Multiplying and Dividing Radical Expressions

5.4 Multiplying and Dividing Radical Expressions

Learning Objectives

Multiply radical expressions.
Divide radical expressions.
Rationalize the denominator.

Multiplying Radical Expressions

When multiplying radical expressions with the same index, we use the product rule for radicals. Given real numbers $\sqrt[n]{A}$ and $\sqrt[n]{B}$ ,

$\sqrt[n]{A} \cdot \sqrt[n]{B} = \sqrt[n]{A \cdot B}$

Example 1

Multiply: $\sqrt[3]{12} \cdot \sqrt[3]{6} .$

Solution:

Apply the product rule for radicals, and then simplify.

$\begin{array}{l} \sqrt[3]{12} \cdot \sqrt[3]{6} & = & \sqrt[3]{12 \cdot 6} & M u l t i p l y t h e r a d i c a n d s . \\ = & \sqrt[3]{72} & S i m p l i f y . \\ = & \sqrt[3]{2^{3} \cdot 3^{2}} \\ = & 2 \sqrt[3]{3^{2}} \\ = & 2 \sqrt[3]{9} \end{array}$

Answer: $2 \sqrt[3]{9}$

Often, there will be coefficients in front of the radicals.

Example 2

Multiply: $3 \sqrt{6} \cdot 5 \sqrt{2}$

Solution:

Using the product rule for radicals and the fact that multiplication is commutative, we can multiply the coefficients and the radicands as follows.

$\begin{array}{l} 3 \sqrt{6} \cdot 5 \sqrt{2} & = & 3 \cdot 5 \cdot \sqrt{6} \cdot \sqrt{2} & M u l t i p l i c a t i o n i s c o m m u t a t i v e . \\ = & 15 \cdot \sqrt{12} & M u l t i p l y t h e c o e f f i c i e n t s a n d \\ t h e r a d i c a n d s . \\ = & 15 \sqrt{4 \cdot 3} & S i m p l i f y . \\ = & 15 \cdot 2 \cdot \sqrt{3} \\ = & 30 \sqrt{3} \end{array}$

Typically, the first step involving the application of the commutative property is not shown.

Answer: $30 \sqrt{3}$

Example 3

Multiply: $- 3 \sqrt[3]{4 y^{2}} \cdot 5 \sqrt[3]{16 y} .$

Solution:

$\begin{array}{l} − 3 \sqrt[3]{4 y^{2}} \cdot 5 \sqrt[3]{16 y} & = & − 15 \sqrt[3]{64 y^{3}} & M u l t i p l y t h e c o e f f i c i e n t s a n d t h e n m u l t i p l y t h e r a d i c a n d s . \\ = & − 15 \sqrt[3]{4^{3} y^{3}} & S i m p l i f y . \\ = & − 15 \cdot 4 y \\ = & − 60 y \end{array}$

Answer: $− 60 y$

Use the distributive property when multiplying rational expressions with more than one term.

Example 4

Multiply: $5 \sqrt{2 x} (3 \sqrt{x} - \sqrt{2 x})$ .

Solution:

Apply the distributive property and multiply each term by $5 \sqrt{2 x} .$

$\begin{array}{l} 5 \sqrt{2 x} (3 \sqrt{x} - \sqrt{2 x}) & = & 5 \sqrt{2 x} \cdot 3 \sqrt{x} - 5 \sqrt{2 x} \cdot \sqrt{2 x} & D i s t r i b u t e . \\ = & 15 \sqrt{2 x^{2}} - 5 \sqrt{4 x^{2}} & S i m p l i f y . \\ = & 15 x \sqrt{2} - 5 \cdot 2 x \\ = & 15 x \sqrt{2} - 10 x \end{array}$

Answer: $15 x \sqrt{2} - 10 x$

Example 5

Multiply: $\sqrt[3]{6 x^{2} y} (\sqrt[3]{9 x^{2} y^{2}} - 5 \cdot \sqrt[3]{4 x y}) .$

Solution:

Apply the distributive property, and then simplify the result.

$\begin{matrix} \sqrt[3]{6 x^{2} y} (\sqrt[3]{9 x^{2} y^{2}} - 5 \cdot \sqrt[3]{4 x y}) & = & \sqrt[3]{6 x^{2} y} \cdot \sqrt[3]{9 x^{2} y^{2}} - \sqrt[3]{6 x^{2} y} \cdot 5 \sqrt[3]{4 x y} \\ = & \sqrt[3]{54 x^{4} y^{3}} - 5 \sqrt[3]{24 x^{3} y^{2}} \\ = & \sqrt[3]{27 \cdot 2 \cdot x \cdot x^{3} \cdot y^{3}} - 5 \sqrt[3]{8 \cdot 3 \cdot x^{3} \cdot y^{2}} \\ = & 3 x y \sqrt[3]{2 x} - 5 \cdot 2 x \sqrt[3]{3 y^{2}} \\ = & 3 x y \sqrt[3]{2 x} - 10 x \sqrt[3]{3 y^{2}} \end{matrix}$

Answer: $3 x y \sqrt[3]{2 x} - 10 x \sqrt[3]{3 y^{2}}$

The process for multiplying radical expressions with multiple terms is the same process used when multiplying polynomials. Apply the distributive property, simplify each radical, and then combine like terms.

