Simplify the following.5x(4+x)\sqrt{5x}(4+\sqrt{x})

10x+10x10\sqrt{x}+\sqrt{10x}10x+10x

Rationalize the denominator.6+x−x\frac{6+\sqrt{x}}{-\sqrt{x}}

6x−x−1\frac{6\sqrt{x}}{-x}-1

Rationalize the denominator and simplify the radical expression.75−6\frac{\sqrt7}{5-\sqrt6}

57+4219\frac{5\sqrt7+\sqrt{42}}{19}

57+42−11\frac{5\sqrt7+\sqrt{42}}{-11}

57+4221\frac{5\sqrt7+\sqrt{42}}{21}

9. Roots, Radicals, and Complex Numbers

Multiplying, Dividing, and Rationalizing Radicals

9. Roots, Radicals, and Complex Numbers

Multiplying, Dividing, and Rationalizing Radicals: Videos & Practice Problems

Video Lessons Worksheet

Topic summary

Rationalizing the denominator involves eliminating radicals from the denominator of a fraction to express it in standard form. For a single-term denominator like 13, multiply numerator and denominator by the radical to create a perfect square, e.g., 33. For binomial denominators such as 12+3, multiply by the conjugate (flip the sign) to apply the difference of squares, removing radicals and yielding a rational denominator. This process uses FOIL to expand and simplify expressions effectively.

Concept

Multiplication of Radical Expressions

Video duration:

Multiplication of Radical Expressions Video Summary

Multiplying and simplifying radical expressions involves applying key properties of radicals and multiplication. When multiplying expressions that include radicals, numbers, and variables, the product rule of radicals is essential. This rule states that for radicals with the same index, you can combine them under a single radical by multiplying the values inside. For example, multiplying $\sqrt{5}$ and $\sqrt{10}$ results in $\sqrt{50}$ because $\sqrt{5} \times \sqrt{10} = \sqrt{5 \times 10} = \sqrt{50}$.

After combining radicals, it is important to simplify the resulting radical expression by factoring out perfect squares. Since 25 is a perfect square and divides 50, $\sqrt{50}$ can be rewritten as $\sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}$. This simplification makes the expression easier to work with and more concise.

When multiplying coefficients and variables outside the radicals, multiply them as usual. For instance, multiplying \$2a$ and \(8$ gives \)16a$. Combining this with the simplified radical, the product $2a \sqrt{5} \times 8 \sqrt{10}$ simplifies to $16a \times 5 \sqrt{2} = 80a \sqrt{2}$.

In cases where a radical term multiplies a binomial expression, the distributive property is used. For example, multiplying $7\sqrt{2}$ by $(3 + \sqrt{2})$ requires distributing $7\sqrt{2}$ to both terms inside the parentheses:

\[7\sqrt{2} \times 3 + 7\sqrt{2} \times \sqrt{2}\]

The first term simplifies to $21\sqrt{2}$ by multiplying the coefficient 7 and 3. The second term involves multiplying two radicals with the same index, which can be combined under one radical:

\[7 \times \sqrt{2 \times 2} = 7 \times \sqrt{4}\]

Since $\sqrt{4} = 2$, this becomes $7 \times 2 = 14$. Thus, the entire expression simplifies to:

\[21\sqrt{2} + 14\]

Note that terms involving radicals and those without radicals cannot be combined further because they are not like terms.

These principles extend to more complex expressions involving variables and multiple terms, where the FOIL method (First, Outer, Inner, Last) and distributive property help in expanding products. Understanding how to multiply radicals, simplify them by extracting perfect squares, and apply distribution ensures accurate and simplified results in radical expressions.

Study Smarter with Worksheets.

