Problem 75 - Algebra for College Students

Question

Rationalize the denominator and simplify. All variables represent positive real numbers. \(\frac{3 \sqrt{y}}{2 \sqrt{x}-3 \sqrt{y}}\)

Step-by-Step Solution

Answer

\(\frac{6 \sqrt{xy} + 9y}{4x - 9y}\) is the expression with a rationalized denominator.

Step 1: Identify the Conjugate

To rationalize the denominator, we will use the conjugate of the denominator. The denominator is \(2 \sqrt{x} - 3 \sqrt{y}\), so its conjugate is \(2 \sqrt{x} + 3 \sqrt{y}\).

Step 2: Multiply by the Conjugate

Multiply the numerator and the denominator of the expression by the conjugate of the denominator. This gives us: \[\frac{3 \sqrt{y}}{2 \sqrt{x} - 3 \sqrt{y}} \times \frac{2 \sqrt{x} + 3 \sqrt{y}}{2 \sqrt{x} + 3 \sqrt{y}} = \frac{3 \sqrt{y} \cdot (2 \sqrt{x} + 3 \sqrt{y})}{(2 \sqrt{x} - 3 \sqrt{y})(2 \sqrt{x} + 3 \sqrt{y})}.\]

Step 3: Simplify the Numerator

Multiply \(3 \sqrt{y}\) by the conjugate: \[3 \sqrt{y} \cdot (2 \sqrt{x} + 3 \sqrt{y}) = 3 \sqrt{y} \cdot 2 \sqrt{x} + 3 \sqrt{y} \cdot 3 \sqrt{y} = 6 \sqrt{xy} + 9y.\]

Step 4: Simplify the Denominator

Use the difference of squares formula to simplify the denominator: \[(2 \sqrt{x} - 3 \sqrt{y})(2 \sqrt{x} + 3 \sqrt{y}) = (2 \sqrt{x})^2 - (3 \sqrt{y})^2 = 4x - 9y.\]

Step 5: Combine Simplified Components

Combine the results from Steps 3 and 4 to form the new fraction: \[\frac{6 \sqrt{xy} + 9y}{4x - 9y}.\] This is the simplified version of the expression with a rationalized denominator.

Key Concepts

Conjugate Method

The conjugate method is a valuable tool for rationalizing denominators that contain square roots or irrational numbers. When faced with an expression like \( \frac{3 \sqrt{y}}{2 \sqrt{x} - 3 \sqrt{y}} \), the goal is to eliminate the irrational denominator by multiplying by its conjugate.

The conjugate of a binomial expression \( a - b \) is \( a + b \). Thus, the conjugate of our denominator \( 2 \sqrt{x} - 3 \sqrt{y} \) is \( 2 \sqrt{x} + 3 \sqrt{y} \).

To apply the conjugate method, multiply both the numerator and the denominator by the conjugate. By doing so, the denominator becomes rational through the difference of squares method, while the value of the expression remains unchanged. Remember, it's like multiplying by one in a clever way that simplifies the problem.

• Identify the conjugate.
• Multiply the numerator and denominator by it.
• Simplify the result.

Simplifying Expressions

Simplifying expressions involves breaking down a complex expression into a simpler, more streamlined form. This process often involves distributing and combining like terms.

In our example, after multiplying by the conjugate, the numerator becomes \( 3 \sqrt{y} \cdot (2 \sqrt{x} + 3 \sqrt{y}) \). By using the distributive property, multiply each term separately:

• First, \( 3 \sqrt{y} \times 2 \sqrt{x} = 6 \sqrt{xy} \).
• Next, \( 3 \sqrt{y} \times 3 \sqrt{y} = 9y \), because \( \sqrt{y} \times \sqrt{y} = y \).

Combine these results to rewrite the numerator: \( 6 \sqrt{xy} + 9y \).

Don't forget, this simplification helps us to work with easier expressions and makes it possible to further manipulate or compare them. A step-by-step approach will ensure no detail is overlooked.

Difference of Squares

The difference of squares formula is a crucial concept to understand when rationalizing expressions. It's particularly helpful in simplifying the denominator after applying the conjugate method.

The formula is \((a - b)(a + b) = a^2 - b^2\). For our previous example's denominator \( (2 \sqrt{x} - 3 \sqrt{y})(2 \sqrt{x} + 3 \sqrt{y}) \):

• Set \( a = 2 \sqrt{x} \) and \( b = 3 \sqrt{y} \).

Applying the difference of squares formula, we have:

• \((2 \sqrt{x})^2 - (3 \sqrt{y})^2\)
• Resulting in: \(4x - 9y\)

By transforming our denominator into a simpler expression, it not only eliminates the square roots but also creates a cleaner rational form.

Understanding this powerful algebraic identity can simplify complex expressions significantly and is widely used across many areas in math. Use it whenever you see a pair of conjugates in multiplication!