Problem 29 - Algebra 2 and Trigonometry

Question

Rationalize the denominator and write each fraction in simplest form. All variables represent positive numbers. \(\frac{\sqrt{10}-1}{\sqrt{10}+1}\)

Step-by-Step Solution

Answer

The fraction simplified and rationalized is \( \frac{11 - 2\sqrt{10}}{9} \).

Step 1: Identify the Conjugate

The given expression is \( \frac{\sqrt{10}-1}{\sqrt{10}+1} \). To rationalize the denominator, identify the conjugate of the denominator \( \sqrt{10}+1 \). This is \( \sqrt{10}-1 \).

Step 2: Multiply by the Conjugate

Multiply both the numerator and the denominator by the conjugate \( \sqrt{10}-1 \). This gives: \( \frac{(\sqrt{10}-1)(\sqrt{10}-1)}{(\sqrt{10}+1)(\sqrt{10}-1)} \).

Step 3: Simplify the Denominator

Use the difference of squares formula \((a+b)(a-b) = a^2 - b^2\) to simplify the denominator: \( (\sqrt{10})^2 - 1^2 = 10 - 1 = 9 \).

Step 4: Expand the Numerator

Expand the numerator \( (\sqrt{10}-1)(\sqrt{10}-1) \) using the distributive property: \( (\sqrt{10})^2 - 2\sqrt{10} + 1^2 = 10 - 2\sqrt{10} + 1 \).

Step 5: Simplify the Numerator

Combine like terms in the numerator: \( 10 - 2\sqrt{10} + 1 = 11 - 2\sqrt{10} \).

Step 6: Write in Simplest Form

Combine the results of Steps 3 and 5. The fraction is \( \frac{11 - 2\sqrt{10}}{9} \), which is already in its simplest form.

Key Concepts

Conjugates

When rationalizing a denominator, the use of conjugates is key. A conjugate in this context refers to changing the sign between two terms in a binomial. For a denominator of the form \( a + b \), its conjugate is \( a - b \). This transformation is fundamental because it helps eliminate square roots from the denominator. When the numerator and denominator are both multiplied by the conjugate of the denominator, we transform the problem into one involving the difference of squares.

Difference of Squares

The difference of squares is a simple yet powerful algebraic identity: \((a+b)(a-b) = a^2 - b^2\). In the context of simplifying terms involving square roots, this property is invaluable. By multiplying a binomial by its conjugate, you effectively apply this identity.

• It converts expressions containing roots to a simpler form.
• It turns denominators with roots into whole numbers.

This simplification comes in handy when you want to manage expressions more effectively, particularly when eliminating roots from denominators in fractions.

Distributive Property

The distributive property allows us to expand expressions like \((a-b)^2\). It states \( a(b + c) = ab + ac \). Within our example, it was the tool to expand the numerator \((\sqrt{10}-1)^2\). By distributing each term:\[(\sqrt{10}) \cdot (\sqrt{10}) - 2 \cdot \sqrt{10} \cdot 1 + 1 \cdot 1 = 10 - 2\sqrt{10} + 1\].

• This principle is essential for rearranging and simplifying expressions.
• It helps combine and manage terms seamlessly.

This property ensures that all components are considered and allows for efficient simplification.

Square Roots

Square roots can often complicate expressions, especially in denominators. To simplify, especially when rationalizing, understanding the nature of square roots is important. Square roots essentially "undo" squaring, so if you have \(\sqrt{x}\), squaring it gives you back the original number, \(x\).

• This characteristic is exploited when dealing with conjugates.
• The goal is often to remove roots from denominators to make expressions clearer.

Understanding this concept is crucial for manipulating algebraic expressions and achieving a simplified form.