Problem 69 - Elementary and Intermediate Algebra

Question

Rationalize each denominator. All variables represent positive real numbers. $$ \frac{\sqrt{48 x^{2}}}{\sqrt{8 x^{2} y}} $$

Step-by-Step Solution

Answer

The rationalized form is \( \frac{\sqrt{6y}}{y} \).

Step 1: Simplify the Numerator

Begin by simplifying the numerator \( \sqrt{48x^2} \). Break down \( 48 \) into its prime factors: \( 48 = 16 \times 3 = 4^2 \times 3 \). Thus, \( \sqrt{48x^2} = \sqrt{4^2 \times 3 \times x^2} = 4x\sqrt{3} \).

Step 2: Simplify the Denominator

Next, simplify the denominator \( \sqrt{8x^2y} \). Break \( 8 \) into its prime factors: \( 8 = 4 \times 2 = 2^2 \times 2 \). Thus, \( \sqrt{8x^2y} = \sqrt{4 \times 2 \times x^2 \times y} = 2x\sqrt{2y} \).

Step 3: Simplify the Fraction

After simplifying both the numerator and the denominator, the expression becomes:\[ \frac{4x\sqrt{3}}{2x\sqrt{2y}} \]Cancel out the \( x \) terms and simplify the fraction \( \frac{4}{2} \) which reduces to \( 2 \), leaving:\[ \frac{2\sqrt{3}}{\sqrt{2y}} \].

Step 4: Rationalize the Denominator

To remove the radical from the denominator, multiply both the numerator and the denominator by \( \sqrt{2y} \):\[ \frac{2\sqrt{3} \times \sqrt{2y}}{\sqrt{2y} \times \sqrt{2y}} = \frac{2\sqrt{6y}}{2y} \].br>The denominator becomes \( 2y \) and cancel the 2, giving:\[ \frac{\sqrt{6y}}{y} \].

Key Concepts

Simplifying Square Roots

When dealing with square roots, one of the key steps to simplifying them is to first break down the number under the square root into its prime factors. This makes it easier to identify which parts of the number can be removed from underneath the square root.

For example:

• If you have \( \sqrt{48} \), you can break it down into its prime factors: \( 48 = 4^2 \times 3 \).
• This means \( \sqrt{48} \) becomes \( \sqrt{4^2 \times 3} \).
• Since the square root of \( 4^2 \) is \( 4 \), you can take \( 4 \) out of the square root, leaving you with \( 4\sqrt{3} \).

This method helps simplify the expression by extracting the largest perfect square factor. Instead of dealing with a large number under the radical, you work with a smaller, simpler number.

Prime Factorization

Prime factorization involves breaking down a composite number into its prime components, which are numbers that can't be divided any further except by 1 or themselves. This concept is crucial when dealing with square roots and simplifying expressions.

In the provided exercise, factoring out the prime components of \( 48 \) was a vital step. Here’s a quick refresher on how this works:

• Divide the number by the smallest prime number (2). Continue until it no longer divides evenly.
• Proceed with the next smallest prime (3, then 5, etc.).
• For \( 48 \), start with \( 2 \):\( 48 \div 2 = 24 \), then \( 24 \div 2 = 12 \), then \( 12 \div 2 = 6 \), and \( 6 \div 2 = 3 \). Once you have \( 3 \), which is also a prime, you stop.
• The result is \( 2^4 \times 3 \).

Prime factorization is a key step in simplifying algebraic expressions, especially those involving square roots, because it helps in pulling out perfect squares, making the simplification straightforward.

Algebraic Fractions

An algebraic fraction is simply a division of two algebraic expressions. Simplifying algebraic fractions involves reducing the fraction to its simplest form. For the best results, ensure that both the numerator and the denominator are fully simplified before proceeding.

In the example given, both the numerator \( \sqrt{48x^2} \) and the denominator \( \sqrt{8x^2y} \) were initially complicated, but through factorization and simplification, they were reduced to simpler forms. Here’s how you manage algebraic fractions:

• Simplify both the numerator and the denominator by removing any perfect squares and common factors.
• Cancel out any common factors between the numerator and denominator.
• For instance, if \( x \) appears in both the numerator and the denominator, they can cancel each other out.

Remember, the goal is to simplify while ensuring both parts of the fraction remain in a form that is easy to work with and understand. Rationalizing the denominator, as done in the exercise by multiplying through by \( \sqrt{2y} \), is often a necessary step to ensure the final result is in its simplest, most conventional form.