Follow along with each video using our printable worksheets

Problem

Simplify the following.
$(2 \sqrt{5}) (4 \sqrt{7})$

$8 the square root of 2$

$8 the square root of 13$

$6 the square root of 35$

$8$\sqrt{35}$$

Problem

Simplify the following.
$(4 \sqrt{x}) (3 \sqrt{x y})$

$12x$\sqrt{y}$$

$7x$\sqrt{y}$$

$x^2y$\sqrt{12}$$

$12x^2y$

Problem

Simplify the following.
$\sqrt{3} (2 - \sqrt{6})$

$2 \sqrt{3} - 3$

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

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

$3 - 2 \sqrt{3}$

Problem

Simplify the following.
$\sqrt{5 x} (4 + \sqrt{x})$

$4 \sqrt{5 x} + \sqrt{10 x}$

$4$\sqrt{5x}$+x$\sqrt$5$

$10$\sqrt{x}$+x$\sqrt$5$

$10$\sqrt{x}$+$\sqrt{10x}$$

Example

Multiplication of Radical Expressions Example 1

Video duration:

Multiplication of Radical Expressions Example 1 Video Summary

When simplifying expressions involving binomials with radicals, it is essential to apply the FOIL method carefully. Consider the expression (4 - 6√2y)(3 + √2y). Both parentheses are binomials, meaning they each contain two terms. The FOIL acronym guides the multiplication process: multiply the First terms, then the Outer terms, followed by the Inner terms, and finally the Last terms.

Starting with the first terms, multiply 4 by 3 to get 12. For the outer terms, multiply 4 by √2y, which results in 4√2y. The inner terms involve multiplying -6√2y by 3, yielding -18√2y. Lastly, multiply the last terms: -6√2y times √2y. Here, the product rule for radicals applies, which states that √a × √a = a. Thus, √2y × √2y equals 2y, and multiplying by -6 gives -12y.

Next, combine like terms. The middle terms, 4√2y and -18√2y, share the same radical part √2y, so they can be combined as coefficients: 4 - 18 = -14, resulting in -14√2y. The simplified expression becomes:

\[12 - 14\sqrt{2y} - 12y\]

No further simplification is possible since the terms involve different radicals or variables. This process highlights the importance of recognizing like terms, applying the FOIL method accurately, and using properties of radicals such as the product rule. Understanding how to multiply and simplify expressions with radicals is fundamental in algebra and helps build a strong foundation for more advanced mathematical concepts.

Problem

Simplify the following.
$(4 \sqrt{3} - 5)^{2}$

$73 - 40 \sqrt{3}$

$- 9 - 20 \sqrt{3}$

$41 - 20 \sqrt{3}$

$23 - 40 \sqrt{3}$

Example

Multiplication of Radical Expressions Example 2

Video duration:

Concept

Rationalizing Denominators

Video duration:

Rationalizing Denominators Video Summary

In mathematics, it is essential to understand that radicals should not be left in the denominator of a fraction. This practice is known as rationalizing the denominator, which is a straightforward process that allows us to eliminate radicals from the bottom of a fraction. For instance, while expressions like $ \frac{1}{\sqrt{2}} $ or $ \frac{\sqrt{2}}{\sqrt{8}} $ can sometimes simplify to perfect squares, there are cases where this is not possible, such as $ \frac{1}{\sqrt{3}} $.

To rationalize the denominator, you multiply both the numerator and the denominator by the radical present in the denominator. For example, to rationalize $ \frac{1}{\sqrt{3}} $, you would multiply by $ \sqrt{3} $ over itself, resulting in:

\[\frac{1 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}} = \frac{\sqrt{3}}{3}\]

This transformation is valid because multiplying by $ \frac{\sqrt{3}}{\sqrt{3}} $ is equivalent to multiplying by 1, thus preserving the value of the expression. The denominator now becomes a rational number, specifically $ 3 $, while the numerator remains as $ \sqrt{3} $.

To verify the equivalence of the two expressions, you can use a calculator. Inputting $ \frac{1}{\sqrt{3}} $ yields approximately $ 0.57 $, which matches the result of $ \frac{\sqrt{3}}{3} $. This demonstrates that both forms represent the same value, but rationalizing the denominator provides a more conventional representation in mathematics.

Problem

Rationalize the denominator.
$- \frac{5}{2 \sqrt{7}}$

$- \frac{5}{14}$

$- \frac{5 \sqrt{7}}{2}$

$- \frac{5 \sqrt{7}}{14}$

$- \frac{10 \sqrt{7}}{14}$

Problem

Rationalize the denominator.
$\frac{6 + \sqrt{x}}{- \sqrt{x}}$

$\frac{6 \sqrt{x}}{- x} - 1$

$6 \sqrt{x} + x$

$\frac{6 \sqrt{x}}{x} + 1$

$\frac{7 \sqrt{x}}{- x}$

Example

Rationalizing Denominators Example 3

Video duration:

Rationalizing Denominators Example 3 Video Summary

When simplifying quotients involving radicals, the goal is to express the result in its lowest terms by breaking down complex expressions and applying properties of radicals and division. For example, consider the expression $$\frac{\sqrt{18}$ + 6}{3}$ . To simplify, first separate the terms as $$\frac{\sqrt{18}$}{3} + $\frac{6}{3}$$ . The fraction $$\frac{6}{3}$$ simplifies to 2, so the expression becomes $$\frac{\sqrt{18}$}{3} + 2$ . Next, apply the product rule for square roots: $$\sqrt{18}$ = $\sqrt{9 \times 2}$ = $\sqrt{9}$ $\times$ $\sqrt{2}$ = 3$\sqrt{2}$$ . Substituting back, the expression is $$\frac{3\sqrt{2}$}{3} + 2$ , and the 3's cancel, leaving $$\sqrt{2}$ + 2$ as the fully simplified form.

In another example, consider $$\frac{2 \times \sqrt[3]{16}$ - 4}{4}$ . Begin by splitting the numerator over the denominator: $2 $\times$ $\frac{\sqrt[3]{16}$}{4} - $\frac{4}{4}$$ . The term $$\frac{4}{4}$$ simplifies to 1, so the expression becomes $2 $\times$ $\frac{\sqrt[3]{16}$}{4} - 1$ . Using the product rule for cube roots, rewrite $$\sqrt$[3]{16} = $\sqrt$[3]{8 $\times$ 2} = $\sqrt$[3]{8} $\times$ $\sqrt$[3]{2} = 2 $\times$ $\sqrt$[3]{2}$ . Substituting this back gives $2 $\times$ $\frac{2 \times \sqrt[3]{2}$}{4} - 1$ . Multiplying the constants in the numerator yields $$\frac{4 \times \sqrt[3]{2}$}{4} - 1$ , and the 4's cancel, leaving $$\sqrt$[3]{2} - 1$ as the simplified quotient.

These examples highlight the importance of separating terms in a quotient, simplifying fractions, and applying radical properties such as the product rule to identify perfect squares or cubes. This approach ensures expressions are reduced to their simplest form, making them easier to interpret and use in further calculations.

Example

Rationalizing Denominators Example 4

Video duration:

Rationalizing Denominators Example 4 Video Summary

When simplifying quotients involving square roots, it is important to express the result in lowest terms and avoid having radicals in the denominator. For example, consider the expression $$\frac{\sqrt{81}}{\sqrt{9}}$$ . Both 81 and 9 are perfect squares, so their square roots simplify directly: $$\sqrt{81} = 9$$ and $$\sqrt{9} = 3$$ . This reduces the expression to $$\frac{9}{3} = 3$$ . This approach eliminates the radical in the denominator without needing to multiply by a conjugate.

Alternatively, the quotient rule for radicals can be applied, which states that $$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$$ for positive values of $a$ and $b$ . Using this, the original expression becomes $$\sqrt{\frac{81}{9}} = \sqrt{9} = 3$$ , confirming the same simplified result.

In another example, consider $$\frac{\sqrt{49x^2}}{\sqrt{x^2}}$$ . Recognizing that the square root of $x^2$ is $x$ (assuming $x $\geq$ 0$ ), the denominator simplifies to $x$ . The numerator simplifies as $$\sqrt{49x^2}$ = 7x$ . Thus, the expression reduces to $$\frac{7x}{x}$ = 7$ , after canceling the common factor $x$ . Using the quotient rule again, this expression can be rewritten as $$\sqrt{\frac{49x^2}{x^2}} = \sqrt{49} = 7$$ , yielding the same result.

These examples highlight the importance of first checking for straightforward simplifications before resorting to more complex methods like rationalizing denominators by multiplying by conjugates. Understanding and applying the properties of square roots and the quotient rule can streamline the simplification process and ensure expressions are written in their simplest form.

Concept

Rationalizing Denominators Using Conjugates

Video duration:

Rationalizing Denominators Using Conjugates Video Summary

Rationalizing the denominator is an important technique in algebra, particularly when dealing with fractions that contain radicals. When you encounter a fraction like 1 over √3, you can eliminate the radical by multiplying both the numerator and the denominator by √3. This process effectively removes the radical from the denominator, resulting in a simplified expression.