Example 6

Multiply: ${(\sqrt{x} - 5 \sqrt{y})}^{2} .$

Solution:

${(\sqrt{x} - 5 \sqrt{y})}^{2} = (\sqrt{x} - 5 \sqrt{y}) (\sqrt{x} - 5 \sqrt{y})$

Begin by applying the distributive property.

$\begin{matrix} = & \sqrt{x} \cdot \sqrt{x} + \sqrt{x} (− 5 \sqrt{y}) + (− 5 \sqrt{y}) \sqrt{x} + (- 5 \sqrt{y}) (− 5 \sqrt{y}) \\ = & \sqrt{x^{2}} - 5 \sqrt{x y} - 5 \sqrt{x y} + 25 \sqrt{y^{2}} \\ = & x - 10 \sqrt{x y} + 25 y \end{matrix}$

Answer: $x - 10 \sqrt{x y} + 25 y$

The binomials $(a + b)$ and $(a - b)$ are called conjugatesThe factors $(a + b)$ and $(a - b)$ are conjugates.. When multiplying conjugate binomials the middle terms are opposites and their sum is zero.

Example 7

Multiply: $(\sqrt{10} + \sqrt{3}) (\sqrt{10} - \sqrt{3}) .$

Solution:

Apply the distributive property, and then combine like terms.

\begin{array}{l} (\sqrt{10} + \sqrt{3}) (\sqrt{10} - \sqrt{3}) & = & \sqrt{10} \cdot \sqrt{10} + \sqrt{10} (- \sqrt{3}) + \sqrt{3} \cdot \sqrt{10} + \sqrt{3} (- \sqrt{3}) & D i s t r i b u t e . \\ = & \sqrt{100} - \sqrt{30} + \sqrt{30} - \sqrt{9} & S i m p l i f y . \\ = & 10 - \sqrt{30} + \sqrt{30} - 3 & o p p o s i t e s a d d t o 0 \\ = & 10 - 3 \\ = & 7 \end{array}

Answer: 7

It is important to note that when multiplying conjugate radical expressions, we obtain a rational expression. This is true in general

$\begin{matrix} (\sqrt{x} + \sqrt{y}) (\sqrt{x} - \sqrt{y}) & = & \sqrt{x^{2}} - \sqrt{x y} + \sqrt{x y} - \sqrt{y^{2}} \\ = & x - y \end{matrix}$

Alternatively, using the formula for the difference of squares we have,

$\begin{matrix} (a + b) (a - b) & = & a^{2} - b^{2} & D i f f e r e n c e o f s q u a r e s . \\ (\sqrt{x} + \sqrt{y}) (\sqrt{x} - \sqrt{y}) & = & {(\sqrt{x})}^{2} - {(\sqrt{y})}^{2} \\ = & x - y \end{matrix}$

Try this! Multiply: $(3 - 2 \sqrt{y}) (3 + 2 \sqrt{y}) .$ (Assume y is positive.)

Answer: $9 - 4 y$

(click to see video)

Dividing Radical Expressions

To divide radical expressions with the same index, we use the quotient rule for radicals. Given real numbers $\sqrt[n]{A}$ and $\sqrt[n]{B}$ ,

$\frac{\sqrt[n]{A}}{\sqrt[n]{B}} = \sqrt[n]{\frac{A}{B}}$

Example 8

Divide: $\frac{\sqrt[3]{96}}{\sqrt[3]{6}}$ .

Solution:

In this case, we can see that 6 and 96 have common factors. If we apply the quotient rule for radicals and write it as a single cube root, we will be able to reduce the fractional radicand.

$\begin{array}{l} \frac{\sqrt[3]{96}}{\sqrt[3]{6}} & = & \sqrt[3]{\frac{96}{6}} & A p p l y t h e q u o t i e n t r u l e f o r r a d i c a l s a n d r e d u c e t h e r a d i c a n d . \\ = & \sqrt[3]{16} & S i m p l i f y . \\ = & \sqrt[3]{8 \cdot 2} \\ = & 2 \sqrt[3]{2} \end{array}$

Answer: $2 \sqrt[3]{2}$

Example 9

Divide: $\frac{\sqrt{50 x^{6} y^{4}}}{\sqrt{8 x^{3} y}}$ .

