Int-Alg Working with Radicals

Section 6.4 Working with Radicals

Sometimes radical notation is more convenient to use than exponents. In these cases, we usually simplify radical expressions algebraically before using a calculator to obtain decimal approximations.

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Subsection 6.4.1 Properties of Radicals

Because

\sqrt[n]{a} = a^{1 / n},

we can use the laws of exponents to derive two important properties that are useful in working with radicals.

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Product Rule for Radicals.

\sqrt[n]{a b} = \sqrt[n]{a} \sqrt[n]{b}, for a, b \geq 0

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Quotient Rule for Radicals.

\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}}, for a \geq 0, b > 0

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As examples, you can verify that

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\sqrt{36} = \sqrt{4} \sqrt{9} and \sqrt[3]{\frac{1}{8}} = \frac{\sqrt[3]{1}}{\sqrt[3]{8}}

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Caution 6.4.1.

In general, it is not true that

\sqrt[n]{a + b}

is equivalent to

\sqrt[n]{a} + \sqrt[n]{b},

or that

\sqrt[n]{a - b}

is equivalent to

\sqrt[n]{a} - \sqrt[n]{b} .

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For example, you can check that

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\begin{aligned} \sqrt{9 + 16} & \neq \sqrt{9} + \sqrt{16} \\ and \sqrt[3]{27 - 8} & \neq \sqrt[3]{27} - \sqrt[3]{8} \end{aligned}

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Example 6.4.2.

Which of the following are true?

[TK]

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Is $\sqrt{36 + 64} = \sqrt{36} + \sqrt{64} ?$
Is $\sqrt[3]{8 (64)} = \sqrt[3]{8} \sqrt[3]{64} ?$
Is $\sqrt{x^{2} + 4} = x + 2 ?$
Is $\sqrt[3]{8 x^{3}} = 2 x ?$

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Solution.

The statements in (b) and (d) are true, and both are examples of the first property of radicals.

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Statements (a) and (c) are false.

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Checkpoint 6.4.3. QuickCheck 1.

Which of the following are true?

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$\sqrt[3]{8 a^{3} - b^{3}} = 2 a - b ?$
$\sqrt[5]{x^{3} y^{4}} = \sqrt[5]{x^{3}} \sqrt[5]{y^{4}} ?$
$\sqrt[4]{\frac{1}{w^{4}}} = \frac{1}{w},$ for $w > 0 ?$
$\sqrt{1 + x^{2}} = 1 + x ?$

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Subsection 6.4.2 Simplifying Radicals

Each time we use a calculator to approximate a radical, we lose accuracy, and in the course of a long calculation, the error accumulates. To avoid this problem, we simplify radicals by factoring out any perfect powers from the radicand.

[TK]

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Example 6.4.4.

Simplify

\sqrt[3]{108}

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Solution.

We look for perfect cubes that divide evenly into

108 .

The easiest way to do this is to try the perfect cubes in order:

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1, 8, 27, 64, 125, \dots

and so on, until we find one that is a factor. For this example, we find that

108 = 27 \cdot 4 .

Applying the Product Rule, we write

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\begin{aligned} \sqrt[3]{108} & = \sqrt[3]{27} \sqrt[3]{4} & Simplify: \sqrt[3]{27} = 3. \\ = 3 \sqrt[3]{4} \end{aligned}

This expression is considered simpler than the original radical because the new radicand,

4,

is smaller than the original,

108 .

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Caution 6.4.5.

Finding a decimal approximation for a radical is not the same as simplifying the radical. In the Example above, we can use a calculator to find

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\sqrt[3]{108} \approx 4.762

but

4.762

is not the exact value for

\sqrt[3]{108} .

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In long calculations, too much error may be introduced by approximating each radical. However,

3 \sqrt[3]{4}

is equal to

\sqrt[3]{108},

so their values are exactly the same. We can replace one expression by the other without losing accuracy.

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Checkpoint 6.4.6. Practice 1.

Simplify

\sqrt[4]{80}

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Solution.