However, when the denominator consists of two terms, such as 2 + √3, simply multiplying by the same radical will not suffice. In this case, you need to use the concept of conjugates. The conjugate of a binomial is formed by changing the sign between the two terms. For example, the conjugate of 2 + √3 is 2 - √3.

To rationalize a denominator with two terms, multiply both the numerator and the denominator by the conjugate. This means that for the expression 1 over 2 + √3, you would multiply by 2 - √3. The multiplication in the denominator will yield a difference of squares:

\[ (2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \]

In the numerator, you multiply straight across:

\[ 1 \cdot (2 - \sqrt{3}) = 2 - \sqrt{3} \]

Thus, the expression simplifies to:

\[ \frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3} \]

This process effectively rationalizes the denominator, resulting in a rational number. Remember, when dealing with a single-term denominator, multiply by that term, and when faced with a two-term denominator, always use the conjugate to eliminate the radical. This technique is essential for simplifying expressions and solving equations in algebra.

Problem

Rationalize the denominator and simplify the radical expression.
$\frac{\sqrt{7}}{5 - \sqrt{6}}$

$\frac{\sqrt{7}}{19}$

$\frac{5 \sqrt{7} + \sqrt{42}}{- 11}$

$\frac{5 \sqrt{7} + \sqrt{42}}{21}$

$\frac{5 \sqrt{7} + \sqrt{42}}{19}$

Problem

Rationalize the denominator and simplify the radical expression.
$\frac{2 - \sqrt{3}}{2 + \sqrt{3}}$

$7 - 4 \sqrt{3}$

$\frac{1}{7 + 4 \sqrt{3}}$

$7 + 4 \sqrt{3}$

$\frac{1}{7 - 4 \sqrt{3}}$

Example

Rationalizing Denominators Using Conjugates Example 5

Video duration:

Rationalizing Denominators Using Conjugates Example 5 Video Summary

Rationalizing the denominator is a process used to eliminate radicals from the denominator of a fraction, making the expression easier to work with. When a denominator contains a square root or other radical, multiplying by the conjugate is an effective method to remove it. The conjugate of a binomial expression involving radicals is formed by changing the sign between the two terms. For example, the conjugate of $\sqrt{x} + 1$ is $\sqrt{x} - 1$.

Consider the expression $\frac{x + 2}{\sqrt{x + 1}}$. To rationalize the denominator, multiply both the numerator and denominator by the conjugate of the denominator. Since the denominator is a single term $\sqrt{x + 1}$, its conjugate is $\sqrt{x + 1}$ itself, but to rationalize, we multiply numerator and denominator by $\sqrt{x + 1}$:

\[\frac{x + 2}{\sqrt{x + 1}} \times \frac{\sqrt{x + 1}}{\sqrt{x + 1}} = \frac{(x + 2) \sqrt{x + 1}}{(\sqrt{x + 1})^2} = \frac{(x + 2) \sqrt{x + 1}}{x + 1}\]

In cases where the denominator is a binomial with radicals, such as $\frac{1}{\sqrt{x} + \sqrt{y}}$, the conjugate is formed by changing the plus sign to a minus sign: $\sqrt{x} - \sqrt{y}$. Multiplying numerator and denominator by this conjugate yields:

\[\frac{1}{\sqrt{x} + \sqrt{y}} \times \frac{\sqrt{x} - \sqrt{y}}{\sqrt{x} - \sqrt{y}} = \frac{\sqrt{x} - \sqrt{y}}{(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})}\]

Expanding the denominator using the difference of squares formula, $(a + b)(a - b) = a^2 - b^2$, gives:

\[ (\sqrt{x})^2 - (\sqrt{y})^2 = x - y \]

Thus, the expression simplifies to:

\[ \frac{\sqrt{x} - \sqrt{y}}{x - y} \]

This process effectively removes radicals from the denominator, making the expression more manageable for further algebraic manipulation or evaluation. Understanding how to multiply by conjugates and apply the difference of squares formula is essential for rationalizing denominators involving radicals.