Solution:

Write as a single square root and cancel common factors before simplifying.

$\begin{array}{l} \frac{\sqrt{50 x^{6} y^{4}}}{\sqrt{8 x^{3} y}} & = & \sqrt{\frac{50 x^{6} y^{4}}{8 x^{3} y}} & A p p l y t h e q u o t i e n t r u l e f o r r a d i c a l s a n d c a n c e l . \\ = & \sqrt{\frac{25 x^{3} y^{3}}{4}} & S i m p l i f y . \\ = & \frac{\sqrt{25 x^{3} y^{3}}}{\sqrt{4}} \\ = & \frac{5 x y \sqrt{x y}}{2} \end{array}$

Answer: $\frac{5 x y \sqrt{x y}}{2}$

Rationalizing the Denominator

When the denominator (divisor) of a radical expression contains a radical, it is a common practice to find an equivalent expression where the denominator is a rational number. Finding such an equivalent expression is called rationalizing the denominatorThe process of determining an equivalent radical expression with a rational denominator..

$\begin{array}{c} R a d i c a l e x p r e s s i o n & R a t i o n a l d e n o m i n a t o r \\ \frac{1}{\sqrt{2}} & = & \frac{\sqrt{2}}{2} \end{array}$

To do this, multiply the fraction by a special form of 1 so that the radicand in the denominator can be written with a power that matches the index. After doing this, simplify and eliminate the radical in the denominator. For example:

$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{4}} = \frac{\sqrt{2}}{2}$

Remember, to obtain an equivalent expression, you must multiply the numerator and denominator by the exact same nonzero factor.

Example 10

Rationalize the denominator: $\frac{\sqrt{2}}{\sqrt{5 x}} .$

Solution:

The goal is to find an equivalent expression without a radical in the denominator. The radicand in the denominator determines the factors that you need to use to rationalize it. In this example, multiply by 1 in the form $\frac{\sqrt{5 x}}{\sqrt{5 x}} .$

$\begin{array}{l} \frac{\sqrt{2}}{\sqrt{5 x}} & = & \frac{\sqrt{2}}{\sqrt{5 x}} \cdot \frac{\sqrt{5 x}}{\sqrt{5 x}} & M u l t i p l y b y \frac{\sqrt{5 x}}{\sqrt{5 x}} . \\ = & \frac{\sqrt{10 x}}{\sqrt{25 x^{2}}} & S i m p l i f y . \\ = & \frac{\sqrt{10 x}}{5 x} \end{array}$

Answer: $\frac{\sqrt{10 x}}{5 x}$

Sometimes, we will find the need to reduce, or cancel, after rationalizing the denominator.

Example 11

Rationalize the denominator: $\frac{3 a \sqrt{2}}{\sqrt{6 a b}} .$

Solution:

In this example, we will multiply by 1 in the form $\frac{\sqrt{6 a b}}{\sqrt{6 a b}} .$

$\begin{array}{l} \frac{3 a \sqrt{2}}{\sqrt{6 a b}} & = & \frac{3 a \sqrt{2}}{\sqrt{6 a b}} \cdot \frac{\sqrt{6 a b}}{\sqrt{6 a b}} \\ = & \frac{3 a \sqrt{12 a b}}{\sqrt{36 a^{2} b^{2}}} & S i m p l i f y . \\ = & \frac{3 a \sqrt{4 \cdot 3 a b}}{6 a b} \\ = & \frac{6 a \sqrt{3 a b}}{6 a b} & C a n c e l . \\ = & \frac{\sqrt{3 a b}}{b} \end{array}$

Notice that b does not cancel in this example. Do not cancel factors inside a radical with those that are outside.

Answer: $\frac{\sqrt{3 a b}}{b}$

Try this! Rationalize the denominator: $\sqrt{\frac{9 x}{2 y}} .$

Answer: $\frac{3 \sqrt{2 x y}}{2 y}$

(click to see video)

Up to this point, we have seen that multiplying a numerator and a denominator by a square root with the exact same radicand results in a rational denominator. In general, this is true only when the denominator contains a square root. However, this is not the case for a cube root. For example, $\frac{1}{\sqrt[3]{x}} \cdot \frac{\sqrt[3]{x}}{\sqrt[3]{x}} = \frac{\sqrt[3]{x}}{\sqrt[3]{x^{2}}}$ Note that multiplying by the same factor in the denominator does not rationalize it. In this case, if we multiply by 1 in the form of $\frac{\sqrt[3]{x^{2}}}{\sqrt[3]{x^{2}}}$ , then we can write the radicand in the denominator as a power of 3. Simplifying the result then yields a rationalized denominator.

