Problem 98 - Elementary and Intermediate Algebra

Question

Rationalize each denominator. All variables represent positive real numbers. $$ \frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Conjugate

To rationalize the denominator, we need to multiply both the numerator and the denominator by the conjugate of the denominator. The current denominator is \( \sqrt{x} - \sqrt{y} \), so the conjugate is \( \sqrt{x} + \sqrt{y} \).

Step 2: Multiply by the Conjugate

Multiply the original expression by \( \frac{\sqrt{x} + \sqrt{y}}{\sqrt{x} + \sqrt{y}} \). This gives: \[ \frac{(\sqrt{x} + \sqrt{y})(\sqrt{x} + \sqrt{y})}{(\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})} \]

Step 3: Expand the Numerator

Expand the numerator using the distributive property: \[ (\sqrt{x} + \sqrt{y})(\sqrt{x} + \sqrt{y}) = \sqrt{x}\sqrt{x} + \sqrt{x}\sqrt{y} + \sqrt{y}\sqrt{x} + \sqrt{y}\sqrt{y} \]This simplifies to: \[ x + 2\sqrt{xy} + y \]

Step 4: Expand the Denominator

Use the difference of squares formula for the denominator: \[ (\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y}) = (\sqrt{x})^2 - (\sqrt{y})^2 \]This simplifies to: \[ x - y \]

Step 5: Formulate the Rationalized Expression

Combine the expanded numerator and denominator: \[ \frac{x + 2\sqrt{xy} + y}{x - y} \] This final expression has a rationalized denominator.

Key Concepts

Conjugates in Algebra

In algebra, the concept of conjugates plays a vital role, especially when dealing with irrational numbers in denominators. A conjugate is simply formed by changing the sign between two terms. For example, given the expression \( \sqrt{x} - \sqrt{y} \), the conjugate is \( \sqrt{x} + \sqrt{y} \). The use of conjugates helps in rationalizing denominators, essentially removing any radicals from the denominator.When dealing with expressions involving radicals, multiplying by the conjugate leads to a simpler, often more manageable result. This process involves multiplying both the numerator and the denominator of the expression by this conjugate. By doing so, we are essentially rearranging the expression into a more standard form that's easier to work with. Remember:

• Changing the sign creates the conjugate.
• Multiplying by the conjugate helps remove radicals from the denominator.

Understanding conjugates in algebra is a fundamental step in simplifying complex expressions and preparing them for further algebraic manipulation.

Difference of Squares

The "difference of squares" is a specific algebraic pattern represented by \((a - b)(a + b) = a^2 - b^2\). This formula is incredibly useful when working with conjugates as seen in the given exercise. By applying this pattern to the denominator \( (\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y}) \), you can simplify it into \( x - y \).Let's break it down:

• \((\sqrt{x})^2\) simplifies to \(x\).
• \((\sqrt{y})^2\) simplifies to \(y\).
• Thus, \(x - y\) removes any radicals from the denominator.

This simplification significantly streamlines the expression, making operations like addition, subtraction, or further simplifications far less cumbersome. The difference of squares is one of those quintessential algebraic identities that provides a shortcut to simplify and solve equations or expressions.

Simplifying Radicals

Simplifying radicals is another cornerstone concept in ensuring expressions are as straightforward as possible. Radicals, or expressions containing a square root, often appear complex, but simplifying them can make them manageable.In the numerator \((\sqrt{x} + \sqrt{y})(\sqrt{x} + \sqrt{y})\), we apply the distributive property to expand and simplify:

• \(\sqrt{x}\sqrt{x} = x\)
• \(\sqrt{y}\sqrt{y} = y\)
• The middle terms: \(\sqrt{x}\sqrt{y} + \sqrt{y}\sqrt{x} = 2\sqrt{xy}\)

After expansion, combining these terms gives \(x + 2\sqrt{xy} + y\).Knowing how to simplify these radicals is not only useful for rationalization but also in performing more advanced algebraic operations. Simplification often involves recognizing patterns and applying known identities, like the conjugates and difference of squares. These skills, honed through practice, transform complex expressions into their simplest form, easing further calculation or analysis.