Problem 42 - College Algebra Essentials

Question

In Exercises \(29-44\), perform the indicated operations and write the result in standard form. $$ \sqrt{-12}(\sqrt{-4}-\sqrt{2}) $$

Step-by-Step Solution

Answer

The simplified form of the given expression in standard form is \(-4\sqrt{3} - 2\sqrt{6}i\).

Step 1: Identify Negative Square Roots

First, recognize that the square root of a negative number is an imaginary number. The square root of -1 is denoted as 'i', so the square root of -12 can be written as \( \sqrt{12} \cdot i \) and the square root of -4 as \( \sqrt{4} \cdot i \). The expression becomes \( \sqrt{12}i(\sqrt{4}i - \sqrt{2}) \).

Step 2: Simplify Each Square Root

\(\sqrt{12}\) simplifies to \(2\sqrt{3}\) (since 12 is \(4 \cdot 3\)), and \(\sqrt{4}\) simplifies to 2. Our expression now reads: \( 2\sqrt{3}i(2i - \sqrt{2}) \).

Step 3: Distribute \(2\sqrt{3}i\)

Next, distribute \(2\sqrt{3}i\) across (2i - \sqrt{2}): \(2\sqrt{3}i \cdot 2i = 4\sqrt{3}i^2\), \(2\sqrt{3}i \cdot -\sqrt{2} = -2\sqrt{6}i\).

Step 4: Replace \(i^2\) with -1

Since \(i^2\) is -1, we can replace it in the formula to get a standard form of a complex number: \(4\sqrt{3}i^2\) becomes \(-4\sqrt{3}\) and \(-2\sqrt{6}i\) remains the same. The standard form of a complex number is \(a + bi\). So our final answer is: \(-4\sqrt{3} - 2\sqrt{6}i\).

Key Concepts

Imaginary Numbers

When we work with complex numbers, the concept of imaginary numbers is key to understanding the mathematical universe beyond real numbers. Imaginary numbers are a class of numbers that when squared give a negative result, which is not possible in the system of real numbers.

Consider, for instance, the square root of -1, denoted as 'i'. It is the basic unit of imaginary numbers, just like 1 is the basic unit of real numbers. When we encounter a square root of a negative number in calculations, like \( \sqrt{-12} \), we can rewrite it in terms of 'i' to make it manageable. The process involves taking the square root of the positive counterpart and multiplying by 'i', transforming \( \sqrt{-12} \) into \( \sqrt{12} \cdot i \).

This opens doors to simplifying expressions involving square roots of negative numbers, making them usable in further operations, such as addition, subtraction, multiplication, and division, which are foundational operations in algebra involving complex numbers.

Simplifying Square Roots

Simplifying square roots is a vital skill when dealing with radicals and complex numbers. The goal is to break down the square root into its simplest form, ideally with no square factors left under the radical sign.

Let's explore this with the example \( \sqrt{12} \). Since 12 is a product of 4 (a perfect square) and 3, we can simplify \( \sqrt{12} \) to \( 2\sqrt{3} \) by taking the square root of 4 out of the radical. This process is based on the property that \( \sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b} \). Simplifying roots reduces the complexity of mathematical expressions and prepares them for further operations, such as multiplication or division by other terms.

Mastering this technique ensures cleaner results and sometimes reveals further opportunities for simplification, particularly crucial when combining terms in the realm of complex numbers.

Distributive Property

The distributive property is a fundamental property used to simplify algebraic expressions, especially when they involve parentheses. It states that \( a(b + c) = ab + ac \). This property allows us to multiply a single term by each term inside a set of parentheses.

In the context of the problem with complex numbers, we must apply the distributive property skillfully. For the given expression \( 2\sqrt{3}i(2i - \sqrt{2}) \), we distribute \( 2\sqrt{3}i \) over the \( 2i \) and the \( -\sqrt{2} \) separately; this gives us two terms: \( 4\sqrt{3}i^2 \) and \( -2\sqrt{6}i \). Recognizing that \( i^2 = -1 \) further simplifies the expression to its standard form. The distributive property is powerful not only in arithmetic but also in handling complex numbers, allowing us to expand and simplify polynomial expressions, and is a staple in solving algebraic equations.