\sqrt[4]{80} = \sqrt[4]{16 \cdot 5} = \sqrt[4]{16} \sqrt[4]{5} = 2 \sqrt[4]{5}

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We can also simplify radicals containing variables. If the exponent on the variable is a multiple of the index, we can extract the variable from the radical. For instance,

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\sqrt[3]{12} = x^{12 / 3} = x^{4}

(You can verify this by noting that

(x^{4})^{3} = x^{12} .

)

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Caution 6.4.7.

Do not confuse

\sqrt{a^{16}}

with

\sqrt{16} .

Compare the two radicals:

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\sqrt{16} = 4 but \sqrt{a^{16}} = a^{8}

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Example 6.4.8.

Simplify

\sqrt[3]{x^{11}}

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Solution.

Because the exponent on the variable, 11, is not a multiple of the index, 3, we factor out the highest power that is a multiple.

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\begin{aligned} \sqrt[3]{x^{11}} & = \sqrt[3]{x^{9} \cdot x^{2}} & Apply the Product Rule. \\ = \sqrt[3]{x^{9}} \cdot \sqrt[3]{x^{2}} & Simplify \sqrt[3]{x^{9}} = x^{9 / 3} . \\ = x^{3} \sqrt[3]{x^{2}} \end{aligned}

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Checkpoint 6.4.9. Practice 2.

Simplify

\sqrt[4]{a^{15}}

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Solution.

a^{3} \sqrt[4]{a^{3}}

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To simplify a root of a monomial, we factor the coefficient and each power of a variable separately.

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Example 6.4.10.

Simplify each radical.

[TK]

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$\sqrt{18 x^{5}}$
$\sqrt[3]{24 x^{6} y^{8}}$

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Solution.

The index of the radical is $2,$ so we look for perfect square factors of $18 x^{5} .$ The factor $9$ is a perfect square, and $x^{4}$ has an exponent divisible by $2 .$ Thus,

$\begin{aligned} \sqrt{18 x^{5}} & = \sqrt{9 x^{4} \cdot 2 x} & Apply the Product Rule. \\ = \sqrt{9 x^{4}} \sqrt{2 x} & Take square roots. \\ = 3 x^{2} \sqrt{2 x} \end{aligned}$
The index of the radical is $3,$ so we look for perfect cube factors of $24 x^{6} y^{8} .$ The factor $8$ is a perfect cube, and $x^{6}$ and $y^{6}$ have exponents divisible by $3 .$ Thus,

$\begin{aligned} \sqrt[3]{24 x^{6} y^{8}} & = \sqrt[3]{8 x^{6} y^{6} \cdot 3 y^{2}} & Apply the Product Rule. \\ = \sqrt[3]{8 x^{6} y^{6}} \sqrt[3]{3 y^{2}} & Take cube roots. \\ = 2 x^{2} y^{2} \sqrt[3]{3 y^{2}} \end{aligned}$

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Checkpoint 6.4.11. Practice 3.

Simplify

\sqrt[3]{250 b^{7}}

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Hint: Look for perfect cube factors of

250 b^{7} .

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Apply the product Rule.

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Take cube roots.

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Solution.

5 b^{2} \sqrt[3]{2 b}

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Caution 6.4.12.

It is worth stating again that the Product Rule applies only to products under the radical, not to sums or differences. For example,

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\sqrt{4 \cdot 9} = \sqrt{4} \sqrt{9} = 2 \cdot 3, but \sqrt{4 + 9} \neq \sqrt{4} + \sqrt{9}

and

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\sqrt[3]{x^{3} y^{6}} = \sqrt[3]{x^{3}} \sqrt[3]{y^{6}} = x y^{2}, but \sqrt[3]{x^{3} - y^{6}} \neq x - y^{2}

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Checkpoint 6.4.13. QuickCheck 2.

True or false.

