Problem 105 - College Algebra with Modeling and Visualization

Question

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

Step-by-Step Solution

Answer

The rationalized expression is \( \sqrt{10} - 2\sqrt{2} \).

Step 1: Identify the Expression

The expression we need to rationalize is \( \frac{\sqrt{2}}{\sqrt{5}+2} \). Our goal is to eliminate the square root in the denominator.

Step 2: Multiply by the Conjugate

To remove the square root, multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of \(\sqrt{5}+2\) is \(\sqrt{5}-2\). Thus, the expression becomes: \[\frac{\sqrt{2}}{\sqrt{5}+2} \times \frac{\sqrt{5}-2}{\sqrt{5}-2}.\]

Step 3: Apply the Difference of Squares

When you multiply the denominator by its conjugate, use the difference of squares formula: \((a+b)(a-b) = a^2 - b^2\). Therefore, the denominator becomes: \((\sqrt{5})^2 - 2^2 = 5 - 4 = 1\).

Step 4: Simplify the Denominator

Since the denominator equals 1 after applying the difference of squares, our expression now simplifies to: \[\frac{\sqrt{2}(\sqrt{5} - 2)}{1} = \sqrt{2}(\sqrt{5} - 2).\]

Step 5: Distribute in the Numerator

Distribute \(\sqrt{2}\) across \(\sqrt{5} - 2\):\[\sqrt{2} \cdot \sqrt{5} - \sqrt{2} \cdot 2 = \sqrt{10} - 2\sqrt{2}.\]

Step 6: Write the Final Rationalized Expression

Our final expression, with the denominator rationalized, is:\[\sqrt{10} - 2\sqrt{2}.\]

Key Concepts

Difference of Squares

The concept of the difference of squares is a powerful tool in algebra, particularly when simplifying expressions that involve polynomials and radicals. The principle states that \((a+b)(a-b) = a^2 - b^2\). This formula represents a relation between products and differences of squares.

When rationalizing the denominator, especially one containing radicals, this principle helps eliminate the radicals effectively. In our original problem, the denominator \(\sqrt{5}+2\) becomes simple to handle using its conjugate, resulting in \(5 - 4\) after applying the difference of squares, which is equal to 1.

By reducing the denominator to a rational number, the fraction's "irrational" aspect is moved to the numerator. This action makes expressions easier to understand, compare, and use in further calculations, as is common in mathematics.

Conjugate

Conjugates play a significant role in simplifying radical expressions and rationalizing denominators. They are pairs like \(a + b\) and \(a - b\), which multiply to yield a difference of squares. When a radical is present in a denominator, multiplying by its conjugate ensures that the radical component disappears.

Here's a simple rundown of how to use conjugates:

• Identify the expression with the radical in the denominator.
• Write down the conjugate of the denominator – if the denominator is \(a + b\), its conjugate is \(a - b\).
• Multiply both the numerator and the denominator of the fraction by this conjugate. This step modifies the expression but doesn't change its overall value because you're multiplying by 1 \(\frac{a-b}{a-b}\).

In our exercise, using \(\sqrt{5} - 2\) as a conjugate cleared the radical from the denominator and simplified the expression greatly, making further operations straightforward.

Simplifying Radical Expressions

The essence of simplifying radical expressions is to make them easier to work with, which often involves removing radicals from the denominator of a fraction. Simplifying effectively often leads to cleaner, more manageable expressions, as shown in our example.

Here's a simple guide to simplifying these expressions:

• Identify whether there's a radical in the denominator.
• If there is, look for opportunities to use conjugates or rewrite terms in a simpler form.
• Apply operations such as distribution to consolidate terms and reduce the expression to its simplest form.

By following these steps, as we did in the exercise with \(\frac{\sqrt{2}(\sqrt{5} - 2)}{1}\), we transform complex radical expressions into simplified forms like \(\sqrt{10} - 2\sqrt{2}\). This method not only makes calculations easier but also helps in recognizing patterns and results in algebraic operations.