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# 8.4: Multiplying and Dividing Radical Expressions

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Skills to Develop

• 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. If a and b represent positive real numbers,

$\sqrt[n]{a} \cdot \sqrt[n]{b}=\sqrt[n]{a \cdot b}$

Example $$\PageIndex{1}$$

Multiply:

$$\sqrt{2} \cdot \sqrt{6}$$

Solution:

This problem is a product of two square roots. Apply the product rule for radicals and then simplify.

\begin{aligned} \sqrt{2} \cdot \sqrt{6} &=\sqrt{2 \cdot 6}\qquad\color{Cerulean}{Multiply\:the\:radicands.} \\ &=\sqrt{12}\qquad\:\:\:\color{Cerulean}{Simplify.} \\ &=\sqrt{2^{2} \cdot 3} \\ &=2 \sqrt{3} \end{aligned}

Answer:

$$2\sqrt{3}$$

Example $$\PageIndex{2}$$

Multiply:

$$\sqrt[3]{9} \cdot \sqrt[3]{6}$$

Solution:

This problem is a product of cube roots. Apply the product rule for radicals and then simplify.

\begin{aligned} \sqrt[3]{9} \cdot \sqrt[3]{6} &=\sqrt[3]{9 \cdot 6}\qquad\color{Cerulean}{Multiply\:the\:radiands.} \\ &=\sqrt[3]{54}\qquad\:\color{Cerulean}{Simplify.} \\ &=\sqrt[3]{3^{3} \cdot 2} \\ &=3 \sqrt[3]{2} \end{aligned}

Answer:

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

Often there will be coefficients in front of the radicals.

Example $$\PageIndex{3}$$

Multiply:

$$2 \sqrt{3} \cdot 5 \sqrt{2}$$

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{aligned} 2 \sqrt{3} \cdot 5 \sqrt{2} &=2 \cdot 5 \cdot \sqrt{3} \cdot \sqrt{2}\qquad\color{Cerulean}{Multiplication\:is\:commutative.} \\ &=10 \cdot \sqrt{6}\qquad\qquad\:\:\:\color{Cerulean}{Multiply\:the\:coefficients\:and\:the\:radicands.} \\ &=10 \sqrt{6} \end{aligned}

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

Answer:

$$10\sqrt{6}$$

Example $$\PageIndex{4}$$

Multiply:

$$-2 \sqrt[3]{5 x} \cdot 3 \sqrt[3]{25 x^{2}}$$

Solution:

\begin{aligned}-2 \sqrt[3]{5 x} \cdot 3 \sqrt[3]{25 x^{2}} &=-6 \sqrt[3]{125 x^{3}}\qquad\color{Cerulean}{Multiple\:the\:coefficients\:and\:then\:multiply\:the\:radicands.} \\ &=-6 \sqrt[3]{5^{3} x^{3}}\qquad\:\:\:\color{Cerulean}{Simplify.} \\ &=-6.5 x \\ &=-30 x \end{aligned}

Answer:

$$-30x$$

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

Example $$\PageIndex{5}$$

Multiply:

$$4 \sqrt{3}(2 \sqrt{3}-3 \sqrt{6})$$

Solution:

Apply the distributive property and multiply each term by $$4\sqrt{3}$$.

\begin{aligned} 4 \sqrt{3}(2 \sqrt{3}-3 \sqrt{6}) &=\color{Cerulean}{4 \sqrt{3}}\color{black}{ \cdot} 2 \sqrt{3}-\color{Cerulean}{4 \sqrt{3}}\color{black}{ \cdot} 3 \sqrt{6} \qquad\color{Cerulean}{Distribute.}\\ &=8 \sqrt{9}-12 \sqrt{18}\qquad\qquad\qquad\color{Cerulean}{Simplify.} \\ &=8 \sqrt{9}-12 \sqrt{9 \cdot 2} \\ &=8 \cdot 3-12 \cdot 3 \sqrt{2} \\ &=24-36 \sqrt{2} \end{aligned}

Answer:

$$24-36 \sqrt{2}$$

Example $$\PageIndex{6}$$

Multiply:

$$\sqrt[3]{4 x^{2}}\left(\sqrt[3]{2 x}-5 \sqrt[3]{4 x^{2}}\right)$$

Solution:

