Problem 327 - Elementary Algebra

Question

In the following exercises, simplify. $$ \frac{\sqrt{128}}{\sqrt{72}} $$

Step-by-Step Solution

Answer

The simplified form is \(\frac{4}{3}\).

Step 1: Simplify the Numerator

First, simplify the square root in the numerator. Write 128 as a product of its prime factors: \(128 = 2^7\). So, \(\frac{\text{sqrt}{128}}{\text{sqrt}{72}} = \frac{\text{sqrt}{(2^7)}}{\text{sqrt}{72}}\). Simplify \( \sqrt{2^7} = \sqrt{(2^6) \times 2} = \sqrt{(2^6)} \times \sqrt{2} = 2^3 \times \sqrt{2} = 8\sqrt{2} \).

Step 2: Simplify the Denominator

Next, simplify the square root in the denominator. Write 72 as a product of its prime factors: \(72 = 2^3 \times 3^2\). So, \( \sqrt{72} = \sqrt{(2^3 \times 3^2)} = \sqrt{(2^2 \times 2 \times 3^2)} = \sqrt{(2^2)} \times \sqrt{2} \times \sqrt{(3^2)} = 2 \times 3 \sqrt{2} = 6\sqrt{2}\).

Step 3: Simplify the Fraction

Divide the simplified numerator by the simplified denominator: \(\frac{8\sqrt{2}}{6\sqrt{2}}\). The \(\sqrt{2}\) terms cancel out, leaving \(\frac{8}{6}\), which can be simplified to \(\frac{4}{3}\).

Key Concepts

Prime Factorization

Prime factorization involves breaking down a number into its basic building blocks, which are prime numbers. For example, the number 128 can be broken down as follows: \(128 = 2^7\).\Prime numbers are numbers greater than 1 that only have two divisors: 1 and themselves. The prime numbers used for factorization in this exercise are 2 and 3.

• 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2^7
• 72 = 2^3 x 3^2

This breakdown helps us simplify the expressions under the square root, making the overall problem easier to solve. Understanding prime factorization is crucial because it enables us to simplify complex mathematical expressions by reducing them to their simplest form.

Square Root Simplification

Simplifying square roots entails breaking down the expression into smaller parts that are easier to manage. To simplify square roots using prime factorization, we identify and simplify pairs of prime factors.Consider the numerator, \( \text{sqrt}{128} \):

• 128 = 2^7
• \(2^7 = 2^6 \times 2 = (2^3)^2 \times 2 = 8^2 \times 2\)
• Thus, \( \text{sqrt}{2^7} = \text{sqrt}{(8^2 \times 2)} = 8\text{sqrt}{2} \)

Next, consider the denominator, \( \text{sqrt}{72} \):

• 72 = 2^3 \times 3^2
• \(72 = 2^2 \times 2 \times 3^2 \)
• Thus, \( \text{sqrt}{72} = 6\text{sqrt}{2} \)

Using these simplifications, we can clearly see how to handle both the numerator and the denominator in our fraction, thus making it easier to solve the problem.

Rationalization

Rationalization is a technique used to remove square roots from the denominator of a fraction. In this exercise, we do not need to rationalize because \( \text{sqrt}{2} \) terms already cancel out from both numerator and denominator. However, understanding the process is still valuable.Suppose we have a different fraction such as \( \frac{1}{\text{sqrt}{3}} \). To rationalize the denominator, we multiply both the numerator and the denominator by \( \text{sqrt}{3} \).

• \( \frac{1}{\text{sqrt}{3}} \times \frac{\text{sqrt}{3}}{\text{sqrt}{3}} = \frac{\text{sqrt}{3}}{3} \)

By doing so, we convert the denominator into a rational number without square root.Overall, rationalization helps in converting a fraction with an irrational denominator into an equivalent fraction with a rational denominator, making it easier to handle in further mathematical computations.