Problem 24 - Precalculus Student Solutions Manual 5th

Question

Write each number as the product of a real number and i. $$-\sqrt{-80}$$

Step-by-Step Solution

Answer

-4i \sqrt{5}

Step 1: Understand the problem

The goal is to express \text{-} \sqrt{-80} as a product of a real number and the imaginary unit \(i\). Recall that \(i\) represents \(\sqrt{-1}\).

Step 2: Simplify the negative inside the square root

Rewrite \(\sqrt{-80}\) as \(\sqrt{80} \cdot \sqrt{-1}\). Because \(\sqrt{-1} = i\), this becomes \(\sqrt{80} \cdot i\).

Step 3: Simplify the square root of 80

Break down \(\sqrt{80}\) into its prime factors: \(\sqrt{80} = \sqrt{16 \cdot 5}\). Since \(16\) is a perfect square, \(\sqrt{80} = \sqrt{16} \cdot \sqrt{5} = 4 \cdot \sqrt{5}\).

Step 4: Combine the real number with the imaginary unit

Substitute \(4 \cdot \sqrt{5}\) into the expression from Step 2. This gives \(\sqrt{-80} = 4i \cdot \sqrt{5}\).

Step 5: Include the negative sign

Remember the initial problem was \(\text{-} \sqrt{-80}\). Include the negative to get the final answer: \(-4i \cdot \sqrt{5}\).

Key Concepts

Complex Numbers

Complex numbers are an extension of real numbers and include a real part and an imaginary part.
They are written in the form: \(a + bi\) where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit defined as \(i = \sqrt{-1}\).
For example, in the number \(3 + 4i\), \(3\) is the real part, and \(4i\) is the imaginary part.
Complex numbers are especially useful in solving equations that don't have real solutions, like \(x^{2} + 1 = 0\). Simplifying complex numbers follows the same basic math rules.
Operations here include addition, subtraction, multiplication, and division.
For more on complex numbers, keep these points in mind:

• Addition/Subtraction: Combine like terms (\text{i.e., real with real and imaginary with imaginary}).
• Multiplication: Apply distributive property and the fact that \(i^{2} = -1\).
• Division: Multiply numerator and denominator by the conjugate of the denominator.

Understanding complex numbers can simplify problems involving radicals and negative numbers.

Roots of Negative Numbers

Squaring any real number always yields a non-negative result.
This means the square root of a negative number isn't a real number.
Instead, we introduce the imaginary unit \(i\) to represent the square root of \(-1\).
For example, the square root of \(-4\) can be written as:

• \(\sqrt{-4} = \sqrt{4} \cdot \sqrt{-1}\)
• Since \(\sqrt{4} = 2\) and \(\sqrt{-1} = i\), it becomes \(2i\).

This idea extends to any negative number under a square root.
In our problem, we had \(\sqrt{-80}\).

• First, recognize \(\sqrt{-80} = \sqrt{80} \cdot \sqrt{-1}\).
• Then, use \(i\) for \(\sqrt{-1}\) to rewrite it as \(\sqrt{80} \cdot i\).

This method helps in expressing roots of negative numbers in terms of imaginary numbers.

Simplifying Radicals

Simplifying radicals involves breaking the number inside the root into its prime factors.
Consider \(\sqrt{80}\). To simplify this:

• First, identify its prime factors: \(80 = 16 \cdot 5\).
• Next, use the property that \(\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}\).
• This gives \(\sqrt{80} = \sqrt{16} \cdot \sqrt{5}\).
• Recognize that \(16\) is a perfect square, so \(\sqrt{16} = 4\).

Thus, \(\sqrt{80} = 4 \cdot \sqrt{5}\).
In our problem, we substitute this back in to get \(4i \sqrt{5}\), and account for the negative sign to get \(-4i \sqrt{5}\).
Simplifying radicals helps to break down complex expressions into more manageable parts that can be easily combined with imaginary numbers.