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

Rationalizing the denominator involves eliminating radicals from the denominator of a fraction to express it in standard form. For a single-term denominator like $\frac{1}{\sqrt{3}}$, multiply numerator and denominator by the radical to create a perfect square, e.g., $\frac{\sqrt{3}}{3}$. For binomial denominators such as $\frac{1}{2+\sqrt{3}}$, multiply by the conjugate (flip the sign) to apply the difference of squares, removing radicals and simplifying the expression. This process ensures the denominator is rational, facilitating easier computation and comparison.

Rationalizing Denominators

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.

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

• A. $-\frac{5}{14}$
• B. $-\frac{5\sqrt7}{2}$
• C. $-\frac{5\sqrt7}{14}$
• D. $-\frac{10\sqrt7}{14}$

Answer: (C)

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

• A. $\frac{6\sqrt{x}}{-x}-1$
• B. $6\sqrt{x}+x$
• C. $\frac{6\sqrt{x}}{x}+1$
• D. $\frac{7\sqrt{x}}{-x}$

Answer: (A)

Rationalizing Denominators Example 1

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 roots and division. For example, consider the expression √18 + 6 divided by 3. To simplify, first separate the terms as √18 / 3 + 6 / 3. The division 6 / 3 simplifies to 2, so the expression becomes √18 / 3 + 2. Next, apply the product rule for square roots, which states that √(a × b) = √a × √b, to rewrite √18 as √9 × √2. Since √9 = 3, the expression becomes 3√2 / 3 + 2. The 3 in the numerator and denominator cancel out, leaving √2 + 2 as the fully simplified form.

In another example, consider the expression 2 × ∛16 - 4 divided by 4. Begin by splitting the quotient into two parts: 2 × (∛16 / 4) - 4 / 4. The term 4 / 4 simplifies to 1, so the expression becomes 2 × (∛16 / 4) - 1. Using the product rule for cube roots, ∛(a × b) = ∛a × ∛b, rewrite ∛16 as ∛8 × ∛2. Since ∛8 = 2 (because 8 is a perfect cube), the expression is now 2 × (2 × ∛2 / 4) - 1. Multiplying 2 by 2 gives 4, which cancels with the denominator 4, simplifying the expression to ∛2 - 1. This is the fully simplified quotient.

These examples highlight the importance of recognizing perfect squares and cubes within radicals and applying root properties to simplify expressions. By breaking down radicals into products of perfect powers and other factors, and carefully performing division, complex quotients can be reduced to simpler forms without radicals in denominators or unnecessary terms.

Rationalizing Denominators Example 2

When simplifying quotients involving square roots, it is important to first look for straightforward simplifications before attempting to rationalize the denominator. For example, consider the expression $$\backslash (\backslash frac\{\backslash sqrt\{81\}\}\{\backslash sqrt\{9\}\}\backslash )$$. Both 81 and 9 are perfect squares, so their square roots simplify directly: $$\backslash (\backslash sqrt\{81\}\; =\; 9\backslash )$$ and $$\backslash (\backslash sqrt\{9\}\; =\; 3\backslash )$$. This reduces the expression to $$\backslash (\backslash frac\{9\}\{3\}\; =\; 3\backslash )$$. Alternatively, using the quotient rule for radicals, which states $$\backslash (\backslash frac\{\backslash sqrt\{a\}\}\{\backslash sqrt\{b\}\}\; =\; \backslash sqrt\{\backslash frac\{a\}\{b\}\}\backslash )$$, you can rewrite the original expression as $$\backslash (\backslash sqrt\{\backslash frac\{81\}\{9\}\}\; =\; \backslash sqrt\{9\}\; =\; 3\backslash )$$, arriving at the same result.

In another example, simplify $$\backslash (\backslash frac\{\backslash sqrt\{49x^2\}\}\{\backslash sqrt\{x^2\}\}\backslash )$$. Recognize that the square root of $\backslash (x^2\backslash )$ is $\backslash (x\backslash )$, so the denominator simplifies to $\backslash (x\backslash )$. The numerator simplifies as $\backslash (\backslash sqrt\{49x^2\}\; =\; \backslash sqrt\{49\}\; \backslash times\; \backslash sqrt\{x^2\}\; =\; 7x\backslash )$. This allows cancellation of $\backslash (x\backslash )$ in numerator and denominator, leaving the simplified result as 7. Using the quotient rule again, rewrite the expression as $$\backslash (\backslash sqrt\{\backslash frac\{49x^2\}\{x^2\}\}\; =\; \backslash sqrt\{49\}\; =\; 7\backslash )$$, confirming the same answer.