Here's what students ask on this topic:

Rationalizing the denominator means eliminating any radicals (square roots) from the denominator of a fraction. This is important because having radicals in the denominator is generally considered improper form in mathematics. To rationalize, you multiply both the numerator and denominator by a value that will remove the radical from the denominator. For example, if the denominator is a single radical like $\sqrt{3}$ , you multiply numerator and denominator by $\sqrt{3}$ to create a perfect square in the denominator. This process converts the denominator into a rational number, making the expression easier to work with and standardizing its form.

When the denominator has a single radical term, such as $\frac{1}{\sqrt{3}}$ , you rationalize it by multiplying both the numerator and denominator by the same radical. For example, multiply by $\sqrt{3}$ over $\sqrt{3}$ . This gives:

$\frac{1}{\sqrt{3}} \cdot \sqrt{3} \sqrt{3} = \sqrt{3} 3$

The denominator becomes $3$ because $\sqrt{3} \cdot \sqrt{3} = \sqrt{9} = 3$ . This removes the radical from the denominator, leaving a rational number.

The conjugate of a binomial radical expression is formed by changing the sign between the two terms. For example, the conjugate of $2 + \sqrt{3}$ is $2 - \sqrt{3}$ . It is used in rationalizing denominators when the denominator has two terms (a binomial) involving radicals. Multiplying the denominator by its conjugate uses the difference of squares formula:

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

This eliminates the radical because squaring the radical term removes the root, resulting in a rational number. This method is necessary because simply multiplying by the same binomial does not remove the radical.

To rationalize a denominator with two terms, such as $\frac{1}{2 + \sqrt{3}}$ , multiply both numerator and denominator by the conjugate of the denominator, which is $2 - \sqrt{3}$ . This gives:

$\frac{1}{2 + \sqrt{3}} \cdot 2 - \sqrt{3} 2 - \sqrt{3} = \frac{2 - \sqrt{3}}{(2)^2 - (\sqrt{3})^2}$

The denominator simplifies to $4 - 3 = 1$ , so the expression becomes $\frac{2 - \sqrt{3}}{1}$ , which is just $2 - \sqrt{3}$ . This removes the radical from the denominator.

Multiplying the numerator and denominator by the same binomial radical expression, such as $2 + \sqrt{3}$ , does not eliminate the radical in the denominator because it results in a product that still contains radicals. When you multiply a binomial by itself, you get terms like $2 \cdot \sqrt{3}$ twice, which are radicals. This means the denominator still has radicals, which is not the goal. Instead, multiplying by the conjugate (which flips the sign) creates a difference of squares, eliminating the radicals and leaving a rational number in the denominator.

The difference of squares is a special algebraic identity expressed as:

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

When rationalizing denominators with binomials containing radicals, multiplying by the conjugate applies this identity. Squaring the radical term removes the root because $(2sqrt{b}2)^2 = b$ . This results in the denominator becoming a rational number (difference of two squares), effectively eliminating the radical. This method is essential for simplifying expressions with binomial denominators involving radicals.

Multiplying, Dividing, and Rationalizing Radicals: Videos & Practice Problems

Multiplication of Radical Expressions

Multiplication of Radical Expressions Video Summary

Simplify the following.
$(2 \sqrt{5}) (4 \sqrt{7})$

Simplify the following.
$(4 \sqrt{x}) (3 \sqrt{x y})$

Simplify the following.
$\sqrt{3} (2 - \sqrt{6})$

Simplify the following.
$\sqrt{5 x} (4 + \sqrt{x})$

Multiplication of Radical Expressions Example 1

Multiplication of Radical Expressions Example 1 Video Summary

Simplify the following.
$(4 \sqrt{3} - 5)^{2}$

Multiplication of Radical Expressions Example 2

Rationalizing Denominators

Rationalizing Denominators Video Summary

Rationalize the denominator.
$- \frac{5}{2 \sqrt{7}}$

Rationalize the denominator.
$\frac{6 + \sqrt{x}}{- \sqrt{x}}$

Rationalizing Denominators Example 3

Rationalizing Denominators Example 3 Video Summary

Rationalizing Denominators Example 4

Rationalizing Denominators Example 4 Video Summary

Rationalizing Denominators Using Conjugates

Rationalizing Denominators Using Conjugates Video Summary

Rationalize the denominator and simplify the radical expression.
$\frac{\sqrt{7}}{5 - \sqrt{6}}$

Rationalize the denominator and simplify the radical expression.
$\frac{2 - \sqrt{3}}{2 + \sqrt{3}}$

Rationalizing Denominators Using Conjugates Example 5

Rationalizing Denominators Using Conjugates Example 5 Video Summary

Here's what students ask on this topic:

What does it mean to rationalize the denominator in a radical expression?

