
Evaluate right-hand side
S = sqrt(4π) * sqrt((4π)^2)
S = sqrt(4π(4π)^2)
S = sqrt((4π)^3)
S = 4π
Therefore, the surface area of the model Earth is 4π square inches.
Discussion

Irrational Conjugate of an n^\text{th} Root

Some irrational numbers are written as expressions involving n^(th) roots of powers. Previously, the Product Property of Radicals and the n^\text{th} Roots of n^\text{th} Powers Property were used to eliminate the root. Consider the following example. 3sqrt(16) = 3sqrt(4^2) Given an irrational number in this form, it is possible to find another irrational number by changing the power in the radicand.

Concept

Irrational Conjugate of an n^\text{th} Root

Let a, b, and x be rational numbers and n a natural number so that the n^\text{th} root sqrt(b^x) is irrational. The irrational conjugate of an n^\text{th} root asqrt(b^x) is obtained by raising b to the power of n-x in the radicand. ccc Number & & Conjugate [0.15cm] asqrt(b^x) & Change Power & asqrt(b^(n-x)) For example, the conjugate of 4sqrt(7^2) is 4sqrt(7^3). 4sqrt(7^2) Change Power 4sqrt(7^(5-2)) = 4sqrt(7^3)

The irrational conjugate can be used to rationalize a denominator containing an n^\text{th} root.
In the next part of this lesson, the conjugate will be used to rationalize the denominator of a numeric expression involving an n^\text{th} root.
Pop Quiz

Determining the Irrational Conjugate of a Radical Expression

Discussion

Rationalizing a Denominator Involving an n^\text{th} Root

When a numeric expression has a denominator that involves an irrational n^\text{th} root, it is convenient to rewrite the expression so that the denominator is a rational number. This process is known as rationalization. Consider the following example. 4/sqrt(27) Since the denominator involves an n^\text{th} root, the fraction can be rewritten by multiplying both the numerator and denominator by the irrational conjugate of the denominator. The resulting expression can then be simplified.
1
Multiply by the Conjugate of the Denominator
expand_more

First, the radicand should be rewritten as a power. 4/sqrt(27) = 4/sqrt(3^3) To get rid of the radical expression in the denominator, the numerator and the denominator are multiplied by its irrational conjugate. Since the radicand involves the third power and the index of the root is 5, multiply by sqrt(3^(5 - 3)) = sqrt(3^2). 4* sqrt(3^2)/sqrt(3^3)* sqrt(3^2)

2
Simplify the Expression
expand_more
The obtained expression can now be simplified.
4*sqrt(3^2)/sqrt(3^3)* sqrt(3^2)
Simplify
4*sqrt(3^2)/sqrt(3^3* 3^2)
4*sqrt(3^2)/sqrt(3^(3+2))
4*sqrt(3^2)/sqrt(3^5)
4*sqrt(3^2)/3
4sqrt(9)/3
The denominator has now been rationalized because it contains an integer number and no roots.

Notice that this method also works when the denominator is the product of two roots with different indexes. In these cases, the method should be applied twice. However, if the denominator involves a sum of two n^\text{th} roots with different indexes, rationalizing is a more complicated task.

Example

Calculating the Circular Velocity of a Satellite

While reading some research papers on astronomy, Ignacio came across the fact that the circular velocity v of a satellite circling the Earth is given by the following formula. v = sqrt(K/r) In this formula, v is measured in miles per hour, r is the distance in miles from the satellite to the center of the Earth, and K is a constant depending on the gravitational constant and the mass of the Earth.

A satellite circling the Earth
a Rationalize the denominator of the formula for the velocity v of a satellite when r is equal to sqrt(262 528^2). Choose the correct answer.
b A second satellite is orbiting the Earth at a distance (3-sqrt(2))^2 times greater than the distance of the satellite from Part A. Write the velocity v_2 of this second satellite in terms of v. Rationalize the denominator if needed.

Hint

a Rewrite the formula using the Quotient Property of Square Roots. Then, multiply both the numerator and denominator by the irrational conjugate of the n^\text{th} root.
b Write the distance from the second satellite in terms of r. Consider the quotient of the velocities to cancel out common factors.

