Problem 383 - Elementary Algebra

Question

In the following exercises, simplify by rationalizing the denominator. $$ \frac{\sqrt{x}+\sqrt{8}}{\sqrt{x}-\sqrt{8}} $$

Step-by-Step Solution

Answer

\[ \frac{x + 2\sqrt{8x} + 8}{x - 8} \]

Step 1: Identify the Conjugate

To rationalize the denominator, identify the conjugate of the denominator. The conjugate of \( \sqrt{x} - \sqrt{8} \) is \( \sqrt{x} + \sqrt{8} \).

Step 2: Multiply Numerator and Denominator by the Conjugate

Multiply both the numerator and the denominator by the conjugate \( \sqrt{x} + \sqrt{8} \). The expression becomes \[ \frac{(\sqrt{x} + \sqrt{8})(\sqrt{x} + \sqrt{8})}{(\sqrt{x} - \sqrt{8})(\sqrt{x} + \sqrt{8})} \]

Step 3: Simplify the Numerator

Simplify the numerator using the distributive property: \[ (\sqrt{x} + \sqrt{8})(\sqrt{x} + \sqrt{8}) = (\sqrt{x})^2 + 2\sqrt{x}\sqrt{8} + (\sqrt{8})^2 \] which simplifies to \[ x + 2 \sqrt{8x} + 8 \]

Step 4: Simplify the Denominator

Use the difference of squares formula for the denominator: \[ (\sqrt{x} - \sqrt{8})(\sqrt{x} + \sqrt{8}) = (\sqrt{x})^2 - (\sqrt{8})^2 = x - 8 \]

Step 5: Write the Final Simplified Form

Combine the simplified numerator and denominator to write the final simplified form: \[ \frac{x + 2\sqrt{8x} + 8}{x - 8} \]

Key Concepts

Conjugate

A conjugate is a mathematical term to describe a binomial formed by changing the sign between two terms. When rationalizing the denominator of a fraction that involves a square root, you use the conjugate to simplify the expression. For example, the conjugate of \( \sqrt{x} - \sqrt{8} \) is \( \sqrt{x} + \sqrt{8} \).

The goal of using a conjugate is to eliminate the square root in the denominator. This simplifies the expression and makes it easier to work with. When both the numerator and denominator are multiplied by the conjugate of the denominator, the denominator becomes a rational number. This process involves applying the difference of squares as you will see next.

Distributive Property

The distributive property is fundamental in algebra and is used to multiply one term by each term inside a set of parentheses. This property states that \( a(b + c) = ab + ac \). In the problem given, we use the distributive property to simplify both the numerator and denominator after multiplying by the conjugate.

Consider the numerator after multiplying by the conjugate: \( ( \sqrt{x} + \sqrt{8})( \sqrt{x} + \sqrt{8}) \). Applying the distributive property, it becomes:

• \( ( \sqrt{x})( \sqrt{x}) \)
• \( ( \sqrt{x})( \sqrt{8}) \)
• \( ( \sqrt{8})( \sqrt{x}) \)
• \( ( \sqrt{8})( \sqrt{8}) \)

Combining these results, you get \( x + 2\sqrt{8x} + 8 \).

The distributive property helps in breaking down and rearranging the calculations to make simplifications manageable and clear.

Difference of Squares

The difference of squares is a special algebraic property used to simplify expressions when the conjugate is involved. It states that \( (a - b)(a + b) = a^2 - b^2 \). This property allows us to remove the square roots in the denominator.

In the example given, the denominator \( ( \sqrt{x} - \sqrt{8})( \sqrt{x} + \sqrt{8}) \) simplifies to:

• \( ( \sqrt{x})^2 \)
• \( ( \sqrt{8})^2 \)

Using the difference of squares property, this becomes \( x - 8 \).

Applying the difference of squares helps eliminate square roots, resulting in a rational number in the denominator. This is why the final step is straightforward: the numerator remains as \( x + 2\sqrt{8x} + 8 \), while the denominator simplifies to \( x - 8 \). Understanding and applying the difference of squares is crucial for rationalizing fractions involving square roots.