Apply the distributive property and then simplify the result.

\begin{aligned} \sqrt[3]{4 x^{2}}\left(\sqrt[3]{2 x}-5 \sqrt[3]{4 x^{2}}\right) &=\color{Cerulean}{\sqrt[3]{4 x^{2}}} \color{black}{\cdot} \sqrt[3]{2 x}-\color{black}{\left(\color{Cerulean}{\sqrt[3]{4 x^{2}}}\right)} \cdot 5 \sqrt[3]{4 x^{2}} \\ &=\sqrt[3]{8 x^{3}}-5 \sqrt[3]{16 x^{4}} \\ &=\sqrt[3]{2^{3} x^{3}}-5 \sqrt[3]{2^{4} x^{4}} \\ &=2 x-5 \cdot 2 x \sqrt[3]{2 x} \\ &=2 x-10 x \sqrt[3]{2 x} \end{aligned}

Answer:

$$2 x-10 x \sqrt[3]{2 x}$$

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 $$\PageIndex{7}$$

Multiply:

$$(\sqrt{5}+2)(\sqrt{5}-4)$$

Solution:

Begin by applying the distributive property.

\begin{aligned}(\sqrt{5}+2)(\sqrt{5}-4) &=\color{Cerulean}{\sqrt{5}}\color{black}{ \cdot} \sqrt{5}-\color{Cerulean}{\sqrt{5}}\color{black}{ \cdot} 4+\color{OliveGreen}{2}\color{black}{ \cdot} \sqrt{5}-\color{OliveGreen}{2}\color{black}{ \cdot} 4\qquad\color{Cerulean}{Distribute.} \\ &=\sqrt{25}-4 \sqrt{5}+2 \sqrt{5}-8\qquad\qquad\qquad\:\color{Cerulean}{Simplify.} \\ &=5-4 \sqrt{5}+2 \sqrt{5}-8 \\ &=5-8 -4 \sqrt{5}+2 \sqrt{5}\qquad\qquad\qquad\quad\:\:\color{Cerulean}{Combine\:like\:terms.} \\ &=-3-2 \sqrt{5} \end{aligned}

Answer:

$$-3-2\sqrt{5}$$

Example $$\PageIndex{8}$$

Multiply:

$$(3 \sqrt{x}-\sqrt{y})^{2}$$

Solution:

\begin{aligned}(3 \sqrt{x}-\sqrt{y})^{2} &=(3 \sqrt{x}-\sqrt{y})(3 \sqrt{x}-\sqrt{y}) \\ &=\color{Cerulean}{3 \sqrt{x}}\color{black}{ \cdot} 3 \sqrt{x}-\color{Cerulean}{3 \sqrt{x}}\color{black}{ \cdot} \sqrt{y}-\color{OliveGreen}{\sqrt{y}}\color{black}{ \cdot} 3 \sqrt{x}+\color{OliveGreen}{\sqrt{y}}\color{black}{ \cdot} \sqrt{y}\qquad\color{Cerulean}{Distribute.} \\ &=9 \sqrt{x^{2}}-3 \sqrt{x y}-3 \sqrt{x y}+\sqrt{y^{2}}\qquad\qquad\qquad\qquad\:\:\:\color{Cerulean}{Simplify.} \\ &=9 x-6 \sqrt{x y}+y \end{aligned}

Answer:

$$9x-6\sqrt{xy}+y$$

Exercise $$\PageIndex{1}$$

Multiply:

$$(2\sqrt{3}+5\sqrt{2})(\sqrt{3}-2\sqrt{6})$$

Answer

$$6-12\sqrt{2}+5\sqrt{6}-20\sqrt{3}$$

The expressions $$(a+b)$$ and $$(a−b)$$ are called conjugates. When multiplying conjugates, the sum of the products of the inner and outer terms results in 0.