Solution

a First, notice that the formula for the circular velocity v can be rewritten using the Quotient Property of Square Roots.
v = sqrt(K/r) = sqrt(K)/sqrt(r) It is given that the distance from the Earth to the satellite is sqrt(262 528^2) miles. Therefore, r = sqrt(262 528^2) will be substituted into the formula and the denominator rationalized.
v = sqrt(K)/sqrt(r)
v = sqrt(K)/sqrt(sqrt(262 528^2))
Rewrite
v = sqrt(K)* sqrt(sqrt(262 528^2))/sqrt(sqrt(262 528^2))* sqrt(sqrt(262 528^2))
v = sqrt(K)* sqrt(sqrt(262 528^2))/sqrt(262 528^2)
v = sqrt(K)* sqrt(sqrt(262 528^2))* sqrt(262 528^1)/sqrt(262 528^2) * sqrt(262 528^1)
v = sqrt(K)* sqrt(sqrt(262 528^2))* sqrt(262 528^1)/sqrt(262 528^2 *262 528^1)
v = sqrt(K)* sqrt(sqrt(262 528^2))* sqrt(262 528^1)/sqrt(262 528^3)
v = sqrt(K)* sqrt(sqrt(262 528^2))* sqrt(262 528^1)/262 528
Now that the denominator has been rationalized, the numerator of the obtained expression will be further simplified.
v = sqrt(K)* sqrt(sqrt(262 528^2))* sqrt(262 528^1)/262 528
Simplify right-hand side
v = sqrt(K)* ( sqrt(262 528^2))^()12* sqrt(262 528^1)/262 528
v = sqrt(K)* ( ( 262 528^2 )^()13)^()12* sqrt(262 528^1)/262 528

(a^m)^n= (a^n)^m

v = sqrt(K)* ( ( 262 528^2 )^()12)^()13* sqrt(262 528^1)/262 528
v = sqrt(K)* ( 262 528^(2* 12) )^()13* sqrt(262 528^1)/262 528

a * 1/a=1

v = sqrt(K)* ( 262 528^1 )^()13* sqrt(262 528^1)/262 528
v = sqrt(K)* ( 262 528 )^()13* sqrt(262 528)/262 528
v = sqrt(K)* sqrt(262 528) * sqrt(262 528)/262 528
v = sqrt(K)* sqrt(262 528 * 262 528)/262 528
v = sqrt(K)* sqrt(262 528^2)/262 528
Now that the denominator contains only an integer number, it has been successfully rationalized. The calculated velocity corresponds to option B.
b It is given that the distance r_2 of the second satellite to the center of the Earth is (3-sqrt(2))^2 times greater than the distance r of the first satellite considered in Part A.
r_2 = (3-sqrt(2))^2 r To write v_2 in terms of v, start with the formula for v_2 by using the given formula. v_2 = sqrt(K/r_2) Next, the Division Property of Equality will be applied to divide both sides of the equation by v. This will allow the expression to be written in terms of v. Then, v=sqrt(Kr) can be substituted on the right-hand side. v_2/v = sqrt(Kr_2)/v ⇒ v_2/v = sqrt(Kr_2)/sqrt(Kr) Now, substitute (3-sqrt(2))^2 r for r_2 in the obtained quotient and evaluate the right-hand side.
v_2/v = sqrt(Kr_2)/sqrt(Kr)
v_2/v = sqrt(K(3-sqrt(2))^2 r)/sqrt(Kr)
Evaluate right-hand side
v_2/v = sqrt(K)sqrt((3-sqrt(2))^2 r)/sqrt(K)sqrt(r)
v_2/v = sqrt(K)/sqrt((3-sqrt(2))^2 r) * sqrt(r)/sqrt(K)
v_2/v = sqrt(K) * sqrt(r)/sqrt((3-sqrt(2))^2 r) * sqrt(K)
v_2/v = sqrt(K) * sqrt(r)/sqrt((3-sqrt(2))^2) * sqrt(r) * sqrt(K)
v_2/v = sqrt(K) * sqrt(r)/sqrt((3-sqrt(2))^2) * sqrt(r) * sqrt(K)
v_2/v = 1/sqrt((3-sqrt(2))^2)
v_2/v = 1/| 3-sqrt(2)|
v_2/v = 1/3-sqrt(2)
v_2/v = 1(3+sqrt(2))/(3-sqrt(2)) (3+sqrt(2))
v_2/v = 3+sqrt(2)/(3-sqrt(2)) (3+sqrt(2))

(a-b)(a+b)=a^2-b^2

v_2/v = 3+sqrt(2)/3^2 - ( sqrt(2) )^2
v_2/v = 3+sqrt(2)/9 - ( sqrt(2) )^2
v_2/v = 3+sqrt(2)/9-2
v_2/v = 3+sqrt(2)/7
v_2 = 3+sqrt(2)/7 v
Finally, it can concluded that v_2 is equal to 3+sqrt(2)7 v.
Discussion

Adding and Subtracting Like Radical Expressions and Roots

Concept

Like Radical Expressions

Radical expressions are called like radical expressions or like radicals if both the index and the radicand of the corresponding roots are identical.