$\frac{1}{\sqrt[3]{x}} = \frac{1}{\sqrt[3]{x}} \cdot \frac{\sqrt[3]{x^{2}}}{\sqrt[3]{x^{2}}} = \frac{\sqrt[3]{x^{2}}}{\sqrt[3]{x^{3}}} = \frac{\sqrt[3]{x^{2}}}{x}$

Therefore, to rationalize the denominator of a radical expression with one radical term in the denominator, begin by factoring the radicand of the denominator. The factors of this radicand and the index determine what we should multiply by. Multiply the numerator and denominator by the nth root of factors that produce nth powers of all the factors in the radicand of the denominator.

Example 12

Rationalize the denominator: $\frac{\sqrt[3]{2}}{\sqrt[3]{25}} .$

Solution:

The radical in the denominator is equivalent to $\sqrt[3]{5^{2}} .$ To rationalize the denominator, we need: $\sqrt[3]{5^{3}} .$ To obtain this, we need one more factor of 5. Therefore, multiply by 1 in the form of $\frac{\sqrt[3]{5}}{\sqrt[3]{5}} .$

$\begin{array}{l} \frac{\sqrt[3]{2}}{\sqrt[3]{25}} & = & \frac{\sqrt[3]{2}}{\sqrt[3]{5^{2}}} \cdot \frac{\sqrt[3]{5}}{\sqrt[3]{5}} & M u l t i p l y b y t h e c u b e r o o t o f f a c t o r s t h a t r e s u l t i n p o w e r s o f 3. \\ = & \frac{\sqrt[3]{10}}{\sqrt[3]{5^{3}}} & S i m p l i f y . \\ = & \frac{\sqrt[3]{10}}{5} \end{array}$

Answer: $\frac{\sqrt[3]{10}}{5}$

Example 13

Rationalize the denominator: $\sqrt[3]{\frac{27 a}{2 b^{2}}}$ .

Solution:

In this example, we will multiply by 1 in the form $\frac{\sqrt[3]{2^{2} b}}{\sqrt[3]{2^{2} b}}$ .

$\begin{array}{l} \sqrt[3]{\frac{27 a}{2 b^{2}}} & = & \frac{\sqrt[3]{3^{3} a}}{\sqrt[3]{2 b^{2}}} & A p p l y t h e q u o t i e n t r u l e f o r r a d i c a l s . \\ = & \frac{3 \sqrt[3]{a}}{\sqrt[3]{2 b^{2}}} \cdot \frac{\sqrt[3]{2^{2} b}}{\sqrt[3]{2^{2} b}} & M u l t i p l y b y t h e c u b e r o o t o f f a c t o r s t h a t r e s u l t i n p o w e r s o f 3. \\ = & \frac{3 \sqrt[3]{2^{2} a b}}{\sqrt[3]{2^{3} b^{3}}} & S i m p l i f y . \\ = & \frac{3 \sqrt[3]{4 a b}}{2 b} \end{array}$

Answer: $\frac{3 \sqrt[3]{4 a b}}{2 b}$

Example 14

Rationalize the denominator: $\frac{2 x \sqrt[5]{5}}{\sqrt[5]{4 x^{3} y}}$ .

Solution:

In this example, we will multiply by 1 in the form $\frac{\sqrt[5]{2^{3} x^{2} y^{4}}}{\sqrt[5]{2^{3} x^{2} y^{4}}}$ .

$\begin{array}{l} \frac{2 x \sqrt[5]{5}}{\sqrt[5]{4 x^{3} y}} & = & \frac{2 x \sqrt[5]{5}}{\sqrt[5]{2^{2} x^{3} y}} \cdot \frac{\sqrt[5]{2^{3} x^{2} y^{4}}}{\sqrt[5]{2^{3} x^{2} y^{4}}} & M u l t i p l y b y t h e f i f t h r o o t o f f a c t o r s t h a t r e s u l t i n p o w e r s o f 5. \\ = & \frac{2 x \sqrt[5]{5 \cdot 2^{3} x^{2} y^{4}}}{\sqrt[5]{2^{5} x^{5} y^{5}}} & S i m p l i f y . \\ = & \frac{2 x \sqrt[5]{40 x^{2} y^{4}}}{2 x y} \\ = & \frac{\sqrt[5]{40 x^{2} y^{4}}}{y} \end{array}$

Answer: $\frac{\sqrt[5]{40 x^{2} y^{4}}}{y}$

When two terms involving square roots appear in the denominator, we can rationalize it using a very special technique. This technique involves multiplying the numerator and the denominator of the fraction by the conjugate of the denominator. Recall that multiplying a radical expression by its conjugate produces a rational number.