Example $$\PageIndex{9}$$

Multiply:

$$(\sqrt{2}+\sqrt{5})(\sqrt{2}-\sqrt{5})$$

Solution:

Apply the distributive property and then combine like terms.

\begin{aligned} (\sqrt{2}+\sqrt{5})(\sqrt{2}-\sqrt{5})&=\color{Cerulean}{\sqrt{2}}\color{black}{\cdot}\sqrt{2}-\color{Cerulean}{\sqrt{2}}\color{black}{\cdot}\sqrt{5}+\color{OliveGreen}{\sqrt{5}}\color{black}{\cdot}\sqrt{2}-\color{OliveGreen}{\sqrt{5}}\color{black}{\cdot}\sqrt{5}\qquad\color{Cerulean}{Distribute.}\\&=\sqrt{4}-\sqrt{10}+\sqrt{10}-\sqrt{25}\qquad\qquad\qquad\qquad\:\:\:\color{Cerulean}{Simplify.} \\&= 2 \underbrace{-\sqrt{10}+\sqrt{10}}_{\color{Cerulean}{\text { opposites add to 0}} }-5\\&=2-5\\&=-3 \end{aligned}

Answer:

$$-3$$

It is important to note that when multiplying conjugate radical expressions, we obtain a rational expression. This is true in general and is often used in our study of algebra.

\begin{aligned}(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b}) &=\sqrt{a^{2}}-\sqrt{a b}+\sqrt{a b}-\sqrt{b^{2}} \\ &=a-b \end{aligned}

Therefore, for nonnegative real numbers a and b, we have the following property:

$(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=a-b$

## Dividing Radical Expressions (Rationalizing the Denominator)

To divide radical expressions with the same index, we use the quotient rule for radicals. If a and b represent nonnegative numbers, where $$b≠0$$, then we have

$\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}}$

Example $$\PageIndex{10}$$

Divide:

$$\frac{\sqrt{80}}{\sqrt{10}}$$

Solution:

In this case, we can see that 10 and 80 have common factors. If we apply the quotient rule for radicals and write it as a single square root, we will be able to reduce the fractional radicand.

\begin{aligned} \frac{\sqrt{80}}{\sqrt{10}} &=\sqrt{\frac{80}{10}}\qquad\color{Cerulean}{Apply\:the\:quotient\:rule\:for\:radicals\:and\:reduce\:the\:radicand.} \\ &=\sqrt{8}\qquad\quad\color{Cerulean}{Simplify.} \\ &=\sqrt{4 \cdot 2} \\ &=2 \sqrt{2} \end{aligned}

Answer:

$$2\sqrt{2}$$

Example $$\PageIndex{11}$$

Divide:

$$\frac{\sqrt{16 x^{5} y^{4}}}{\sqrt{2 x y}}$$

Solution:

\begin{aligned} \frac{\sqrt{16 x^{5} y^{4}}}{\sqrt{2 x y}} &=\sqrt{\frac{16 x^{5} y^{4}}{2 x y}} \qquad\qquad\qquad\color{Cerulean}{Apply\:the\:quotient\:rule\:for\:radicals\:and\:cancel.}\\ &=\sqrt{8 x^{4} y^{3}} \qquad\qquad\qquad\quad\color{Cerulean}{Simplify.}\\ &=\sqrt{4 \cdot 2 \cdot\left(x^{2}\right)^{2} \cdot y^{2} \cdot y} \\ &=2 x^{2} y \sqrt{2 y} \end{aligned}

Answer:

$$2 x^{2} y \sqrt{2 y}$$

Example $$\PageIndex{12}$$

Divide:

$$\frac{\sqrt[3]{54 a^{3} b^{5}}}{\sqrt[3]{16 a^{2} b^{2}}}$$

Solution:

\begin{aligned}\frac{\sqrt[3]{54a^{3}b^{5}}}{\sqrt[3]{16a^{2}b^{2}}} &=\sqrt[3]{\frac{54 \cdot a^{3} b^{5}}{16 a^{2} b^{2}}}\qquad\color{Cerulean}{Apply\:the\:quotient\:rule\:for\:radicals\:and\:then\:cancel.} \\ &=\sqrt[3]{\frac{27 a b^{3}}{8}}\qquad\:\:\:\color{Cerulean}{Replace\:27\:and\:8\:with\:their\:prime\:factorizations.} \\&=\sqrt[3]{\frac{3^{3}ab^{3}}{2^{3}}}\qquad\:\:\:\color{Cerulean}{Simplify.} \\&=\frac{\sqrt[3]{3^{3}ab^{3}}}{\sqrt[3]{2^{3}}} \\&=\frac{3b\sqrt[3]{a}}{2}\end{aligned}