Examples of like and unlike radical expressions

Use the Distributive Property to add or subtract like radical expressions.


asqrt(x) + bsqrt(x) = (a + b) sqrt(x)
asqrt(x) - bsqrt(x) = (a - b) sqrt(x)

Rule

Adding and Subtracting Radicals

A numeric or algebraic expression that contains two or more radical terms with the same radicand and the same index — called like radical expressions — can be simplified by adding or subtracting the corresponding coefficients.


asqrt(x)± bsqrt(x)=(a± b)sqrt(x)

Here, a,b, and x are real numbers and n is a natural number. If n is even, then x must be greater than or equal to zero.

Proof

If n is an even number, then x is greater than or equal to zero. If n is odd, then x can be any real number. In both cases, sqrt(x) is a real number. This means that the Distributive Property can be used to factor out sqrt(x).
asqrt(x)± bsqrt(x)
(a± b)sqrt(x)
Method

Adding and Subtracting Radicals

Two radicals can be added or subtracted if it is possible to rewrite them as like radical expressions. Use the Distributive Property to add or subtract like radicals.


asqrt(x) + bsqrt(x) = ( a + b) sqrt(x)
asqrt(x) - bsqrt(x) = ( a - b) sqrt(x)

For example, consider the following radical expressions. sqrt(x^5 y^3) and sqrt(16xy^7) In order to add these expressions, there are three steps to follow. Note that subtraction of the radicals can be performed by applying the same three steps, only instead of adding the like radicals, they will be subtracted.
1
Simplify the Radicals to Have the Same Index and Radicand
expand_more
Start by rewriting the expressions as like radicals. To do so, split the radicands into factors and use the properties of radicals.
sqrt(x^5 y^3) + sqrt(16xy^7)
Rewrite
sqrt(x^5 * y^3) + sqrt(16* x* y^7)
sqrt(x^(4+1) * y^3) + sqrt(16* x* y^(4+3))
sqrt(x^4 * x^1 * y^3) + sqrt(16* x* y^4* y^3)
sqrt(x^4 * x^1 * y^3) + sqrt(16 * y^4* x* y^3)
sqrt(x^4) * sqrt(x^1 * y^3) + sqrt(16 * y^4)* sqrt(x* y^3)
sqrt(x^4) * sqrt(x * y^3) + sqrt(16 * y^4)* sqrt(x* y^3)

sqrt(a^4)=|a|

|x| * sqrt(x * y^3) + sqrt(16 * y^4)* sqrt(x* y^3)
|x| * sqrt(x * y^3) + sqrt(2^4 * y^4)* sqrt(x* y^3)
|x| * sqrt(x * y^3) + sqrt((2y)^4)* sqrt(x* y^3)

sqrt(a^4)=|a|

|x| * sqrt(x * y^3) + |2y|* sqrt(x* y^3)
|x| sqrt(xy^3) + |2y| sqrt(xy^3)
Since both the index and the radicand of the roots are identical, they are like radical expressions.
2
Add or Subtract Like Radicals
expand_more

Now, like radical expressions can by added by using the Distributive Property. |x| sqrt(xy^3) + |2y| sqrt(xy^3) [0.3em] ⇕ [0.3em] (|x| + |2y|)sqrt(xy^3)

3
Simplify the Result as Necessary
expand_more

The signs of the x- and y-variables are not given. This means that the expression in parentheses cannot be further simplified and the result of the addition is as follows. (|x| + |2y|)sqrt(xy^3)

Example

The Area of Triangles on a Door

The last step in designing the observatory is to come up with a new logo. Ignacio has sketched the following prototype of his logo.

The logo of an astronomical observatory
The logo is made of equilateral triangles. The side of a large triangle is 6 inches and the side of a small triangle is sqrt(32) inches. Find the exact total area of the design.

Hint

How many large and small triangles form the logo? Use the formula for the area of a triangle using sine. Also, keep in mind that sin 60^(∘) = sqrt(3)2.

