Problem 94 - Algebra and Trigonometry

Question

Rationalize the denominator. $$ \frac{2(x-y)}{\sqrt{x}-\sqrt{y}} $$

Step-by-Step Solution

Answer

The rationalized form is \( 2\sqrt{x} + 2\sqrt{y} \).

Step 1: Identify the Problem

We need to rationalize the denominator of the expression \( \frac{2(x-y)}{\sqrt{x}-\sqrt{y}} \). Rationalizing the denominator means getting rid of the square roots in the denominator.

Step 2: Choose the Conjugate

The denominator is \( \sqrt{x} - \sqrt{y} \). To eliminate the square roots, multiply the numerator and both terms of the denominator by the conjugate of the denominator: \( \sqrt{x} + \sqrt{y} \).

Step 3: Multiply the Numerator and Denominator

The expression becomes: \[ \frac{2(x-y)(\sqrt{x} + \sqrt{y})}{(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})}. \] The numerator needs to be expanded, and the denominator uses the difference of squares formula.

Step 4: Apply the Difference of Squares

The denominator \( (\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y}) \) simplifies using the difference of squares to \( x - y \).

Step 5: Simplify the Numerator

Expand the numerator: \[ 2(x-y)(\sqrt{x} + \sqrt{y}) = 2(x-y)\sqrt{x} + 2(x-y)\sqrt{y}. \]

Step 6: Cancel and Simplify

Simplify the expression:\[ \frac{2(x-y)\sqrt{x} + 2(x-y)\sqrt{y}}{x-y}. \]The factor \(x-y\) cancels out from the numerator and denominator, assuming \(x eq y\), resulting in:\[ 2\sqrt{x} + 2\sqrt{y}. \]

Step 7: Final Simplified Expression

The expression is now rationalized, and the final answer is \[ 2\sqrt{x} + 2\sqrt{y}. \]

Key Concepts

Square Roots

Square roots are a foundational concept in mathematics, used to find a number that, when multiplied by itself, gives the original number. For example, the square root of 4 is 2 because 2 times 2 equals 4. When dealing with algebraic expressions involving square roots, it’s common to find them in denominators, like in our original expression \( \frac{2(x-y)}{\sqrt{x}-\sqrt{y}} \).
We often aim to "rationalize" the denominator, meaning we want to remove any square roots from it, as this makes the expression easier to interpret and work with mathematically.
This involves manipulating the expression using specific techniques, such as the conjugate method, which we'll explore further.

Conjugate Method

The conjugate method is a powerful tool for rationalizing denominators containing square roots. It involves multiplying both the numerator and denominator by the conjugate of the denominator. The conjugate of an expression \( a-b \) is \( a+b \), meaning you flip the sign between these terms.
In our exercise, we have \( \sqrt{x} - \sqrt{y} \). To apply the conjugate, we multiply by \( \sqrt{x} + \sqrt{y} \).

• This step alters the expression while keeping its value unchanged, as it’s equivalent to multiplying by 1 (the numerator and denominator are the same).
• The rationalization process then becomes \( \frac{2(x-y)(\sqrt{x} + \sqrt{y})}{(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})} \).

Applying this can transform complex expressions into simpler forms, preparing them for the next steps in solving or simplifying.

Difference of Squares Formula

The difference of squares formula is an essential algebraic identity expressed as \( a^2 - b^2 = (a-b)(a+b) \). This formula provides a quick way to factor or multiply expressions, particularly in problems involving conjugates.
In this exercise, the denominator \( (\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y}) \) is effectively a difference of squares situation. By applying \( a=\sqrt{x} \) and \( b=\sqrt{y} \), we simplify it to \( x-y \).

• Using this formula eliminates the square roots in the denominator, giving us a much simpler expression to work with.
• This simplification is vital as it helps move the problem towards a fully rationalized and simplified form.

The difference of squares formula, combined with the conjugate method, is a crucial step in dealing with and understanding rationalizing problems.

Simplification in Algebra

Simplification in algebra makes expressions easier to read and solve. After applying the conjugate method and the difference of squares, the expression \( \frac{2(x-y)(\sqrt{x} + \sqrt{y})}{x-y} \) is far simpler.
You notice that \( (x-y) \) appears in both the numerator and the denominator, allowing for cancellation given \( x eq y \), which leads to:
\[ 2\sqrt{x} + 2\sqrt{y} \]

• Cancelling is often the final step in simplification - it reduces fractions to their simplest forms and clears redundant factors.
• Additionally, thoroughly simplifying ensures the expression is precise and avoids complex calculations in future steps.

Proper simplification supports better mathematical reasoning, providing clarity and demonstrating deeper insights into the relationships between algebraic components.