Problem 76 - Intermediate Algebra

Question

Rationalize each numerator. See Example 7. $$ \frac{\sqrt{5}+2}{\sqrt{2}} $$

Step-by-Step Solution

Answer

The rationalized fraction is \( \frac{1}{\sqrt{2}(\sqrt{5} - 2)} \).

Step 1: Identify the Numerator

The numerator of the given fraction is \( \sqrt{5} + 2 \). Our goal is to rationalize this numerator.

Step 2: Multiply by Conjugate

To rationalize the numerator, we will multiply both the numerator and the denominator by the conjugate of the numerator, \( \sqrt{5} - 2 \). This will help remove the square root from the numerator: \[ \frac{(\sqrt{5} + 2)(\sqrt{5} - 2)}{\sqrt{2}(\sqrt{5} - 2)} \]

Step 3: Apply the Difference of Squares Formula

The product \( (\sqrt{5} + 2)(\sqrt{5} - 2) \) is a difference of squares, which simplifies to \( \sqrt{5}^2 - 2^2 = 5 - 4 = 1 \). So, the numerator becomes 1.

Step 4: Simplify Denominator

The denominator simplifies to \( \sqrt{2}(\sqrt{5} - 2) \), which is left in its factored form as no further simplification is required.

Step 5: Write Final Answer

After simplification, the expression becomes \( \frac{1}{\sqrt{2}(\sqrt{5} - 2)} \). However, if required, you can distribute, yet it's common to leave it as this, unless specified otherwise in a particular context.

Key Concepts

Conjugate Multiplication

Rationalizing numerators often employs a technique known as conjugate multiplication. The conjugate of a binomial with a square root is the same binomial but with the opposite sign between the terms. In this exercise, the conjugate of the numerator \( \sqrt{5} + 2 \) is \( \sqrt{5} - 2 \). This clever tactic plays a crucial role in eliminating irrational numbers from the numerator. To rationalize the numerator, multiply both the numerator and the denominator by this conjugate. Doing so doesn't change the value of the expression; it simply alters the form to make further simplifications possible. This strategic multiplication helps convert the numerator into a difference of squares, leading to a simple rational number.

Difference of Squares

A vital algebraic identity used in rationalizing numerators is the difference of squares. This formula states that \( (a + b)(a - b) = a^2 - b^2 \). When applying this to the exercise, the product \( (\sqrt{5} + 2)(\sqrt{5} - 2) \) simplifies to \( (\sqrt{5})^2 - 2^2 \). This process simplifies irrational binomials effectively:

• Begin by squaring \( \sqrt{5} \), resulting in 5.
• Next, square 2 to get 4.
• The subtraction of these squares, \( 5 - 4 \), results in 1.

Now, the numerator is rationalized to 1, which is a significantly simplified form. This transformation is possible due to the difference of squares formula.

Simplifying Fractions

Upon rationalizing the numerator, it's essential to address the simplification of the entire fraction. In the initial step, we obtained a new fraction form: \( \frac{1}{\sqrt{2}(\sqrt{5} - 2)} \). Simplifying fractions involves a few key principles:

• Check if there's any need to distribute or further simplify the remaining denominator.
• Consider separating or manipulating any remaining radicals, if needed.

In this case, while the denominator \( \sqrt{2}(\sqrt{5} - 2) \) is in a clean rationalized state, sometimes instructions might further require adjustments or standardization. However, unless specified otherwise, this fraction form is often left unaltered.

Intermediate Algebra

Intermediate algebra covers a variety of methods and concepts including operations with radicals and rational expressions. Rationalizing numerators is a common activity and serves as a practice in manipulation skills involving radicals. This problem adds a layer of complexity by requiring a combination of clever algebraic transformations. It strengthens skills like:

• Understanding and applying conjugates.
• Using identity formulas like the difference of squares.
• Working step-by-step to achieve simplified expressions.

Mastering these skills promotes accurate simplifications and prepares students for more complex algebra challenges. Engaging with exercises like this one enhances your ability to handle and refine expressions in algebraic settings. Intermediate algebra builds a strong foundation for advancing in mathematics.