Solution

It is given that the logo is made of two types of equilateral triangles. To find the area of the design, start by determining the number of triangles of each type.
Counting the number of the logo of an astronomical observatory
There are 6 large and 4 small triangles. Now, the area of a single triangle of each type will be found. Recall the formula for the area of a triangle involving the sine ratio. Start with the area A_1 of a large triangle. Area = 1/2absin C The length of all sides is 6 inches and the measure of all angles in the triangle is 60^(∘). This means that a= 6, b= 6, and C = 60^(∘) can be substituted into the formula to find the area.
A_1 = 1/2absin C
A_1 = 1/2( 6)( 6)sin 60^(∘)
Evaluate right-hand side
A_1 = 1/2(36) sin 60^(∘)
A_1 = 36/2 sin 60^(∘)

\ifnumequal{60}{0}{\sin\left(0^\circ\right)=0}{}\ifnumequal{60}{30}{\sin\left(30^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{60}{45}{\sin\left(45^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{60}{60}{\sin\left(60^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{60}{90}{\sin\left(90^\circ\right)=1}{}\ifnumequal{60}{120}{\sin\left(120^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{60}{135}{\sin\left(135^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{60}{150}{\sin\left(150^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{60}{180}{\sin\left(180^\circ\right)=0}{}\ifnumequal{60}{210}{\sin\left(210^\circ\right)=\text{-} \dfrac 1 2}{}\ifnumequal{60}{225}{\sin\left(225^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{60}{240}{\sin\left(240^\circ\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{60}{270}{\sin\left(270^\circ\right)=\text{-}1}{}\ifnumequal{60}{300}{\sin\left(300^\circ\right)=\text{-}\dfrac {\sqrt 3}2}{}\ifnumequal{60}{315}{\sin\left(315^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{60}{330}{\sin\left(330^\circ\right)=\text{-} \dfrac 1 2}{}\ifnumequal{60}{360}{\sin\left(360^\circ\right)=0}{}

A_1 = 36/2 ( sqrt(3)/2 )
A_1 = 36sqrt(3)/2(2)
A_1 = 36sqrt(3)/4
A_1 = 9sqrt(3)
Similarly, the area A_2 of a small triangle can be calculated. In this case, the side length is sqrt(32) inches and the measure of the included angle is 60^(∘) as previously.
A_2 = 1/2absin C
A_2 = 1/2( sqrt(32))( sqrt(32))sin 60^(∘)
Evaluate right-hand side
A_2 = 1/2( sqrt(1024) ) sin 60^(∘)
A_2 = sqrt(1024)/2 sin 60^(∘)

\ifnumequal{60}{0}{\sin\left(0^\circ\right)=0}{}\ifnumequal{60}{30}{\sin\left(30^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{60}{45}{\sin\left(45^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{60}{60}{\sin\left(60^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{60}{90}{\sin\left(90^\circ\right)=1}{}\ifnumequal{60}{120}{\sin\left(120^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{60}{135}{\sin\left(135^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{60}{150}{\sin\left(150^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{60}{180}{\sin\left(180^\circ\right)=0}{}\ifnumequal{60}{210}{\sin\left(210^\circ\right)=\text{-} \dfrac 1 2}{}\ifnumequal{60}{225}{\sin\left(225^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{60}{240}{\sin\left(240^\circ\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{60}{270}{\sin\left(270^\circ\right)=\text{-}1}{}\ifnumequal{60}{300}{\sin\left(300^\circ\right)=\text{-}\dfrac {\sqrt 3}2}{}\ifnumequal{60}{315}{\sin\left(315^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{60}{330}{\sin\left(330^\circ\right)=\text{-} \dfrac 1 2}{}\ifnumequal{60}{360}{\sin\left(360^\circ\right)=0}{}

A_2 = sqrt(1024)/2 ( sqrt(3)/2 )
A_2 = sqrt(1024)* sqrt(3)/4
In this expression, it is important not to make a mistake by using the Product Property of Radicals. The indices of roots are not equal, so they cannot be combined. Next, the total area A can be found by adding 6 times the area of a large triangle to 4 times the area of a small triangle. A = 6A_1 + 4A_2 Finally, substitute the simplified areas of a single triangle of each type and evaluate the area by adding like radical expressions.
A = 6A_1 + 4A_2
A = 6( 9sqrt(3) ) + 4( sqrt(1024)* sqrt(3)/4 )
Evaluate right-hand side
A = 54sqrt(3) + 4( sqrt(1024)* sqrt(3)/4 )
A = 54sqrt(3) + sqrt(1024)* sqrt(3)
A = ( 54 + sqrt(1024)) sqrt(3)
A = ( 54 + sqrt(512* 2)) sqrt(3)
A = ( 54 + sqrt(512)* sqrt(2)) sqrt(3)
A = ( 54 + 8sqrt(2)) sqrt(3)
Therefore, the total area of the logo is ( 54 + 8sqrt(2)) sqrt(3) square inches.
Closure

Finding the Length of a Fencing

In the challenge presented at the beginning of this lesson, the dimensions of Ignacio's garden were given. He wants to fence in a triangular area of the garden in which to build his observatory. Notice that some side lengths are missing in the diagram. They can be calculated by using the given lengths. Also, unknown side lengths of an interior triangles will be marked.

