Problem 37 - Algebra A Combined Function

Question

Rationalize each denominator. Assume that all variables represent positive real numbers. \(\frac{\sqrt{a}+1}{2 \sqrt{a}-\sqrt{b}}\)

Step-by-Step Solution

Answer

Rationalized form is \(\frac{2a + \sqrt{a}\sqrt{b} + 2\sqrt{a} + \sqrt{b}}{4a - b}\).

Step 1: Identify the Rationalizing Factor

To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of the denominator \(2\sqrt{a} - \sqrt{b}\) is \(2\sqrt{a} + \sqrt{b}\).

Step 2: Multiply by the Conjugate

Multiply the numerator and the denominator by the conjugate: \(\frac{\sqrt{a}+1}{2\sqrt{a}-\sqrt{b}} \times \frac{2\sqrt{a}+\sqrt{b}}{2\sqrt{a}+\sqrt{b}}\).

Step 3: Simplify the Numerator

Expand the numerator: \((\sqrt{a}+1)(2\sqrt{a}+\sqrt{b}) = \sqrt{a} \cdot 2\sqrt{a} + \sqrt{a} \cdot \sqrt{b} + 1 \cdot 2\sqrt{a} + 1 \cdot \sqrt{b}\).

Step 4: Simplify the Denominator Using Algebraic Identity

Apply the difference of squares formula to simplify the denominator: \((2\sqrt{a})^2 - (\sqrt{b})^2 = 4a - b\).

Step 5: Combine and Simplify Terms

Combine and simplify the expanded terms from Step 3:\[\frac{2a + \sqrt{a}\sqrt{b} + 2\sqrt{a} + \sqrt{b}}{4a - b}\].

Step 6: Final Structure

Ensure that the expression\(\frac{2a + \sqrt{a}\sqrt{b} + 2\sqrt{a} + \sqrt{b}}{4a - b}\) is simplified and cannot be further reduced.

Key Concepts

Conjugate Method

In the realm of algebra, rationalizing denominators is a common process, and using a conjugate is key to this method. When you have a denominator that consists of a binomial involving roots, you're often required to multiply both the numerator and denominator by the conjugate of the denominator.
The conjugate of a binomial expression like \( a - b \) is \( a + b \). By multiplying by the conjugate, you eliminate the root from the denominator. This is because when a binomial and its conjugate are multiplied, it results in a difference of squares, which is a square number. This is a neat algebraic trick to clear radicals and simplify expressions.

• Identify the conjugate: For \( 2\sqrt{a} - \sqrt{b} \), its conjugate is \( 2\sqrt{a} + \sqrt{b} \).
• Multiply both numerator and denominator by the conjugate.

By achieving a binomial result in the denominator, any further complexities from square roots should be removed, thus simplifying the expression significantly.

Difference of Squares

The expression \((a - b)(a + b)\) illustrates the powerful concept known as the "difference of squares." This algebraic identity is a crucial part of many algebraic simplifications and is foundational in rationalizing denominators.
When you multiply two conjugates, you apply the difference of squares formula: \((a-b)(a+b) = a^2 - b^2\). This process transforms the denominator into a simpler form without radicals. For instance, if you multiply \( 2\sqrt{a} - \sqrt{b} \) by its conjugate \( 2\sqrt{a} + \sqrt{b} \), you use the difference of squares formula to get: \[ (2\sqrt{a})^2 - (\sqrt{b})^2 = 4a - b \].
The difference of squares method is vital for clearing complex numbers from the denominator, making expressions more digestible and easier to work with further in calculations.

Simplifying Expressions in Algebra

Simplification in algebra involves transforming expressions into a form that is easier to understand or act upon. This is achieved by combining like terms, using mathematical identities, and canceling out elements that can be simplified.
After multiplying by the conjugate and applying the difference of squares, the next step is to carefully expand both the numerator and denominator. For example, when expanding \((\sqrt{a}+1)(2\sqrt{a}+\sqrt{b})\), distribute each term to simplify:

• \( \sqrt{a} \cdot 2\sqrt{a} = 2a \)
• \( \sqrt{a} \cdot \sqrt{b} = \sqrt{a}\sqrt{b} \)
• \( 1 \cdot 2\sqrt{a} = 2\sqrt{a} \)
• \( 1 \cdot \sqrt{b} = \sqrt{b} \)

Afterward, these terms are combined to yield an expression \( 2a + \sqrt{a}\sqrt{b} + 2\sqrt{a} + \sqrt{b} \), which can then be placed over the simplified result of the denominator \(4a - b\).
This method of simplification allows for clearer readability and ease of further computation, which is fundamental to problem-solving in algebra.