Problem 36 - A Graphical Approach to College Algebra

Question

Multiply or divide as indicated. Simplify each answer. $$\sqrt{-5} \cdot \sqrt{-15}$$

Step-by-Step Solution

Answer

The simplified product is \(-5\sqrt{3}\).

Step 1: Understanding the Problem Statement

We are asked to multiply two complex square roots: \(\sqrt{-5}\) and \(\sqrt{-15}\). To do this, we must first convert them into complex numbers, since we cannot take square roots of negative numbers in the real number system.

Step 2: Express in Terms of Imaginary Numbers

Recall that \(\sqrt{-1} = i\), where \(i\) is the imaginary unit. Thus, \(\sqrt{-5} = \sqrt{5} \cdot i\) and \(\sqrt{-15} = \sqrt{15} \cdot i\).

Step 3: Multiply the Expressions

Multiply the two expressions: \((\sqrt{5} \cdot i) \cdot (\sqrt{15} \cdot i) = (\sqrt{5} \cdot \sqrt{15}) \cdot i^2\).

Step 4: Simplify the Product of Square Roots

Combine the square roots as follows: \(\sqrt{5} \cdot \sqrt{15} = \sqrt{5 \cdot 15} = \sqrt{75}\). Further simplify \(\sqrt{75}\) to \(\sqrt{25 \cdot 3} = \sqrt{25} \cdot \sqrt{3} = 5\sqrt{3}\).

Step 5: Simplify Using \(i^2 = -1\)

From Step 3, use \(i^2 = -1\) to simplify: \(5 \sqrt{3} \cdot i^2 = 5 \sqrt{3} \cdot (-1) = -5 \sqrt{3}\).

Step 6: Final Answer

The expression simplifies to \(-5\sqrt{3}\). Therefore, \(\sqrt{-5} \cdot \sqrt{-15} = -5\sqrt{3}\).

Key Concepts

Imaginary Numbers

Imaginary numbers are a fascinating part of mathematics that extend the real number system. Whenever we encounter the square root of a negative number, such as \( \sqrt{-5} \), we are stepping into the domain of imaginary numbers. These numbers are crucial for deep mathematical concepts and real-world applications, despite being termed "imaginary".

The core idea of imaginary numbers revolves around the imaginary unit \( i \), defined by \( i = \sqrt{-1} \). This allows us to express square roots of negative numbers in terms of \( i \). For example, \( \sqrt{-5} \) can be rewritten as \( \sqrt{5} \cdot i \).

Once you grasp this concept, it becomes natural to handle operations that involve negative square roots, converting them into a form involving \( i \) before proceeding with multiplication, division, or any further simplification.

Square Roots

Understanding square roots is essential when dealing with complex numbers because they appear frequently, especially when multiplying complex expressions. A square root finds the value that, when multiplied by itself, gives the original number. For positive numbers, we compute square roots as expected. However, for negative numbers, we shift to imaginary numbers.

To compute \( \sqrt{-5} \) in our exercise, you start by recognizing it as \( \sqrt{5} \cdot i \). This method of isolating the negative sign allows us to manage the square root of positive numbers and handle the imaginary part separately. Also, remember when multiplying square roots, the product \( \sqrt{a} \times \sqrt{b} \) simplifies to \( \sqrt{a \cdot b} \). This property is particularly useful in simplifying expressions involving radicals.

Multiplication and Division of Radicals

When dealing with radicals, especially in multiplication and division, a few key principles simplify the process. First, it's essential to remember that multiplying square roots is straightforward when they share the same root symbol. For instance, \( \sqrt{-5} \times \sqrt{-15} \) can initially seem complex.

Instead, express each term with \( i \): \( \sqrt{5} \cdot i \) and \( \sqrt{15} \cdot i \). Multiply the terms: \( (\sqrt{5} \cdot i) \times (\sqrt{15} \cdot i) \) combines as \( \sqrt{5 \times 15} \cdot i^2 \).

Dividing radicals follows similar rules and often involves multiplying by a conjugate to simplify fractions. Always remember that radicals adhere to the same basic arithmetic rules, which makes their manipulation much easier once they're well understood.

Simplification of Expressions

The simplification of expressions involving radicals and imaginary numbers is vital to reaching a clear and accurate final answer. The process demands careful attention to detail, following defined mathematical properties.

After initially simplifying square roots in the exercise, \( \sqrt{75} \) is broken down to \( \sqrt{25 \times 3} \), which further simplifies to \( 5\sqrt{3} \). Meanwhile, using the property \( i^2 = -1 \), the term \( i^2 \) simplifies the expression by effectively negating the term, as seen when the product \( 5\sqrt{3} \cdot i^2 \) becomes \( -5\sqrt{3} \).

Precise simplification not only provides the correct result but also enhances comprehension of how different mathematical tools—like \( i \) and square roots—interact to reshape and distill the output into its simplest, most meaningful form.