Problem 32 - Intermediate Algebra

Question

Write each of the following in terms of \(i\) and simplify. $$ \sqrt{-\frac{64}{36}} $$

Step-by-Step Solution

Answer

\(\frac{4}{3}i\)

Step 1: Simplify the Fraction Inside the Square Root

First, simplify the fraction inside the square root: \(-\frac{64}{36}\). Divide both the numerator and the denominator by 4: \(-\frac{64 \div 4}{36 \div 4} = -\frac{16}{9}\).

Step 2: Identify the Negative and Break it Apart

Recognize that the expression \(\sqrt{-\frac{16}{9}}\) contains a negative symbol. This can be rewritten as \(\sqrt{-1 \times \frac{16}{9}}\). We know that \(\sqrt{-1}\) is equal to \(i\).

Step 3: Apply the Property of Square Roots

Use the property \(\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\) to break it into \(\sqrt{-1} \times \sqrt{\frac{16}{9}}\). Simplify further: \(i \times \sqrt{\frac{16}{9}}\).

Step 4: Simplify the Square Root of the Fraction

Find the square root of \(\frac{16}{9}\). This can be rewritten as \(\frac{\sqrt{16}}{\sqrt{9}}\). Calculating these gives: \(\frac{4}{3}\) since \(\sqrt{16} = 4\) and \(\sqrt{9} = 3\).

Step 5: Combine the Results

Now multiply \(i\) by \(\frac{4}{3}\): \(i \times \frac{4}{3} = \frac{4}{3}i\). This is the simplified form in terms of \(i\).

Key Concepts

Simplifying Fractions

Simplifying fractions is an important concept, especially when dealing with complex numbers. To simplify a fraction, you simply divide both the numerator and the denominator by their greatest common divisor (GCD). For example, in the problem above, we started with the fraction \(-\frac{64}{36}\).

• Finding the GCD of 64 and 36, which is 4.
• Dividing both the numerator and the denominator by 4 results in \(-\frac{16}{9}\).

This process reduces the fraction to its simplest form, making further calculations easier. Remember, even if the numbers inside the square root are negative or fractions, the process of simplification remains the same.

Imaginary Unit

The imaginary unit \(i\) is a fundamental component when working with complex numbers. Imaginary numbers arise when taking the square root of a negative number since in real numbers, this is not possible.

• The imaginary unit is defined as \(i = \sqrt{-1}\).
• It is used to express numbers like \(\sqrt{-x}\) as \(i\sqrt{x}\).

In our example, the problem requires finding \(\sqrt{-\frac{16}{9}}\). Recognize the "\(-1\)" allows you to express it with the imaginary unit: \(\sqrt{-1} \times \sqrt{\frac{16}{9}} = i \times \sqrt{\frac{16}{9}}\). Utilizing \(i\) here helps in performing further operations easily.

Square Roots

Understanding square roots is essential, especially when they appear with negative signs or fractions. The square root of a number \(x\) is a number \(y\) such that \(y^2 = x\).

• With positive numbers, the process is straightforward: \(\sqrt{9} = 3\).
• For fractions like \(\frac{16}{9}\), separate the square roots: \(\sqrt{16} = 4\) and \(\sqrt{9} = 3\).
• The square root of the fraction becomes: \(\frac{4}{3}\).

This method allows you to handle square roots of fractions easily. By applying this property, you can also address the negative components by using \(i\), the imaginary unit, as demonstrated in this exercise.

Algebraic Expressions

Algebraic expressions, including those involving square roots and imaginary numbers, require careful handling. They form the building blocks for solving equations involving complex numbers.

• First, simplify terms wherever possible, such as fractions under square roots.
• Identify components like negative square roots that introduce the imaginary unit \(i\).
• Keep expressions clear and simplified, e.g., \(i \times \frac{4}{3}\) results in \(\frac{4}{3}i\).

In algebra, always break complex expressions into simpler parts. Doing this systematically will help you easily navigate even the most complicated problems, ensuring clarity in calculations.