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To simplify a radical means to find a decimal approximation.
$\sqrt{x^{16}} = x^{4}$
We cannot simplify $\sqrt[3]{x^{10}} .$
$\sqrt{83^{5}} = 83^{2} \sqrt{83}$

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Subsection 6.4.3 Sums and Differences of Radicals

How can we add or subtract radicals? Keep in mind that, in general,

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\sqrt[n]{a} + \sqrt[n]{b} \neq \sqrt[n]{a + b}

We cannot add or subtract expressions that occur under a radical. For example, you can check that

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\sqrt{16} - \sqrt{4} \neq \sqrt{12} and \sqrt{7} + \sqrt{7} \neq \sqrt{14}

However, if two roots have the same index and identical radicands, they are said to be like radicals. We can add or subtract like radicals in the same way that we add or subtract like terms, namely by adding or subtracting their coefficients. For example, we know that

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2 r + 3 r = (2 + 3) r = 5 r

where

r

is a variable that can stand for any real number. In particular, if

r = \sqrt{x},

we have

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2 \sqrt{x} + 3 \sqrt{x} = (2 + 3) \sqrt{x} = 5 \sqrt{x}

So we may add like radicals by adding their coefficients. The same idea applies to subtraction.

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Like Radicals.

To add or subtract like radicals, we add or subtract their coefficients. We do not change the index or the radicand.

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Example 6.4.14.

$3 \sqrt{3} + 4 \sqrt{3} = (3 + 4) \sqrt{3} = 7 \sqrt{3}$
$4 \sqrt[3]{y} - 6 \sqrt[3]{y} = (4 - 6) \sqrt[3]{y} = - 2 \sqrt[3]{y}$

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Caution 6.4.15.

In Example 6.4.14a, $3 \sqrt{3} + 4 \sqrt{3} \neq 7 \sqrt{6} .$ Only the coefficients are added; the radicand does not change.
Sums of radicals with different radicands or different indices cannot be combined. Thus,

$\begin{aligned} \sqrt{11} + \sqrt{5} & \neq \sqrt{16} & Radicands are not the same. \\ \sqrt[3]{10 x} - \sqrt[3]{2 x} & \neq \sqrt[3]{8 x} & Radicands are not the same. \\ \sqrt[3]{7} + \sqrt{7} & \neq \sqrt[5]{7} & Indices are not the same. \end{aligned}$

None of the expressions above can be simplified.

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Checkpoint 6.4.16. Practice 4.

Simplify

8 \sqrt[3]{5 b} + 4 \sqrt[3]{5 b}

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Solution.

12 \sqrt[3]{5 b}

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Checkpoint 6.4.17. QuickCheck 3.

True or false.

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We combine like radicals the same way we combine like terms: by adding or subtracting their coefficients.
To add radicals with different indices, we multiply the indices.
$3 \sqrt{5} + 6 \sqrt{5} = 9 \sqrt{10}$
Like radicals must have identical coefficients.

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Sometimes we must simplify the roots in a sum or difference before we can recognize like radicals.

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Example 6.4.18.

Simplify

\sqrt[3]{40 x^{2}} - 3 \sqrt[3]{16 x^{2}} + \sqrt[3]{54 x^{2}}

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Solution.

We simplify each cube root by factoring perfect cubes from the radicals.

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\begin{aligned} \sqrt[3]{40 x^{2}} & - 3 \sqrt[3]{16 x^{2}} + \sqrt[3]{54 x^{2}} = \\ = \sqrt[3]{8 \cdot 5 x^{2}} - 3 \sqrt[3]{8 \cdot 2 x^{2}} + \sqrt[3]{27 \cdot 2 x^{2}} & Extract roots. \\ = 2 \sqrt[3]{5 x^{2}} - 3 \cdot 2 \sqrt[3]{2 x^{2}} + 3 \sqrt[3]{2 x^{2}} \\ = 2 \sqrt[3]{5 x^{2}} - 6 \sqrt[3]{2 x^{2}} + 3 \sqrt[3]{2 x^{2}} & Combine like radicals. \\ = 2 \sqrt[3]{5 x^{2}} - 3 \sqrt[3]{2 x^{2}} \end{aligned}

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Checkpoint 6.4.19. Practice 5.

Simplify

\sqrt{12 x} + \sqrt{8 x} - 4 \sqrt{18 x}

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Solution.