Added Measures to Ignacios Garden
The length of fencing required to build the observatory is the perimeter of an inscribed triangle. Each side of this triangle is the hypotenuse of a right triangle. Since the length of all these triangle's legs are known, their hypotenuse can be found by applying the Pythagorean Theorem one at a time.
(l_1)^2 = (10-x)^2 + ( sqrt(40x) )^2
Solve for l_1
(l_1)^2 = 10^2-2(10)x+x^2 + ( sqrt(40x) )^2
(l_1)^2 = 100-20x+x^2 + ( sqrt(40x) )^2
(l_1)^2 = 100-20x+x^2 + 40x
(l_1)^2 = 100+20x+x^2
(l_1)^2 = 10^2+20x+x^2
(l_1)^2 = 10^2+2(10)x+x^2
(l_1)^2 = (10+x)^2
sqrt((l_1)^2) = sqrt((10+x)^2)
|l_1| = |10+x|
l_1 = |10+x|
Since x is one of side lengths, it must be positive. Consequently, 10+x is also positive and l_1 is equal to 10+x. Next, l_2 will be calculated.
(l_2)^2 = 10^2 + ( sqrt(360x) - sqrt(40x) )^2
Solve for l_2
(l_2)^2 = 100 + ( sqrt(360x) - sqrt(40x) )^2
(l_2)^2 = 100 + ( sqrt(36* 10x) - sqrt(4* 10x) )^2
(l_2)^2 = 100 + ( sqrt(36)* sqrt(10x) - sqrt(4)* sqrt(10x) )^2
(l_2)^2 = 100 + ( 6sqrt(10x) - 2sqrt(10x) )^2
(l_2)^2 = 100 + ( 4sqrt(10x) )^2
(l_2)^2 = 100 + 4^2( sqrt(10x))^2
(l_2)^2 = 100 + 16( sqrt(10x))^2
(l_2)^2 = 100 + 16(10x)
(l_2)^2 = 100 + 160x
sqrt((l_2)^2) = sqrt(100 + 160x)
|l_2| = sqrt(100+160x)
l_2 = sqrt(100+160x)
l_2 = sqrt(4(25+40x))
l_2 = sqrt(4)*sqrt(25+40x)
l_2 = 2sqrt(25+40x)
Now, the last side length will be calculated.
(l_3)^2 = x^2 + ( sqrt(360x) )^2
Solve for l_3
(l_3)^2 = x^2 + 360x
sqrt((l_3)^2) = sqrt(x^2 + 360x)
|l_3| = sqrt(x^2 + 360x)
l_3 = sqrt(x^2 + 360x)
The length of the fencing required is given by the perimeter of the inscribed triangle. The perimeter L(x) will be calculated by summing up the side lengths. L(x) = l_1 + l_2 + l_3 ⇓ L(x) = 10 + x + 2sqrt(25 + 40x) + sqrt(x^2 + 360x) Unfortunately, since there are no like radicals, the obtained expression cannot be further simplified. Finally, substitute x= 5 into the expression to find L(5).
L(x) = 10 + x + 2sqrt(25 + 40x) + sqrt(x^2 + 360x)
L( 5) = 10 + 5 + 2sqrt(25 + 40( 5)) + sqrt(5^2 + 360( 5))
Evaluate right-hand side
L(5) = 10 + 5 + 2sqrt(25+200) + sqrt(25 + 1800)
L(5) = 15 + 2sqrt(225) + sqrt(1825)
L(5) = 15 + 2(15) + sqrt(1825)
L(5) = 15 + 30 + sqrt(1825)
L(5) = 45 + sqrt(1825)
L(5) = 45 + sqrt(25* 73)
L(5) = 45 + sqrt(25)* sqrt(73)

Calculate root and product

L(5) = 45 + 5sqrt(73)
To conclude, if x is equal to 5 meters, the length of the fencing required for the observatory is 45+5sqrt(73) meters.



Operations With Radical Expressions
Exercises
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