Answer:

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

When the 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 denominator.

\begin{aligned}\color{Cerulean}{ Radical} &\:\color{Cerulean}{expression} &\color{Cerulean}{ Rational}& \:\color{Cerulean}{denominator} \\ &\frac{1}{\sqrt{3}} &=&\frac{\sqrt{3}}{3} \end{aligned}

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,

$$\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}} \cdot \color{Cerulean}{\frac{\sqrt{3}}{\sqrt{3}}}\color{black}{=}\frac{\sqrt{3}}{\sqrt{3^{2}}}=\frac{\sqrt{3}}{3}$$

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

Example $$\PageIndex{13}$$

Rationalize the denominator:

$$\frac{\sqrt{3}}{\sqrt{2}}$$

Solution:

The goal is to find an equivalent expression without a radical in the denominator. In this example, multiply by 1 in the form $$\color{Cerulean}{\frac{\sqrt{2}}{\sqrt{2}}}$$.

\begin{aligned} \frac{\sqrt{3}}{\sqrt{2}} &=\frac{\sqrt{3}}{\sqrt{2}} \cdot\color{Cerulean}{ \frac{\sqrt{2}}{\sqrt{2}}}\qquad\color{Cerulean}{Multiply\:by\:\frac{\sqrt{2}}{\sqrt{2}}.} \\ &=\frac{\sqrt{6}}{\sqrt{2^{2}}}\qquad\qquad\color{Cerulean}{Simplify.} \\ &=\frac{\sqrt{6}}{2}\qquad\qquad\:\color{Cerulean}{Rational\:denominator.} \end{aligned}

Answer:

$$\frac{\sqrt{6}}{2}$$

Example $$\PageIndex{14}$$

Rationalize the denominator:

$$\frac{1}{2 \sqrt{3 x}}$$

Solution:

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 $$\frac{1}{2 \sqrt{3 x}}$$.

\begin{aligned} \frac{1}{2 \sqrt{3 x}} &=\frac{1}{2 \sqrt{3 x}} \cdot \color{Cerulean}{ \frac{\sqrt{3 x}}{\sqrt{3 x}}}\qquad\color{Cerulean}{Multiply\:by\:\frac{\sqrt{3x}}{\sqrt{3x}}.} \\ &=\frac{\sqrt{3 x}}{2 \sqrt{3^{2} x^{2}}}\qquad\qquad\color{Cerulean}{Simplify.} \\ &=\frac{\sqrt{3 x}}{2 \cdot 3 x} \\ &=\frac{\sqrt{3 x}}{6 x} \end{aligned}

Answer:

$$\frac{\sqrt{3 x}}{6 x}$$

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

Example $$\PageIndex{15}$$

Rationalize the denominator:

$$\frac{5 \sqrt{2}}{\sqrt{5 a b}}$$

Solution

In this example, we will multiply by 1 in the form $$\frac{\sqrt{5 a b}}{\sqrt{5 a b}}$$.

\begin{aligned} \frac{5 \sqrt{2}}{\sqrt{5 a b}} &=\frac{5 \sqrt{2}}{\sqrt{5 a b}} \cdot \color{Cerulean}{\frac{\sqrt{5 a b}}{\sqrt{5 a b}}} \\ &=\frac{5 \sqrt{10 a b}}{\sqrt{25 a^{2} b^{2}}}\qquad\quad\color{Cerulean}{Simplify.} \\ &=\frac{5 \sqrt{10 a b}}{5 a b}\qquad\quad\:\color{Cerulean}{Cancel.} \\ &=\frac{\sqrt{10 a b}}{a b} \end{aligned}

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

Answer:

$$\frac{\sqrt{10 a b}}{a b}$$

Exercise $$\PageIndex{2}$$

Rationalize the denominator:

$$\sqrt{\frac{4a}{3b}}$$

Answer

$$\frac{2\sqrt{3ab}}{3b}$$

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,

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

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. For example,

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

Therefore, to rationalize the denominator of radical expressions 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 numerator and denominator by the th root of factors that produce th powers of all the factors in the radicand of the denominator.

