Problem 50 - Algebra for College Students

Question

Write each of the following in terms of \(i\), perform the indicated operations, and simplify. For example, $$ \begin{aligned} \sqrt{-3} \sqrt{-8} &=(i \sqrt{3})(i \sqrt{8}) \\ &=i^{2} \sqrt{24} \\ &=(-1) \sqrt{4} \sqrt{6} \\ &=-2 \sqrt{6} \end{aligned} $$ $$\sqrt{-2} \sqrt{-20}$$

Step-by-Step Solution

Answer

The simplified expression is \(-2\sqrt{10}\).

Step 1: Express in terms of i

First, express each square root with a negative number inside in terms of the imaginary unit \(i\). Recall that \(i = \sqrt{-1}\), so for any negative number \(-x\), the square root can be written as \(i\sqrt{x}\). Thus, \(\sqrt{-2} = i\sqrt{2}\) and \(\sqrt{-20} = i\sqrt{20}\).

Step 2: Multiply the expressions

Next, multiply the expressions involving \(i\). We have \((i\sqrt{2}) \times (i\sqrt{20}) = i^2 \times \sqrt{2} \times \sqrt{20}\).

Step 3: Simplify \(i^2\)

Simplify \(i^2\) using the fact that \(i^2 = -1\). Thus, the expression becomes \(-1 \times \sqrt{2} \times \sqrt{20}\).

Step 4: Simplify the square roots

Simplify \(\sqrt{2} \times \sqrt{20}\). Since \(\sqrt{a} \times \sqrt{b} = \sqrt{a \cdot b}\), this becomes \(\sqrt{2 \times 20} = \sqrt{40}\).

Step 5: Simplify \(\sqrt{40}\)

Simplify \(\sqrt{40}\) further. Note that \(\sqrt{40} = \sqrt{4 \times 10}\), which can be split into \(\sqrt{4} \times \sqrt{10} = 2 \sqrt{10}\).

Step 6: Final simplification

Multiply the result from Step 5 by the \(-1\) left from \(i^2\). Thus, \(-1 \times 2 \sqrt{10} = -2\sqrt{10}\). This is the simplified form of the expression.

Key Concepts

Imaginary Unit

In mathematics, dealing with negative square roots can be tricky because square rooting a negative number is not possible within the realm of real numbers. This is where the imaginary unit comes into play, represented by the symbol \(i\). The imaginary unit \(i\) is defined such that \(i = \sqrt{-1}\). This clever invention allows us to explore square roots of negative numbers by rewriting them.

• For instance, the square root of \(-2\) is rewritten using the imaginary unit as \(i\sqrt{2}\).
• Similarly, \(\sqrt{-20}\) becomes \(i\sqrt{20}\).

By converting negative square roots into terms of \(i\), we utilize the properties of imaginary numbers to perform further mathematical operations seamlessly. This conversion is the first step in simplifying expressions involving complex numbers.

Simplifying Radicals

The next part of our journey involves simplifying radicals, which is all about making square roots easier to manage. Once radicals are expressed in terms of the imaginary unit, they need to be simplified for clean and easy calculations. Simplifying radicals often involves finding factors of the number inside the square root that are perfect squares.When simplifying \(\sqrt{40}\), for example, we know that \(40\) is the product of \(4\) and \(10\), where \(4\) is a perfect square. Thus:

• We break it down as \(\sqrt{40} = \sqrt{4 \times 10}\).
• This becomes \(\sqrt{4} \times \sqrt{10}\).
• Since \(\sqrt{4} = 2\), the expression simplifies to \(2\sqrt{10}\).

Simplifying radicals is crucial as it makes further multiplication and operations more straightforward. It can also help in achieving the final simplified expression much more efficiently.

Multiplication of Square Roots

Multiplying square roots is straightforward once they are broken down into simpler terms. This process involves combining the radicals together and then simplifying where possible.Given two expressions such as \(i\sqrt{2}\) and \(i\sqrt{20}\), the multiplication process involves the following steps:

• First, multiply the terms including \(i\), resulting in \(i^2\).
• Recognize that \(i^2 = -1\) due to the definition of the imaginary unit.
• Next, handle the multiplication of the square roots by applying the property \(\sqrt{a} \times \sqrt{b} = \sqrt{a \cdot b}\).
• For our expressions, this means \(\sqrt{2} \times \sqrt{20} = \sqrt{40}\).
• Lastly, simplify \(\sqrt{40}\) down to \(2\sqrt{10}\) as discussed in simplifying radicals.

By applying these steps, the entire process is methodical and leads you to a simplified expression smoothly, especially when handling complex numbers and imaginary units.