Example 15

Rationalize the denominator: $\frac{1}{\sqrt{5} - \sqrt{3}} .$

Solution:

In this example, the conjugate of the denominator is $\sqrt{5} + \sqrt{3} .$ Therefore, multiply by 1 in the form $\frac{(\sqrt{5} + \sqrt{3})}{(\sqrt{5} + \sqrt{3})} .$

$\begin{matrix} \frac{1}{\sqrt{5} - \sqrt{3}} & = & \frac{1}{(\sqrt{5} - \sqrt{3})} \frac{(\sqrt{5} + \sqrt{3})}{(\sqrt{5} + \sqrt{3})} & \begin{matrix} M u l t i p l y n u m e r a t o r a n d \\ d e n o m i n a t o r b y t h e c o n j u g a t e \\ o f t h e d e n o m i n a t o r . \end{matrix} \\ = & \frac{\sqrt{5} + \sqrt{3}}{\sqrt{25} + \sqrt{15} - \sqrt{15} - \sqrt{9}} & S i m p l i f y . \\ = & \frac{\sqrt{5} + \sqrt{3}}{5 - 3} \\ = & \frac{\sqrt{5} + \sqrt{3}}{2} \end{matrix}$

Answer: $\frac{\sqrt{5} + \sqrt{3}}{2}$

Notice that the terms involving the square root in the denominator are eliminated by multiplying by the conjugate. We can use the property $(\sqrt{a} + \sqrt{b}) (\sqrt{a} - \sqrt{b}) = a - b$ to expedite the process of multiplying the expressions in the denominator.

Example 16

Rationalize the denominator: $\frac{\sqrt{10}}{\sqrt{2} + \sqrt{6}}$ .

Solution:

Multiply by 1 in the form $\frac{\sqrt{2} - \sqrt{6}}{\sqrt{2} - \sqrt{6}}$ .

$\begin{array}{l} \frac{\sqrt{10}}{\sqrt{2} + \sqrt{6}} & = & \frac{(\sqrt{10})}{(\sqrt{2} + \sqrt{6})} \frac{(\sqrt{2} - \sqrt{6})}{(\sqrt{2} - \sqrt{6})} & M u l t i p l y b y t h e c o n j u g a t e o f t h e d e n o m i n a t o r . \\ = & \frac{\sqrt{20} - \sqrt{60}}{2 - 6} & S i m p l i f y . \\ = & \frac{\sqrt{4 \cdot 5} - \sqrt{4 \cdot 15}}{− 4} \\ = & \frac{2 \sqrt{5} - 2 \sqrt{15}}{− 4} \\ = & \frac{2 (\sqrt{5} - \sqrt{15})}{− 4} \\ = & \frac{\sqrt{5} - \sqrt{15}}{− 2} = - \frac{\sqrt{5} - \sqrt{15}}{2} = \frac{- \sqrt{5} + \sqrt{15}}{2} \end{array}$

Answer: $\frac{\sqrt{15} - \sqrt{5}}{2}$

Example 17

Rationalize the denominator: $\frac{\sqrt{x} - \sqrt{y}}{\sqrt{x} + \sqrt{y}}$ .

Solution:

In this example, we will multiply by 1 in the form $\frac{\sqrt{x} - \sqrt{y}}{\sqrt{x} - \sqrt{y}} .$

$\begin{array}{l} \frac{\sqrt{x} - \sqrt{y}}{\sqrt{x} + \sqrt{y}} & = & \frac{(\sqrt{x} - \sqrt{y})}{(\sqrt{x} + \sqrt{y})} \frac{(\sqrt{x} - \sqrt{y})}{(\sqrt{x} - \sqrt{y})} & M u l t i p l y b y t h e c o n j u g a t e o f t h e d e n o m i n a t o r . \\ = & \frac{\sqrt{x^{2}} - \sqrt{x y} - \sqrt{x y} + \sqrt{y^{2}}}{x - y} & S i m p l i f y . \\ = & \frac{x - 2 \sqrt{x y} + y}{x - y} \end{array}$

Answer: $\frac{x - 2 \sqrt{x y} + y}{x - y}$

Try this! Rationalize the denominator: $\frac{2 \sqrt{3}}{5 - \sqrt{3}}$

Answer: $\frac{5 \sqrt{3} + 3}{11}$

(click to see video)

Key Takeaways

To multiply two single-term radical expressions, multiply the coefficients and multiply the radicands. If possible, simplify the result.
Apply the distributive property when multiplying a radical expression with multiple terms. Then simplify and combine all like radicals.
Multiplying a two-term radical expression involving square roots by its conjugate results in a rational expression.
It is common practice to write radical expressions without radicals in the denominator. The process of finding such an equivalent expression is called rationalizing the denominator.
If an expression has one term in the denominator involving a radical, then rationalize it by multiplying the numerator and denominator by the nth root of factors of the radicand so that their powers equal the index.
If a radical expression has two terms in the denominator involving square roots, then rationalize it by multiplying the numerator and denominator by the conjugate of the denominator.