Example $$\PageIndex{16}$$

Rationalize the denominator:

$$\frac{1}{\sqrt[3]{25}}$$

Solution:

The radical in the denominator is equivalent to $$\sqrt[3]{5^{2}}$$. To rationalize the denominator, it should be $$\sqrt[3]{5^{3}}$$. 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{aligned} \frac{1}{\sqrt[3]{25}} &=\frac{1}{\sqrt[3]{5^{2}}} \cdot\color{Cerulean}{ \frac{\sqrt[3]{5}}{\sqrt[3]{5}}}\qquad\color{Cerulean}{Multiply\:by\:the\:cube\:root\:of\:factors\:that\:result\:in\:powers\:of\:3.} \\ &=\frac{\sqrt[3]{5}}{\sqrt[3]{5^{3}}}\qquad\qquad\:\:\color{Cerulean}{Simplify.} \\ &=\frac{\sqrt[3]{5}}{5} \end{aligned}

Answer:

$$\frac{\sqrt[3]{5}}{5}$$

Example $$\PageIndex{17}$$

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{aligned} \sqrt[3]{\frac{27 a}{2 b^{2}}} &=\frac{\sqrt[3]{3^{3} a}}{\sqrt[3]{2 b^{2}}}\qquad\qquad\quad\color{Cerulean}{Apply\:the\:quotient\:rule\:for\:radicals.} \\ &=\frac{3 \sqrt[3]{a}}{\sqrt[3]{2 b^{2}}} \cdot\color{Cerulean}{ \frac{\sqrt[3]{2^{2} b}}{\sqrt[3]{2^{2} b}}} \quad\:\:\:\color{Cerulean}{Multiply\:by\:the\:cube\:root\:of\:factors\:that\:result\:in\:powers\:of\:3.}\\ &=\frac{3 \sqrt[3]{2^{2} a b}}{\sqrt[3]{2^{3} b^{3}}}\qquad\qquad\color{Cerulean}{Simplify.} \\ &=\frac{3 \sqrt[3]{4 a b}}{2 b} \end{aligned}

Answer:

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

Example $$\PageIndex{18}$$

Rationalize the denominator:

$$\frac{1}{\sqrt[5]{4 x^{3}}}$$

Solution:

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

\begin{aligned} \frac{1}{\sqrt[5]{4 x^{3}}} &=\frac{1}{\sqrt[5]{2^{2} x^{3}}} \cdot \color{Cerulean}{\frac{\sqrt[5]{2^{3} x^{2}}}{\sqrt[5]{2^{3} x^{2}}}}\qquad\color{Cerulean}{Multiply\:by\:the\:fifth\:root\:of\:factors\:that\:result\:in\:powers\:of\:5.} \\ &=\frac{\sqrt[5]{2^{3} x^{5}}}{\sqrt[5]{2^{5} x^{5}}}\qquad\qquad\qquad\color{Cerulean}{Simplify.} \\ &=\frac{\sqrt[5]{8 x^{2}}}{2 x} \end{aligned}

Answer:

$$\frac{\sqrt[5]{8 x^{2}}}{2 x}$$

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 $$\PageIndex{19}$$

Rationalize the denominator:

$$\frac{1}{\sqrt{3}-\sqrt{2}}$$

Solution:

In this example, the conjugate of the denominator is $$\sqrt{3}+\sqrt{2}$$. Therefore, multiply by 1 in the form $$\frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})}$$.