2 \sqrt{3 x} - 10 \sqrt{2 x}

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Subsection 6.4.4 Products and Quotients of Radicals

Because of the Product and Quotient Rules, we can multiply or divide radicals of the same index.

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\begin{aligned} \sqrt[n]{a} \sqrt[n]{b} & = \sqrt[n]{a b} & (a, b \geq 0) \\ \frac{\sqrt[n]{a}}{\sqrt[n]{b}} & = \sqrt[n]{\frac{a}{b}} & (a \geq 0, b > 0) \end{aligned}

These rules tell us that for products and quotients, the radicands do not have to be the same; only the indices must match.

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Example 6.4.20.

Simplify.

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$\sqrt[4]{6 x^{2}} \sqrt[4]{8 x^{3}}$
$\frac{\sqrt[3]{16 y^{5}}}{\sqrt[3]{y^{2}}}$

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Solution.

We apply the Product Rule to write the product as a single radical, then simplify.

$\begin{aligned} \sqrt[4]{6 x^{2}} \sqrt[4]{8 x^{3}} & = \sqrt[4]{48 x^{5}} & Factor out perfect fourth powers. \\ = \sqrt[4]{16 x^{4}} \sqrt[4]{3 x} & Simplify. \\ = 2 x \sqrt[4]{3 x} \end{aligned}$
We apply the Quotient Rule to write the quotient as a single radical, then reduce the fraction under the radical.

$\begin{aligned} \frac{\sqrt[3]{16 y^{5}}}{\sqrt[3]{y^{2}}} & = \sqrt[3]{\frac{16 y^{5}}{y^{2}}} & Reduce. \\ = \sqrt[3]{16 y^{3}} & Simplify: factor out perfect cubes. \\ = \sqrt[3]{8 y^{3}} \sqrt[3]{2} = 2 y \sqrt[3]{2} \end{aligned}$

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Checkpoint 6.4.21. Practice 6.

Simplify.

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$\sqrt{\frac{18 x^{5}}{25 y^{4}}}$
$\sqrt[3]{2 a^{2}} \sqrt[3]{6 a^{2}}$

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Solution.

$\frac{3 x^{2} \sqrt{2 x}}{5 y^{2}}$
$a \sqrt[3]{12 a}$

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For products involving binomials, we apply the distributive law.

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Example 6.4.22.

Expand each product.

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

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Solution.

We multiply each term within the parentheses by $2 \sqrt{3} .$

$\begin{aligned} 2 \sqrt{3} & (\sqrt{2 x} + 5 \sqrt{6}) = \\ = 2 \sqrt{3} \cdot \sqrt{2 x} + 2 \sqrt{3} \cdot 5 \sqrt{6} & Apply the Product Rule. \\ = 2 \sqrt{3 \cdot 2 x} + 2 \cdot 5 \sqrt{3 \cdot 6} & Simplify the radicals. \\ = 2 \sqrt{6 x} + 10 \sqrt{18} \\ = 2 \sqrt{6 x} + 10 \cdot \sqrt{9} \sqrt{2} \\ = 2 \sqrt{6 x} + 30 \sqrt{2} \end{aligned}$
We use the "FOIL" method to expand the product.

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

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Note 6.4.23.

In part (a) of the Example above, observe that

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2 \sqrt{3} \cdot 5 \sqrt{6} = 2 \cdot 5 \cdot \sqrt{3} \cdot \sqrt{6} = 10 \sqrt{18}

We multiply together any expressions outside the radical, and apply the product rule to expressions under the radical.

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Checkpoint 6.4.24. Practice 7.

Expand

(\sqrt{5} - 2 \sqrt{3})^{2}

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Solution.

17 - 4 \sqrt{15}

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Checkpoint 6.4.25. QuickCheck 4.

True or false.

