Problem 97 - 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 expression is \( \frac{x - 2\sqrt{xy} + y}{x - y} \).

Step 1: Identify the problem

The problem requires us to rationalize the denominator of the expression \( \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}} \). Rationalizing the denominator means eliminating any square roots from the denominator.

Step 2: Multiply by the Conjugate

To rationalize, multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of \( \sqrt{x}+\sqrt{y} \) is \( \sqrt{x}-\sqrt{y} \). Hence, multiply the expression by \( \frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}} \).

Step 3: Apply the Distributive Property

Apply the distributive property (i.e., multiply throughout) in both the numerator and the denominator:In the numerator:\( (\sqrt{x} - \sqrt{y})(\sqrt{x} - \sqrt{y}) = (\sqrt{x})^2 - 2\sqrt{x}\sqrt{y} + (\sqrt{y})^2 = x - 2\sqrt{xy} + y \).In the denominator:\( (\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) = (\sqrt{x})^2 - (\sqrt{y})^2 = x - y \).

Step 4: Write the Rationalized Expression

Combine the results from Step 3 into a single expression: The numerator is \( x - 2\sqrt{xy} + y \) and the denominator is \( x - y \). Therefore, the rationalized expression is:\[\frac{x - 2\sqrt{xy} + y}{x - y}\].

Key Concepts

Conjugate in Algebra

In algebra, the conjugate of an expression is highly useful, especially when dealing with square roots. By definition, the conjugate of a binomial expression is simply the same expression with the sign between the two terms switched. For example, the conjugate of \( a + b \) is \( a - b \) and vice versa. The beauty of using the conjugate lies in the property that when you multiply a binomial expression like \( a + b \) with its conjugate \( a - b \), the product eliminates the middle term, rendering a difference of squares.
This is used because the square root terms disappear when multiplying with conjugates. So, terms become straightforward like \( a^2 - b^2 \), without any cross products. In our original problem, multiplying \( \sqrt{x} + \sqrt{y} \) by its conjugate \( \sqrt{x} - \sqrt{y} \) resulted in \( x - y \). This helps to remove the square roots from the denominator, a crucial step in rationalization.

Square Roots

Square roots are fundamental in algebra and indicate a value that, when multiplied by itself, yields the original number. The operation is denoted by the square root symbol \( \sqrt{} \). For instance, the square root of \( x \), written as \( \sqrt{x} \), is a number which, when squared, equals \( x \).
Square roots often appear in expressions and require simplification. They can make calculations unwieldy, particularly in denominators of fractions, leading us to rationalize them.
Understanding how to manage these expressions with square roots is crucial for expanding binomials and simplifying expressions.

• Square roots simplify algebraic expressions.
• They must often be rationalized when in denominators.
• Using conjugates assists in square root elimination.

Distributive Property in Algebra

The distributive property is a fundamental algebraic rule that allows us to multiply a single term by each term within a parenthesis. It is expressed as \( a(b + c) = ab + ac \). This property is particularly useful when dealing with binomials and polynomials.
In the process of rationalizing a denominator, applying the distributive property allows us to multiply terms systematically. In our problem:

• The numerator \((\sqrt{x} - \sqrt{y})\) is multiplied by itself using the distributive property to create \((\sqrt{x})^2 - 2\sqrt{x}\sqrt{y} + (\sqrt{y})^2\).
• Similarly, this property helps simplify the denominator by removing square root terms: \( (\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) = x - y \).

Understanding how to apply this property ensures accurate expansion and simplification of expressions containing multiple terms.