\begin{aligned} \frac{1}{\sqrt{3}-\sqrt{2}} &=\frac{1}{(\sqrt{3}-\sqrt{2})}\color{Cerulean}{ \frac{(\sqrt{3}+\sqrt{2})}{(\sqrt{3}+\sqrt{2})}}\qquad\color{Cerulean}{Multiply\:numerator\:and\:denominator\:by\:the\:conjugate\:of\:the\:denominator.} \\ &=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{9}+\sqrt{6}-\sqrt{6}-\sqrt{4}}\qquad\:\:\:\color{Cerulean}{Simplify.} \\ &=\frac{\sqrt{3}+\sqrt{2}}{3-2} \\ &=\frac{\sqrt{3}+\sqrt{2}}{1} \\ &=\sqrt{3}+\sqrt{2} \end{aligned}

Answer:

$$\sqrt{3}+\sqrt{2}$$

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

Example $$\PageIndex{20}$$

Rationalize the denominator:

$$\frac{\sqrt{2}-\sqrt{6}}{\sqrt{2}+\sqrt{6}}$$

Solution:

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

\begin{aligned}\frac{\sqrt{2}-\sqrt{6}}{\sqrt{2}+\sqrt{6}} &=\frac{(\sqrt{2}-\sqrt{6})}{(\sqrt{2}+\sqrt{6})}\color{Cerulean}{ \frac{(\sqrt{2}-\sqrt{6})}{(\sqrt{2}-\sqrt{6})}}\qquad\quad\color{Cerulean}{Multiply\:by\:the\:conjugate\:of\:the\:denominator.} \\ &=\frac{\sqrt{4}-\sqrt{12}-\sqrt{12}+\sqrt{36}}{2-6}\qquad\color{Cerulean}{Simplify.} \\ &=\frac{2-2 \sqrt{12}+6}{-4} \\ &=\frac{8-2\sqrt{4\cdot3}}{-4}\\&=\frac{8-4\sqrt{3}}{-4}\\&=\frac{8}{-4}-\frac{4\sqrt{3}}{-4}\\&=-2+\sqrt{3} \end{aligned}

Answer:

$$-2+\sqrt{3}$$

Example $$\PageIndex{21}$$

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{aligned} \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}} &=\frac{(\sqrt{x}-\sqrt{y})}{(\sqrt{x}+\sqrt{y})}\color{Cerulean}{ \frac{(\sqrt{x}-\sqrt{y})}{(\sqrt{x}-\sqrt{y})}}\qquad\quad\color{Cerulean}{Multiply\:by\:the\:conjugate\:of\:the\:denominator.} \\ &=\frac{\sqrt{x^{2}}-\sqrt{x y}-\sqrt{x y}+\sqrt{y^{2}}}{x-y}\:\:\:\quad\color{Cerulean}{Simplify.} \\ &=\frac{x-2 \sqrt{x y}+y}{x-y} \end{aligned}

Answer:

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

Exercise $$\PageIndex{3}$$

Rationalize the denominator:

$$\frac{3 \sqrt{5}+5}{2 \sqrt{5}-3}$$

Answer

$$\frac{45+19 \sqrt{5}}{11}$$

## 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 radical expressions 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 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 its conjugate.

Exercise $$\PageIndex{4}$$ multiplying radical expressions

Multiply. (Assume all variables are nonnegative.)

