Problem 43 - A Graphical Approach to College Algebra

Question

Multiply or divide as indicated. Simplify each answer. $$\frac{\sqrt{-6} \cdot \sqrt{-2}}{\sqrt{3}}$$

Step-by-Step Solution

Answer

The simplified result is \( -2 \).

Step 1: Identify Complex Terms

Both \( \sqrt{-6} \) and \( \sqrt{-2} \) contain negative numbers inside the square roots, which means they are complex numbers. We rewrite them using the imaginary unit \( i \), where \( i = \sqrt{-1} \). This gives us \( \sqrt{-6} = \sqrt{6}i \) and \( \sqrt{-2} = \sqrt{2}i \).

Step 2: Multiply the Numerator

Multiply \( \sqrt{6}i \) by \( \sqrt{2}i \). Using the properties of square roots, \( (\sqrt{a}i)(\sqrt{b}i) = \sqrt{ab}i^2 \). So, \((\sqrt{6}i)(\sqrt{2}i) = \sqrt{12}i^2 = \sqrt{12}(-1) = -\sqrt{12}\).

Step 3: Simplify the Numerator

Simplify \( -\sqrt{12} \). \( \sqrt{12} \) can be simplified as \( \sqrt{4 \times 3} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3} \). Thus, \( -\sqrt{12} = -2\sqrt{3} \).

Step 4: Divide by the Denominator

Divide the result from Step 3 by \( \sqrt{3} \), giving \( \frac{-2\sqrt{3}}{\sqrt{3}} \). Cancel out \( \sqrt{3} \), resulting in \( -2 \).

Step 5: Final Result: Simplification

After cancellation, the expression fully simplifies to \( -2 \).

Key Concepts

Imaginary Unit

In mathematics, the imaginary unit is a fundamental concept when dealing with complex numbers. It is represented by the symbol \( i \), defined as the square root of \(-1\). This means that \( i^2 = -1 \). Imaginary numbers arise when taking square roots of negative numbers. For example, the square root of \(-6\) can be expressed as \( \sqrt{6}i \), where \( i \) captures the imaginary component.
Understanding \( i \) is crucial for working with complex numbers. It allows us to convert expressions that involve the square roots of negative values into a form that can be more easily manipulated. When two imaginary numbers are multiplied, such as \( (\sqrt{6}i)(\sqrt{2}i) \), the property \( i^2 = -1 \) plays a key role in simplifying the expression.

Square Roots

Square roots are an essential operation in mathematics, allowing us to find a number that, when multiplied by itself, results in a given value. Square roots of positive numbers are straightforward, but negative numbers introduce complexity. Without the imaginary unit, the square root of a negative number is undefined in the set of real numbers.
Using the imaginary unit \( i \), we transform these square roots into complex numbers. For instance, \( \sqrt{-6} \) becomes \( \sqrt{6}i \), and \( \sqrt{-2} \) becomes \( \sqrt{2}i \). These transformations are based on the principle that \( \sqrt{-1} = i \).

• To simplify further, we can multiply the square roots: \( (\sqrt{6}i) \times (\sqrt{2}i) = \sqrt{12}i^2 = -\sqrt{12} \).
• Simplification often involves reducing expressions by finding patterns or common factors, such as \( \sqrt{12} = 2\sqrt{3} \).

Simplifying Expressions

Simplifying expressions is a key skill in mathematics, making complex calculations more manageable. It involves reducing expressions to their simplest form without changing the value. When dealing with complex numbers, simplification can involve operations like multiplication and division alongside handling imaginary units.
For example, consider the multiplication of two complex terms \( \sqrt{6}i \) and \( \sqrt{2}i \). After performing the operation, the expression \( \sqrt{12}i^2 \) simplifies to \( -\sqrt{12} \) because \( i^2 = -1 \).
Further simplification involves breaking down square roots into more elemental parts, such as converting \( \sqrt{12} \) into \( 2\sqrt{3} \). This step helps us simplify the entire fraction \( \frac{-2\sqrt{3}}{\sqrt{3}} \) to \(-2\) by cancelling out \( \sqrt{3} \) in the numerator and denominator.
Thus, simplification not only makes complex expressions easier to handle but also leads us to more intuitive and readily interpretable results.