How do you rationalize the denominator when it contains a single radical term?

What is the conjugate of a binomial radical expression and why is it used in rationalizing denominators?

How do you rationalize the denominator of a fraction with a binomial denominator like $\frac{1}{2 + \sqrt{3}}$ ?

Why can't you just multiply the numerator and denominator by the same binomial radical expression to rationalize a binomial denominator?

What is the difference of squares and how does it help in rationalizing denominators?

Multiplying, Dividing, and Rationalizing Radicals: Videos & Practice Problems

Multiplication of Radical Expressions

Multiplication of Radical Expressions Video Summary

Simplify the following.(25)(47)(2\(\sqrt\)5)(4\(\sqrt\)7)

Simplify the following.(4x)(3xy)(4\(\sqrt{x}\))(3\(\sqrt{xy}\))

Simplify the following.3(2−6)\(\sqrt\)3(2-\(\sqrt\)6)

Simplify the following.5x(4+x)\(\sqrt{5x}\)(4+\(\sqrt{x}\))

Multiplication of Radical Expressions Example 1

Multiplication of Radical Expressions Example 1 Video Summary

Simplify the following.(43−5)2(4\(\sqrt\)3-5)^2

Multiplication of Radical Expressions Example 2

Rationalizing Denominators

Rationalizing Denominators Video Summary

Rationalize the denominator. −527-\(\frac{5}{2\sqrt7}\)

Rationalize the denominator.6+x−x\(\frac{6+\sqrt{x}\)}{-\(\sqrt{x}\)}

Rationalizing Denominators Example 3

Rationalizing Denominators Example 3 Video Summary

Rationalizing Denominators Example 4

Rationalizing Denominators Example 4 Video Summary

Rationalizing Denominators Using Conjugates

Rationalizing Denominators Using Conjugates Video Summary

Rationalize the denominator and simplify the radical expression.75−6\(\frac{\sqrt7}{5-\sqrt6}\)

Rationalize the denominator and simplify the radical expression.2−32+3\(\frac{2-\sqrt3}{2+\sqrt3}\)

Rationalizing Denominators Using Conjugates Example 5

Rationalizing Denominators Using Conjugates Example 5 Video Summary

Here's what students ask on this topic:

What does it mean to rationalize the denominator in a radical expression?

What does it mean to rationalize the denominator in a radical expression?

How do you rationalize the denominator when it contains a single radical term?

How do you rationalize the denominator when it contains a single radical term?

What is the conjugate of a binomial radical expression and why is it used in rationalizing denominators?

What is the conjugate of a binomial radical expression and why is it used in rationalizing denominators?

How do you rationalize the denominator of a fraction with a binomial denominator like 12+3?

How do you rationalize the denominator of a fraction with a binomial denominator like 12+3?

Why can't you just multiply the numerator and denominator by the same binomial radical expression to rationalize a binomial denominator?

Why can't you just multiply the numerator and denominator by the same binomial radical expression to rationalize a binomial denominator?

What is the difference of squares and how does it help in rationalizing denominators?

What is the difference of squares and how does it help in rationalizing denominators?

Simplify the following.
$(2 \sqrt{5}) (4 \sqrt{7})$

Simplify the following.
$(4 \sqrt{x}) (3 \sqrt{x y})$

Simplify the following.
$\sqrt{3} (2 - \sqrt{6})$

Simplify the following.
$\sqrt{5 x} (4 + \sqrt{x})$

Simplify the following.
$(4 \sqrt{3} - 5)^{2}$

Rationalize the denominator.
$- \frac{5}{2 \sqrt{7}}$

Rationalize the denominator.
$\frac{6 + \sqrt{x}}{- \sqrt{x}}$

Rationalize the denominator and simplify the radical expression.
$\frac{\sqrt{7}}{5 - \sqrt{6}}$

Rationalize the denominator and simplify the radical expression.
$\frac{2 - \sqrt{3}}{2 + \sqrt{3}}$

How do you rationalize the denominator of a fraction with a binomial denominator like $\frac{1}{2 + \sqrt{3}}$ ?