Topic Exercises

Part A: Multiplying Radical Expressions

Multiply. (Assume all variables represent non-negative real numbers.)

$\sqrt{3} \cdot \sqrt{7}$
$\sqrt{2} \cdot \sqrt{5}$
$\sqrt{6} \cdot \sqrt{12}$
$\sqrt{10} \cdot \sqrt{15}$
$\sqrt{2} \cdot \sqrt{6}$
$\sqrt{5} \cdot \sqrt{15}$
$\sqrt{7} \cdot \sqrt{7}$
$\sqrt{12} \cdot \sqrt{12}$
$2 \sqrt{5} \cdot 7 \sqrt{10}$
$3 \sqrt{15} \cdot 2 \sqrt{6}$
${(2 \sqrt{5})}^{2}$
${(6 \sqrt{2})}^{2}$
$\sqrt{2 x} \cdot \sqrt{2 x}$
$\sqrt{5 y} \cdot \sqrt{5 y}$
$\sqrt{3 a} \cdot \sqrt{12}$
$\sqrt{3 a} \cdot \sqrt{2 a}$
$4 \sqrt{2 x} \cdot 3 \sqrt{6 x}$
$5 \sqrt{10 y} \cdot 2 \sqrt{2 y}$
$\sqrt[3]{3} \cdot \sqrt[3]{9}$
$\sqrt[3]{4} \cdot \sqrt[3]{16}$
$\sqrt[3]{15} \cdot \sqrt[3]{25}$
$\sqrt[3]{100} \cdot \sqrt[3]{50}$
$\sqrt[3]{4} \cdot \sqrt[3]{10}$
$\sqrt[3]{18} \cdot \sqrt[3]{6}$
$(5 \sqrt[3]{9}) (2 \sqrt[3]{6})$
$(2 \sqrt[3]{4}) (3 \sqrt[3]{4})$
${(2 \sqrt[3]{2})}^{3}$
${(3 \sqrt[3]{4})}^{3}$
$\sqrt[3]{3 a^{2}} \cdot \sqrt[3]{9 a}$
$\sqrt[3]{7 b} \cdot \sqrt[3]{49 b^{2}}$
$\sqrt[3]{6 x^{2}} \cdot \sqrt[3]{4 x^{2}}$
$\sqrt[3]{12 y} \cdot \sqrt[3]{9 y^{2}}$
$\sqrt[3]{20 x^{2} y} \cdot \sqrt[3]{10 x^{2} y^{2}}$
$\sqrt[3]{63 x y} \cdot \sqrt[3]{12 x^{4} y^{2}}$
$\sqrt{5} (3 - \sqrt{5})$
$\sqrt{2} (\sqrt{3} - \sqrt{2})$
$3 \sqrt{7} (2 \sqrt{7} - \sqrt{3})$
$2 \sqrt{5} (6 - 3 \sqrt{10})$
$\sqrt{6} (\sqrt{3} - \sqrt{2})$
$\sqrt{15} (\sqrt{5} + \sqrt{3})$
$\sqrt{x} (\sqrt{x} + \sqrt{x y})$
$\sqrt{y} (\sqrt{x y} + \sqrt{y})$
$\sqrt{2 a b} (\sqrt{14 a} - 2 \sqrt{10 b})$
$\sqrt{6 a b} (5 \sqrt{2 a} - \sqrt{3 b})$
$\sqrt[3]{6} (\sqrt[3]{9} - \sqrt[3]{20})$
$\sqrt[3]{12} (\sqrt[3]{36} + \sqrt[3]{14})$
$(\sqrt{2} - \sqrt{5}) (\sqrt{3} + \sqrt{7})$
$(\sqrt{3} + \sqrt{2}) (\sqrt{5} - \sqrt{7})$
$(2 \sqrt{3} - 4) (3 \sqrt{6} + 1)$
$(5 - 2 \sqrt{6}) (7 - 2 \sqrt{3})$
${(\sqrt{5} - \sqrt{3})}^{2}$
${(\sqrt{7} - \sqrt{2})}^{2}$
$(2 \sqrt{3} + \sqrt{2}) (2 \sqrt{3} - \sqrt{2})$
$(\sqrt{2} + 3 \sqrt{7}) (\sqrt{2} - 3 \sqrt{7})$
${(\sqrt{a} - \sqrt{2 b})}^{2}$
${(\sqrt{a b} + 1)}^{2}$
What is the perimeter and area of a rectangle with length measuring $5 \sqrt{3}$ centimeters and width measuring $3 \sqrt{2}$ centimeters?
What is the perimeter and area of a rectangle with length measuring $2 \sqrt{6}$ centimeters and width measuring $\sqrt{3}$ centimeters?
If the base of a triangle measures $6 \sqrt{2}$ meters and the height measures $3 \sqrt{2}$ meters, then calculate the area.
If the base of a triangle measures $6 \sqrt{3}$ meters and the height measures $3 \sqrt{6}$ meters, then calculate the area.