1. $$\sqrt{3}\cdot\sqrt{5}$$
2. $$\sqrt{7}\cdot\sqrt{3}$$
3. $$\sqrt{2}\cdot\sqrt{6}$$
4. $$\sqrt{5}\cdot\sqrt{15}$$
5. $$\sqrt{7}\cdot\sqrt{7}$$
6. $$\sqrt{12}\cdot\sqrt{12}$$
7. $$2\sqrt{5}\cdot 7\sqrt{10}$$
8. $$3\sqrt{15}\cdot 2\sqrt{6}$$
9. $$(2\sqrt{5})^{2}$$
10. $$(6\sqrt{2})^{2}$$
11. $$\sqrt{2x}\cdot\sqrt{2x}$$
12. $$\sqrt{5y}\cdot\sqrt{5y}$$
13. $$\sqrt{3a}\cdot\sqrt{12}$$
14. $$\sqrt{3a}\cdot\sqrt{2a}$$
15. $$4\sqrt{2x}\cdot 3\sqrt{6x}$$
16. $$5\sqrt{10y}\cdot 2\sqrt{2y}$$
17. $$\sqrt[3]{5} \cdot \sqrt[3]{25}$$
18. $$\sqrt[3]{4}\cdot\sqrt[3]{2}$$
19. $$\sqrt[3]{4}\cdot\sqrt[3]{10}$$
20. $$\sqrt[3]{18}\cdot\sqrt[3]{6}$$
21. $$(5\sqrt[3]{9})(2\sqrt[3]{6})$$
22. $$(2\sqrt[3]{4})(3\sqrt[3]{4})$$
23. $$(2\sqrt[3]{2})^{3}$$
24. $$(3\sqrt[3]{4})^{3}$$
25. $$\sqrt[3]{3a^{2}}\cdot\sqrt[3]{9a}$$
26. $$\sqrt[3]{7b}\cdot\sqrt[3]{49b^{2}}$$
27. $$\sqrt[3]{6x^{2}}\cdot\sqrt[3]{4x^{2}}$$
28. $$\sqrt[3]{12y}\cdot\sqrt[3]{9y^{2}}$$
29. $$\sqrt[3]{20x^{2}y}\cdot\sqrt[3]{10x^{2}y^{2}}$$
30. $$\sqrt[3]{63xy}\cdot\sqrt[3]{12x^{4}y^{2}}$$
31. $$\sqrt{5}( 3 − \sqrt{5})$$
32. $$\sqrt{2}(\sqrt{3}−\sqrt{2})$$
33. $$3\sqrt{7}(2\sqrt{7} − \sqrt{3})$$
34. $$2\sqrt{5}(6−3\sqrt{10})$$
35. $$\sqrt{6}(\sqrt{3}−\sqrt{2})$$
36. $$\sqrt{15}(\sqrt{5}+\sqrt{3})$$
37. $$\sqrt{x}(\sqrt{x}+\sqrt{xy})$$
38. $$\sqrt{y}(\sqrt{xy}+\sqrt{y})$$
39. $$\sqrt{2ab}(\sqrt{14a}−2\sqrt{10b})$$
40. $$\sqrt{6ab}(5\sqrt{2a}−\sqrt{3b})$$
41. $$(\sqrt{2}−\sqrt{5})(\sqrt{3}+\sqrt{7})$$
42. $$(\sqrt{3}+\sqrt{2})(\sqrt{5}−\sqrt{7})$$
43. $$(2\sqrt{3}−4)(3\sqrt{6}+1)$$
44. $$(5−2\sqrt{6})(7−2\sqrt{3})$$
45. $$(\sqrt{5}−\sqrt{3})^{2}$$
46. $$(\sqrt{7}−\sqrt{2})^{2}$$
47. $$(2\sqrt{3}+\sqrt{2})(2\sqrt{3}−\sqrt{2})$$
48. $$(\sqrt{2}+3\sqrt{7})(\sqrt{2}−3\sqrt{7})$$
49. $$(\sqrt{a}−\sqrt{2b})^{2}$$
50. $$(\sqrt{ab}+1)^{2}$$
51. What are the perimeter and area of a rectangle with length of $$5\sqrt{3}$$ centimeters and width of $$3\sqrt{2}$$ centimeters?
52. What are the perimeter and area of a rectangle with length of $$2\sqrt{6}$$ centimeters and width of $$\sqrt{3}$$ centimeters?
53. If the base of a triangle measures $$6\sqrt{2}$$ meters and the height measures $$3\sqrt{2}$$ meters, then what is the area?
54. If the base of a triangle measures $$6\sqrt{3}$$ meters and the height measures $$3\sqrt{6}$$ meters, then what is the area?
Answer

