Problem 4 - College Algebra with Modeling and Visualization

Question

Find the square roots of the number. Approximate your answers to the nearest hundredth whenever appropriate. $$\frac{64}{81}$$

Step-by-Step Solution

Answer

The square roots of \(\frac{64}{81}\) are \(\frac{8}{9}\) and \(-\frac{8}{9}\), approximated as 0.89 and -0.89.

Step 1: Simplify the Fraction

First, identify the fraction you need to solve: \(\frac{64}{81}\). Both the numerator and the denominator are perfect squares. \(64\) is \(8^2\) and \(81\) is \(9^2\). Thus, we can simplify the fraction to \(\left(\frac{8}{9}\right)^2\).

Step 2: Apply the Square Root to the Entire Fraction

To find the square root of the fraction, apply the square root to both the numerator and the denominator separately. So, you have \(\sqrt{\frac{64}{81}} = \frac{\sqrt{64}}{\sqrt{81}}\).

Step 3: Calculate the Square Roots of the Numerator and Denominator

The square root of the numerator \(\sqrt{64}\) is \(8\) since \(8^2 = 64\). The square root of the denominator \(\sqrt{81}\) is \(9\) since \(9^2 = 81\).

Step 4: Express the Square Root of the Fraction

Combine the results to express the square root, \(\sqrt{\frac{64}{81}} = \frac{8}{9}\).

Step 5: Approximate the Answer

Convert \(\frac{8}{9}\) to a decimal by dividing: \(8 \div 9 = 0.8888\ldots\). Rounding this to the nearest hundredth gives you \(0.89\).

Key Concepts

Simplifying Fractions

When working with fractions, simplifying them can make complex mathematical tasks much easier. Simplifying a fraction means expressing it in its simplest form. This involves reducing the numerator and denominator to their lowest terms. In our exercise, the fraction \(\frac{64}{81}\) is comprised of perfect squares. Here, 64 can be expressed as \(8^2\) and 81 as \(9^2\). Since both numbers are perfect squares, the fraction can be simplified to \(\left(\frac{8}{9}\right)^2\). This new form provides a much cleaner starting point for further calculations, especially when finding square roots.

Perfect Squares

Understanding perfect squares is crucial when working with square roots and simplifying fractions. A perfect square is a number that can be expressed as the product of an integer with itself. For example, 64 is a perfect square because it is \(8 \times 8\), or \(8^2\), and similarly, 81 is \(9 \times 9\), or \(9^2\).

• Finding perfect squares helps in determining the square root easily since the square root of a perfect square is simply the integer that was squared.
• In our problem, identifying that both numerators and denominators are perfect squares (\(64 = 8^2\) and \(81 = 9^2\)) simplifies the process of calculating the square root by breaking it down into simpler terms.

This recognition transforms the original task into an easier problem by applying the square root to these simpler numbers, resulting in \(\frac{8}{9}\).

Decimal Approximation

Once the square root of a fraction is found, it can be practical to express the result as a decimal, especially when exact decimal representation is required for further calculations or comparisons. To approximate the fraction \(\frac{8}{9}\) as a decimal, divide 8 by 9. This gives an endless decimal of 0.8888..., which, when rounded to the nearest hundredth, becomes 0.89.Here are a few guiding principles for decimal approximation:

• Perform division to convert the simplified fraction to a decimal.
• Identify which decimal place you need to approximate to, such as the nearest hundredth, tenth, or others.
• Round the long decimal appropriately by looking at the digit after the required decimal place.

Decimal approximations enable easier integration into other numerical contexts and facilitate smoother communication of results.