Part B: Dividing Radical Expressions

Divide. (Assume all variables represent positive real numbers.)

$\frac{\sqrt{75}}{\sqrt{3}}$
$\frac{\sqrt{360}}{\sqrt{10}}$
$\frac{\sqrt{72}}{\sqrt{75}}$
$\frac{\sqrt{90}}{\sqrt{98}}$
$\frac{\sqrt{90 x^{5}}}{\sqrt{2 x}}$
$\frac{\sqrt{96 y^{3}}}{\sqrt{3 y}}$
$\frac{\sqrt{162 x^{7} y^{5}}}{\sqrt{2 x y}}$
$\frac{\sqrt{363 x^{4} y^{9}}}{\sqrt{3 x y}}$
$\frac{\sqrt[3]{16 a^{5} b^{2}}}{\sqrt[3]{2 a^{2} b^{2}}}$
$\frac{\sqrt[3]{192 a^{2} b^{7}}}{\sqrt[3]{2 a^{2} b^{2}}}$

Part C: Rationalizing the Denominator

Rationalize the denominator. (Assume all variables represent positive real numbers.)

$\frac{1}{\sqrt{5}}$
$\frac{1}{\sqrt{6}}$
$\frac{\sqrt{2}}{\sqrt{3}}$
$\frac{\sqrt{3}}{\sqrt{7}}$
$\frac{5}{2 \sqrt{10}}$
$\frac{3}{5 \sqrt{6}}$
$\frac{\sqrt{3} - \sqrt{5}}{\sqrt{3}}$
$\frac{\sqrt{6} - \sqrt{2}}{\sqrt{2}}$
$\frac{1}{\sqrt{7 x}}$
$\frac{1}{\sqrt{3 y}}$
$\frac{a}{5 \sqrt{a b}}$
$\frac{3 b^{2}}{2 \sqrt{3 a b}}$
$\frac{2}{\sqrt[3]{36}}$
$\frac{14}{\sqrt[3]{7}}$
$\frac{1}{\sqrt[3]{4 x}}$
$\frac{1}{\sqrt[3]{3 y^{2}}}$
$\frac{9 x \sqrt[3]{2}}{\sqrt[3]{9 x y^{2}}}$
$\frac{5 y^{2} \sqrt[3]{x}}{\sqrt[3]{5 x^{2} y}}$
$\frac{3 a}{2 \sqrt[3]{3 a^{2} b^{2}}}$
$\frac{25 n}{3 \sqrt[3]{25 m^{2} n}}$
$\frac{3}{\sqrt[5]{27 x^{2} y}}$
$\frac{2}{\sqrt[5]{16 x y^{2}}}$
$\frac{a b}{\sqrt[5]{9 a^{3} b}}$
$\frac{a b c}{\sqrt[5]{a b^{2} c^{3}}}$
$\sqrt[5]{\frac{3 x}{8 y^{2} z}}$
$\sqrt[5]{\frac{4 x y^{2}}{9 x^{3} y z^{4}}}$
$\frac{3}{\sqrt{10} - 3}$
$\frac{2}{\sqrt{6} - 2}$
$\frac{1}{\sqrt{5} + \sqrt{3}}$
$\frac{1}{\sqrt{7} - \sqrt{2}}$
$\frac{\sqrt{3}}{\sqrt{3} + \sqrt{6}}$
$\frac{\sqrt{5}}{\sqrt{5} + \sqrt{15}}$
$\frac{10}{5 - 3 \sqrt{5}}$
$\frac{− 2 \sqrt{2}}{4 - 3 \sqrt{2}}$
$\frac{\sqrt{3} + \sqrt{5}}{\sqrt{3} - \sqrt{5}}$
$\frac{\sqrt{10} - \sqrt{2}}{\sqrt{10} + \sqrt{2}}$
$\frac{2 \sqrt{3} - 3 \sqrt{2}}{4 \sqrt{3} + \sqrt{2}}$
$\frac{6 \sqrt{5} + 2}{2 \sqrt{5} - \sqrt{2}}$
$\frac{x - y}{\sqrt{x} + \sqrt{y}}$
$\frac{x - y}{\sqrt{x} - \sqrt{y}}$
$\frac{x + \sqrt{y}}{x - \sqrt{y}}$
$\frac{x - \sqrt{y}}{x + \sqrt{y}}$
$\frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}}$
$\frac{\sqrt{a b} + \sqrt{2}}{\sqrt{a b} - \sqrt{2}}$
$\frac{\sqrt{x}}{5 - 2 \sqrt{x}}$
$\frac{1}{\sqrt{x} - y}$
$\frac{\sqrt{x} + \sqrt{2 y}}{\sqrt{2 x} - \sqrt{y}}$
$\frac{\sqrt{3 x} - \sqrt{y}}{\sqrt{x} + \sqrt{3 y}}$
$\frac{\sqrt{2 x + 1}}{\sqrt{2 x + 1} - 1}$
$\frac{\sqrt{x + 1}}{1 - \sqrt{x + 1}}$
$\frac{\sqrt{x + 1} + \sqrt{x - 1}}{\sqrt{x + 1} - \sqrt{x - 1}}$
$\frac{\sqrt{2 x + 3} - \sqrt{2 x - 3}}{\sqrt{2 x + 3} + \sqrt{2 x - 3}}$
The radius of the base of a right circular cone is given by $r = \sqrt{\frac{3 V}{π h}}$ where V represents the volume of the cone and h represents its height. Find the radius of a right circular cone with volume 50 cubic centimeters and height 4 centimeters. Give the exact answer and the approximate answer rounded to the nearest hundredth.
The radius of a sphere is given by $r = \sqrt[3]{\frac{3 V}{4 π}}$ where V represents the volume of the sphere. Find the radius of a sphere with volume 135 square centimeters. Give the exact answer and the approximate answer rounded to the nearest hundredth.

