Problem 31 - Algebra for College Students

Question

Write each of the following in terms of \(i\) and simplify. For example, $$ \sqrt{-20}=i \sqrt{20}=i \sqrt{4} \sqrt{5}=2 i \sqrt{5} $$ $$\sqrt{-\frac{16}{25}}$$

Step-by-Step Solution

Answer

\( \frac{4}{5}i \).

Step 1: Express as imaginary

The square root of a negative number can be expressed in terms of the imaginary unit, where \( i = \sqrt{-1} \). First, let's represent the given expression \( \sqrt{-\frac{16}{25}} \) as an imaginary number. This becomes \( i \sqrt{\frac{16}{25}} \), because \( \sqrt{-\frac{16}{25}} = i \sqrt{\frac{16}{25}} \).

Step 2: Simplify the numerator and denominator

Next, we simplify the expression under the square root. The fraction \( \frac{16}{25} \) is composed of 16, a perfect square, and 25, also a perfect square. We can simplify \( \sqrt{\frac{16}{25}} \) to \( \frac{\sqrt{16}}{\sqrt{25}} \).

Step 3: Calculate the square roots

Now, we compute the square root of the numerator and the denominator separately. \( \sqrt{16} = 4 \) and \( \sqrt{25} = 5 \). So, \( \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5} \).

Step 4: Multiply by the imaginary unit

Finally, we multiply the simplified fraction by the imaginary unit \( i \) from Step 1. Thus, the expression becomes \( i \cdot \frac{4}{5} = \frac{4}{5}i \).

Key Concepts

square roots

When taking square roots, especially of negative numbers, understanding the basics is vital. A square root undoes squaring a number, which means if you square a number and then take its square root, you should get the original value back. However, when dealing with negative numbers under a square root, things change. Since the square of any real number is positive, the square root of a negative number is not a real number. For example, in the expression \( \sqrt{-16} \), the negative sign presents a unique case where we introduce the concept of an imaginary number to solve it. By using imaginary numbers, we can redefine the square root where \( \sqrt{-a} = i \sqrt{a} \), with \( i \) being the imaginary unit.

simplifying expressions

Simplifying expressions involving square roots is a fundamental skill in algebra. It involves breaking down complex expressions into simpler forms that are easier to understand or work with. For example, consider \( \sqrt{\frac{16}{25}} \). To simplify, we separate the square root of the fraction into the square root of the numerator and the square root of the denominator. This yields \( \frac{\sqrt{16}}{\sqrt{25}} \).- First, identify if the terms are perfect squares;- Calculate \( \sqrt{16} \) which is 4, because 4 times 4 equals 16;- Calculate \( \sqrt{25} \) which is 5, because 5 times 5 equals 25.Thus, \( \sqrt{\frac{16}{25}} \) simplifies to \( \frac{4}{5} \). It's always a good idea to first check for perfect squares, as they make the simplification process straightforward.

imaginary unit

The imaginary unit, denoted by \( i \), is a crucial part of understanding complex numbers, especially when working with the square roots of negative numbers. In mathematics, \( i \) is defined as \( \sqrt{-1} \). Its main property is that when you square it, you get \( i^2 = -1 \).- Using \( i \) allows expressions like \( \sqrt{-16} \) to be rewritten as \( i \sqrt{16} \).- This can then be further simplified to \( 4i \) by calculating \( \sqrt{16} \) as 4, leaving \( i \) as the factor responsible for the negativity.Because real numbers cannot have the square root of a negative number, the introduction of \( i \) expands the number system to complex numbers, thereby allowing solutions to equations previously deemed unsolvable in the realm of real numbers.