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We can only simplify products or quotients of like radicals.
$4 (3 \sqrt{5}) = 3 \cdot 4 + 3 \sqrt{5}$
$(\sqrt{3} + \sqrt{5})^{2} = 3 + 5 = 8$
$(3 \sqrt{x})^{2} = 9 x$

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Subsection 6.4.5 Rationalizing the Denominator

It is easier to work with radicals if there are no roots in the denominators of fractions. We can use the fundamental principle of fractions to remove radicals from the denominator. This process is called rationalizing the denominator. For square roots, we multiply the numerator and denominator of the fraction by the radical in the denominator.

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Example 6.4.26.

Rationalize the denominator of each fraction.

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$\sqrt{\frac{1}{3}}$
$\frac{\sqrt{2}}{\sqrt{50 x}}$

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Solution.

First, we apply the Quotient Rule to write the radical as a quotient.

$\begin{aligned} \sqrt{\frac{1}{3}} & = \frac{\sqrt{1}}{\sqrt{3}} = \frac{1}{\sqrt{3}} & Multiply numerator and denominator by \sqrt{3} . \\ = \frac{1 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} \\ = \frac{\sqrt{3}}{3} \end{aligned}$
It is always best to simplify the denominator before rationalizing.

$\frac{\sqrt{2}}{\sqrt{50 x}} = \frac{\sqrt{2}}{5 \sqrt{2 x}}$

Now we can see that we should multiply numerator and denominator by $\sqrt{2 x}$ — not $\sqrt{50 x}!$

$\begin{aligned} \frac{\sqrt{2} \cdot \sqrt{2 x}}{5 \sqrt{2 x} \cdot \sqrt{2 x}} & = \frac{\sqrt{4 x}}{5 (2 x)} & Apply the Product Rule. \\ = \frac{2 \sqrt{x}}{10 x} = \frac{\sqrt{x}}{5 x} & Simplify. \end{aligned}$

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Checkpoint 6.4.27. Practice 8.

Rationalize the denominator of

\frac{- \sqrt{3}}{\sqrt{7}}

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Solution.

\frac{- \sqrt{21}}{7}

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If the denominator of a fraction is a binomial in which one or both terms is a radical, we can use a special building factor to rationalize it. First, recall that

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(p - q) (p + q) = p^{2} - q^{2}

In particular, then,

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(\sqrt{b} - \sqrt{c}) (\sqrt{b} + \sqrt{c}) = (\sqrt{b})^{2} - (\sqrt{c})^{2} = b - c

The product contains no radicals. Each of the two factors

\sqrt{b} - \sqrt{c}

and

\sqrt{b} + \sqrt{c}

is said to be the conjugate of the other.

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Now consider a fraction of the form

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\frac{a}{b + \sqrt{c}}

If we multiply the numerator and denominator of this fraction by the conjugate of the denominator, we get

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\frac{a (b - \sqrt{c})}{(b + \sqrt{c}) (b - \sqrt{c})} = \frac{a b - a \sqrt{c}}{b^{2} - (\sqrt{c})^{2}} = \frac{a b - a \sqrt{c}}{b^{2} - c}

The denominator of the fraction no longer contains any radicals—it has been rationalized.

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Example 6.4.28.

Rationalize the denominator:

\frac{x}{\sqrt{2} + \sqrt{x}} .

[TK]

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Solution.

We multiply numerator and denominator by the conjugate of the denominator,

\sqrt{2} - \sqrt{x} .

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\frac{x (\sqrt{2} - \sqrt{x})}{(\sqrt{2} + \sqrt{x}) (\sqrt{2} - \sqrt{x})} = \frac{x (\sqrt{2} - \sqrt{x})}{2 - x}

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Checkpoint 6.4.29. Practice 9.

Rationalize the denominator of

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

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Solution.

3 + \sqrt{6}

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Checkpoint 6.4.30. QuickCheck 5.

Fill in the blanks.

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We rationalize the denominator to remove from the denominator.
To rationalize a binomial denominator, we multiply by its .
Before rationalizing, it is always best to .
$\frac{5}{\sqrt{5}} =$

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Exercises 6.4.6 Problem Set 6.4

Warm Up

Exercise Group.

Each question in Problems 1–4 is followed by three examples. Use your calculator to decide if the examples are true or false, then circle the correct answer.

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