Problem 16 - Algebra 2 and Trigonometry

Question

Rationalize the denominator and write each fraction in simplest form. All variables represent positive numbers. \(\frac{1}{5-\sqrt{2}}\)

Step-by-Step Solution

Answer

The fraction is \( \frac{5 + \sqrt{2}}{23} \) in simplest form.

Step 1: Identify the Conjugate

To rationalize the denominator of the expression \( \frac{1}{5 - \sqrt{2}} \), we need to multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of \( 5 - \sqrt{2} \) is \( 5 + \sqrt{2} \).

Step 2: Multiply by the Conjugate

Multiply the numerator and the denominator of the fraction by \( 5 + \sqrt{2} \) to eliminate the square root from the denominator. This gives:\[\frac{1}{5 - \sqrt{2}} \times \frac{5 + \sqrt{2}}{5 + \sqrt{2}} = \frac{5 + \sqrt{2}}{(5 - \sqrt{2})(5 + \sqrt{2})}\]

Step 3: Simplify the Denominator

Apply the difference of squares formula to the expression \((5 - \sqrt{2})(5 + \sqrt{2})\). The formula is \((a-b)(a+b) = a^2 - b^2\). Here, \(a = 5\) and \(b = \sqrt{2}\), so:\[(5 - \sqrt{2})(5 + \sqrt{2}) = 5^2 - (\sqrt{2})^2 = 25 - 2 = 23\]

Step 4: Write the Final Expression

Now that we have simplified the denominator, we can write the fraction as:\[\frac{5 + \sqrt{2}}{23}\]This is the fraction with a rationalized denominator and is also in its simplest form.

Key Concepts

Understanding the Conjugate

A conjugate in mathematics refers to changing the sign between two terms within a binomial. For example, if you have a binomial expression like \( 5 - \sqrt{2} \), its conjugate would be \( 5 + \sqrt{2} \). This idea of conjugates is particularly useful when rationalizing denominators that contain square roots. Rationalizing the denominator means removing any irrational numbers (like square roots) from the denominator. This is important because it simplifies calculations and comparisons in mathematics. By multiplying by the conjugate, we take advantage of the algebraic identity that helps eliminate the square root.

• Find the conjugate by flipping the sign in the middle of the binomial.
• Multiply both the numerator and denominator by this conjugate. This does not change the value of the fraction, only its form.

Understanding and identifying the conjugate is a crucial step in solving problems that involve rationalizing denominators.

Difference of Squares Explained

The difference of squares is a handy algebraic formula: \((a-b)(a+b) = a^2 - b^2\). It is an essential concept for rationalizing denominators.When you multiply conjugates (\((5-\sqrt{2})(5+\sqrt{2})\)), you're applying the difference of squares. Here's how it works: - Set \(a = 5\) and \(b = \sqrt{2}\).- Using the formula, \(a^2 - b^2 = 5^2 - (\sqrt{2})^2\).- Calculate \(5^2 = 25\) and \((\sqrt{2})^2 = 2\).- Subtract these values to simplify: \(25 - 2 = 23\). By applying the difference of squares, you successfully transform a complex expression into a more manageable one, making it far easier to work with.

Simplifying Fractions

Simplifying fractions is often one of the final steps in solving algebraic expressions. It involves reducing the fraction to its most basic form so that it is easier to understand and work with.Once you rationalize the denominator and apply the difference of squares, as in the expression \(\frac{5+\sqrt{2}}{23}\), the fraction is already in its simplest form. Here's what you should check:

• Ensure the numerator and denominator have no common factors other than 1.
• The denominator is now a rational number, which is key to rationalizing.

In this case, the fraction \(\frac{5+\sqrt{2}}{23}\) is as simplified as possible, meaning it is ready for use in further calculations or as a final answer. Mastering this simplification process helps in gaining a deeper understanding of fractions and algebraic expressions.