1. $$\sqrt{15}$$

3. $$2\sqrt{3}$$

5. $$7$$

7. $$70\sqrt{2}$$

9. $$20$$

11. $$2x$$

13. $$6\sqrt{a}$$

15. $$24x\sqrt{3}$$

17. $$5$$

19. $$2 \sqrt[3]{5}$$

21. $$30 \sqrt[3]{2}$$

23. $$16$$

25. $$3a$$

27. $$2 \sqrt[3]{3}\left(x^{2}\right)^{\frac{2}{3}}$$

29. $$2xy\sqrt[3]{25x}$$

31. $$3\sqrt{5}−5$$

33. $$42-3 \sqrt{21}$$

35. $$3\sqrt{2}−2\sqrt{3}$$

37. $$x+x\sqrt{y}$$

39. $$\sqrt{2}(\sqrt{14} \sqrt{a}-2 \sqrt{10} \sqrt{b}) \sqrt{a b}$$

41. $$\sqrt{6}+\sqrt{14}-\sqrt{15}-\sqrt{35}$$

43. $$18 \sqrt{2}+2 \sqrt{3}-12 \sqrt{6}-4$$

45. $$8-2 \sqrt{15}$$

47. $$10$$

49. $$(\sqrt{a}-\sqrt{2b} )^{2}$$

51. Perimeter: $$(10\sqrt{3}+6\sqrt{2})$$ centimeters; area: $$15\sqrt{6}$$ square centimeters

53. $$18$$ square meters

Exercise $$\PageIndex{5}$$ dividing radical expressions

Divide.

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

1. $$5$$

3. $$\frac{2 \sqrt{6}}{5}$$

5. $$3 \sqrt{5} x^{2}$$

7. $$9 x^{3} y^{2}$$

9. $$2a$$

Exercise $$\PageIndex{6}$$ dividing radical expressions

Rationalize the denominator

1. $$\sqrt{\frac{1}{5}}$$
2. $$\sqrt{\frac{1}{6}}$$
3. $$\frac{\sqrt{2}}{\sqrt{3}}$$
4. $$\frac{\sqrt{3}}{\sqrt{7}}$$
5. $$\sqrt{\frac{52}{10}}$$
6. $$\sqrt{\frac{3}{56}}$$
7. $$\frac{\sqrt{3}-\sqrt{5}}{\sqrt{3}}$$
8. $$\frac{\sqrt{6}-\sqrt{2}}{\sqrt{2}}$$
9. $$\sqrt{\frac{1}{7 x}}$$
10. $$\sqrt{\frac{1}{3y}}$$
11. $$a \sqrt{\frac{5}{a b}}$$
12. $$\3 b^{2} \sqrt{\frac{23}{a b}}) 13. 236−−√3 14. 147√3 15. 14x−−√3 16. 13y2−−−−√3 17. 9x⋅2√39xy2−−−−−√3 18. 5y2⋅x−−√35x2y−−−−−√3 19. 3a2 3a2b2−−−−−√3 20. 25n3 25m2n−−−−−−√3 21. 327x2y−−−−−√5 22. 216xy2−−−−−−√5 23. ab9a3b−−−−√5 24. abcab2c3−−−−−√5 25. 310−−√−3 26. 26√−2 27. 15√+3√ 28. 17√−2√ 29. 3√3√+6√ 30. 5√5√+15−−√ 31. 105−35√ 32. −22√4−32√ 33. 3√+5√3√−5√ 34. 10−−√−2√10−−√+2√ 35. 23√−32√43√+2√ 36. 65√+225√−2√ 37. x+y√x−y√ 38. x−y√x+y√ 39. a√−b√a√+b√ 40. ab−−√+2√ab−−√−2√ 41. x−−√5−2x−−√ 106. 1x−−√−y Answer 1. \(\frac{\sqrt{5}}{5}$$

3. $$\frac{\sqrt{6}}{3}$$

5. $$\frac{\sqrt{130}}{5}$$

7. $$\frac{3-\sqrt{15}}{3}$$

9. $$\frac{\sqrt{7x}}{7 x}$$

11. $$a \frac{\sqrt{5a b}}{a b}$$

13. 6√33

15. 2x2−−−−√32x

17. 3 6x2y−−−−−√3y

19. 9ab−−−√32b

21. 9x3y4−−−−−−√5xy

23. 27a2b4−−−−−−√53

25. 310−−√+9

27. 5√−3√2

29. −1+2√

31. −5−35√2

33. −4−15−−√

35. 15−76√23

37. x2+2xy√+yx2−y

39. a−2ab−−√+ba−b

41. 5x−−√+2x25−4x

Exercise $$\PageIndex{7}$$ discussion

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

1. Answer may vary