Problem 48 - Algebra for College Students

Question

Write each of the following in terms of \(i\), perform the indicated operations, and simplify. For example, $$ \begin{aligned} \sqrt{-3} \sqrt{-8} &=(i \sqrt{3})(i \sqrt{8}) \\ &=i^{2} \sqrt{24} \\ &=(-1) \sqrt{4} \sqrt{6} \\ &=-2 \sqrt{6} \end{aligned} $$ $$\sqrt{-8} \sqrt{-16}$$

Step-by-Step Solution

Answer

The simplified expression is \(-8 \sqrt{2}\).

Step 1: Express Each Square Root in Terms of i

We start by expressing the square roots of negative numbers in terms of \(i\), where \(i = \sqrt{-1}\). For \(\sqrt{-8}\), this becomes \(i \sqrt{8}\). For \(\sqrt{-16}\), this becomes \(i \sqrt{16}\). Thus, we have:\[\sqrt{-8} = i \sqrt{8} \quad \text{and} \quad \sqrt{-16} = i \sqrt{16}\]

Step 2: Multiply the Expressions

Multiply the expressions obtained in Step 1:\[(i \sqrt{8})(i \sqrt{16})\]

Step 3: Simplify Using Properties of i

Simplify the product by multiplying the \(i\) terms and simplifying the square roots:\[i^2 \sqrt{8} \cdot \sqrt{16} = (-1) \sqrt{8 \cdot 16}\]Here, \(i^2 = -1\).

Step 4: Simplify the Square Root

Calculate the square root of the product:\[\sqrt{8 \times 16} = \sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2}\]

Step 5: Final Simplification

Insert back the simplified square root into the expression that included \(-1\) from \(i^2\):\[(-1) \cdot 8 \sqrt{2} = -8 \sqrt{2}\]

Key Concepts

Imaginary Unit

The imaginary unit, commonly denoted as \(i\), is a fundamental concept in complex numbers. It is defined as the square root of \(-1\). This means that when you square \(i\), you obtain \(-1\), or mathematically \(i^2 = -1\).

Complex numbers themselves are composed of a real part and an imaginary part. The imaginary part is what involves \(i\), and it's this component that allows for the manipulation and use of square roots of negative numbers.

In practical applications, the imaginary unit is often used in engineering and physics, particularly in dealing with wave functions, signal processing, and various equations that require roots of negative numbers.

Multiplication of Complex Numbers

Multiplying complex numbers is a bit different from multiplying regular numbers. When two complex numbers, each with an imaginary component, are multiplied, the imaginary units must be multiplied as well. Using the imaginary unit \(i\), let's consider an example where you multiply \((i \sqrt{8})\) and \((i \sqrt{16})\).

Here's how it's done:
- First, multiply the imaginary units \(i \times i\), which equals \(i^2\).
- Recall that \(i^2 = -1\); hence, the imaginary units simplify to \(-1\).
- Next, multiply the square root parts: \(\sqrt{8} \cdot \sqrt{16}\), which simplifies using multiplication properties of square roots to \(\sqrt{128}\).

Finally, combine these results, leading to the product \((-1) \cdot \sqrt{128}\), which further simplifies based on square root calculations. This step-by-step method ensures that the multiplication properly considers both the real and imaginary components of complex numbers.

Square Roots of Negative Numbers

Handling square roots of negative numbers can initially seem confusing. However, using the imaginary unit, \(i\), simplifies this process. Remember that \(\sqrt{-1} = i\).

When deriving square roots of larger negative numbers, you can separate the number into a product of \(-1\) and a positive number. For example, consider \(\sqrt{-8}\). You can rewrite it as \(\sqrt{-1 \times 8}\), which allows you to utilize the property of \(\sqrt{-1} = i\).

Thus, \(\sqrt{-8}\) becomes \(i \sqrt{8}\), where \(8\) is now a regular non-negative under the square root. This process applies to any negative number and helps to manipulate these numbers effectively when working with complex numbers.

Using these steps makes performing operations like multiplication or simplification on expressions involving square roots of negative numbers much more manageable and structured.