These examples highlight the importance of recognizing perfect squares and applying the quotient rule for radicals to simplify expressions efficiently. Avoid unnecessary steps like multiplying by conjugates when simpler simplifications are available. Mastery of these techniques ensures that expressions with radicals are written in lowest terms, which is a fundamental skill in algebra and precalculus.

Rationalizing Denominators Using Conjugates

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.

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

• A. $\frac{\sqrt7}{19}$
• B. $\frac{5\sqrt7+\sqrt{42}}{-11}$
• C. $\frac{5\sqrt7+\sqrt{42}}{21}$
• D. $\frac{5\sqrt7+\sqrt{42}}{19}$

Answer: (D)

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

• A. $7-4\sqrt3$
• B. $\frac{1}{7+4\sqrt3}$
• C. $7+4\sqrt3$
• D. $\frac{1}{7-4\sqrt3}$$\frac{1}{7-4\sqrt3}$

Answer: (A)

Rationalizing Denominators Using Conjugates Example 1

Rationalizing the denominator is a key algebraic technique used to eliminate radicals from the denominator of a fraction, making expressions easier to work with and simplifying further calculations. When a denominator contains a square root or other radical expressions, multiplying by the conjugate is an effective method to remove these radicals.

For example, consider the expression (x + 2) / √(x + 1). To rationalize the denominator, multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of √(x + 1) is √(x - 1), where the sign between the terms is flipped. This multiplication leverages the difference of squares formula, which states that for any two terms a and b, (a + b)(a - b) = a² - b². Applying this, the denominator becomes:

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

However, in the original example, the conjugate used was √(x - 1), so the denominator simplifies to:

\[x - 1\]

Multiplying the numerator by the same conjugate maintains the value of the expression, resulting in:

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

This expression no longer has a radical in the denominator, achieving the goal of rationalization.

In cases where the denominator contains a sum of two radicals, such as 1 / (√x + √y), the conjugate is √x - √y. Multiplying numerator and denominator by this conjugate uses the difference of squares to eliminate the radicals:

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

This process removes the radicals from the denominator, simplifying the expression and making it more manageable for further algebraic operations.

Understanding how to rationalize denominators by multiplying by conjugates and applying the difference of squares formula is essential for simplifying expressions involving radicals. This technique not only clarifies the expression but also prepares it for subsequent mathematical procedures such as addition, subtraction, or integration.

Here’s what students ask on this topic:

What does it mean to rationalize the denominator in a fraction with radicals?

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 keeps the value of the fraction the same but expresses it in a simpler, more standard form.

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

When the denominator has a single radical term, such as $\frac{1}{\sqrt{3}}$, you rationalize by multiplying both numerator and denominator by the same radical. For example, multiply by $\sqrt{3}$ over $\sqrt{3}$. This gives $\frac{\sqrt{3}}{3}$. The denominator becomes a perfect square because $\sqrt{3}\backslash times\sqrt{3}=\sqrt{9}=3$. This removes the radical from the denominator, making the expression easier to work with.

What is the conjugate and how is it used to rationalize denominators with two terms?

The conjugate of a binomial expression is formed by changing the sign between the two terms. For example, the conjugate of $2+\sqrt{3}$ is $2-\sqrt{3}$. When the denominator has two terms, like $\frac{1}{2}$, multiplying numerator and denominator by the conjugate eliminates the radical in the denominator. This works because multiplying conjugates results in a difference of squares: $(a\; +\; b)(a\; -\; b)\; =\; a^2\; -\; b^2$. The radicals cancel out, leaving a rational number in the denominator.

Why can't you just multiply the numerator and denominator by the same binomial when the denominator has two terms?

If you multiply the numerator and denominator by the same binomial, such as $2\; +\sqrt{3}$, you end up with a radical still in the denominator. This is because multiplying a binomial by itself involves FOIL (First, Outer, Inner, Last), and the middle terms containing radicals do not cancel out. For example, $(2\; +\sqrt{3})^2\; =\; 4\; +\; 4\sqrt{3}+\; 3$, which still has a radical term. Therefore, this does not rationalize the denominator. Instead, you must multiply by the conjugate, which flips the sign and removes the radical.

How do you verify that rationalizing the denominator does not change the value of the expression?

To verify that rationalizing the denominator does not change the value, you can calculate the decimal value of the original expression and the rationalized expression using a calculator. For example, $\frac{1}{\sqrt{3}}$ is approximately 0.577. After rationalizing, it becomes $\frac{\sqrt{3}}{3}$, which also evaluates to approximately 0.577. Since both expressions have the same decimal value, they are equivalent, confirming that rationalizing the denominator only changes the form, not the value.