Part D: Discussion

Research and discuss some of the reasons why it is a common practice to rationalize the denominator.
Explain in your own words how to rationalize the denominator.

Answers

$\sqrt{21}$
$6 \sqrt{2}$
$2 \sqrt{3}$
7
$70 \sqrt{2}$
20
$2 x$
$6 \sqrt{a}$
$24 x \sqrt{3}$
3
$5 \sqrt[3]{3}$
$2 \sqrt[3]{5}$
$30 \sqrt[3]{2}$
16
$3 a$
$2 x \sqrt[3]{3 x}$
$2 x y \sqrt[3]{25 x}$
$3 \sqrt{5} - 5$
$42 - 3 \sqrt{21}$
$3 \sqrt{2} - 2 \sqrt{3}$
$x + x \sqrt{y}$
$2 a \sqrt{7 b} - 4 b \sqrt{5 a}$
$3 \sqrt[3]{2} - 2 \sqrt[3]{15}$
$\sqrt{6} + \sqrt{14} - \sqrt{15} - \sqrt{35}$
$18 \sqrt{2} + 2 \sqrt{3} - 12 \sqrt{6} - 4$
$8 - 2 \sqrt{15}$
10
$a - 2 \sqrt{2 a b} + 2 b$
Perimeter: $(10 \sqrt{3} + 6 \sqrt{2})$ centimeters; area: $15 \sqrt{6}$ square centimeters
$18$ square meters

5
$\frac{2 \sqrt{6}}{5}$
$3 x^{2} \sqrt{5}$
$9 x^{3} y^{2}$
$2 a$

$\frac{\sqrt{5}}{5}$
$\frac{\sqrt{6}}{3}$
$\frac{\sqrt{10}}{4}$
$\frac{3 - \sqrt{15}}{3}$
$\frac{\sqrt{7 x}}{7 x}$
$\frac{\sqrt{a b}}{5 b}$
$\frac{\sqrt[3]{6}}{3}$
$\frac{\sqrt[3]{2 x^{2}}}{2 x}$
$\frac{3 \sqrt[3]{6 x^{2} y}}{y}$
$\frac{\sqrt[3]{9 a b}}{2 b}$
$\frac{\sqrt[5]{9 x^{3} y^{4}}}{x y}$
$\frac{\sqrt[5]{27 a^{2} b^{4}}}{3}$
$\frac{\sqrt[5]{12 x y^{3} z^{4}}}{2 y z}$
$3 \sqrt{10} + 9$
$\frac{\sqrt{5} - \sqrt{3}}{2}$
$− 1 + \sqrt{2}$
$\frac{− 5 - 3 \sqrt{5}}{2}$
$− 4 - \sqrt{15}$
$\frac{15 - 7 \sqrt{6}}{23}$
$\sqrt{x} - \sqrt{y}$
$\frac{x^{2} + 2 x \sqrt{y} + y}{x^{2} - y}$
$\frac{a - 2 \sqrt{a b} + b}{a - b}$
$\frac{5 \sqrt{x} + 2 x}{25 - 4 x}$
$\frac{x \sqrt{2} + 3 \sqrt{x y} + y \sqrt{2}}{2 x - y}$
$\frac{2 x + 1 + \sqrt{2 x + 1}}{2 x}$
$x + \sqrt{x^{2} - 1}$
$\frac{5 \sqrt{6 π}}{2 π}$ centimeters; 3.45 centimeters

Answer may vary