Problem 37 - Intermediate Algebra

Question

Multiply. Write the product in the form \(a+b i .\) See Example 4. $$ (\sqrt{3}+2 i)(\sqrt{3}-2 i) $$

Step-by-Step Solution

Answer

The product is 7 with no imaginary component.

Step 1: Recognize the FOIL Method

The expression to be multiplied is \((\sqrt{3} + 2i)(\sqrt{3} - 2i)\). To multiply these two binomials, we use the FOIL method, which stands for First, Outer, Inner, Last. This approach will help simplify and find the product.

Step 2: Multiply the First Terms

The first terms are \(\sqrt{3}\) and \(\sqrt{3}\). Multiply them:\[\sqrt{3} \cdot \sqrt{3} = 3\]

Step 3: Multiply the Outer Terms

The outer terms are \(\sqrt{3}\) and \(-2i\). Multiply them:\[\sqrt{3} \cdot (-2i) = -2\sqrt{3}i\]

Step 4: Multiply the Inner Terms

The inner terms are \(2i\) and \(\sqrt{3}\). Multiply them:\[2i \cdot \sqrt{3} = 2\sqrt{3}i\]

Step 5: Multiply the Last Terms

The last terms are \(2i\) and \(-2i\). Multiply them:\[2i \cdot (-2i) = -4i^2\]Recall that \(i^2 = -1\). Thus,\(-4i^2 = -4(-1) = 4\).

Step 6: Sum All the Results

Sum the results from Step 2 to Step 5:\[3 + (-2\sqrt{3}i) + 2\sqrt{3}i + 4\]Combine like terms:- The imaginary terms \(-2\sqrt{3}i\) and \(2\sqrt{3}i\) cancel each other out.- Combine the real numbers: \(3 + 4 = 7\).Thus, the product is \(7\).

Step 7: Write Product in the Desired Form

In the form \(a + bi\), we have \(a = 7\) and \(b = 0\), so the final answer is simply \(7\).

Key Concepts

FOIL Method

The FOIL method is a useful tool designed to simplify the multiplication of two binomials. Understanding it helps make the process straightforward. When given two binomials, like \((a+b)(c+d)\), the FOIL method stands for:

• First - Multiply the first terms in each binomial. For example, \(a \cdot c\).
• Outer - Multiply the outer terms in the expression. For example, \(a \cdot d\).
• Inner - Multiply the inner terms, such as \(b \cdot c\).
• Last - Multiply the last terms in each binomial, such as \(b \cdot d\).

After calculating each component from these steps, you add them all together. This method helps ensure you don't miss any part of the multiplication. It's especially handy when dealing with complex numbers, where each term might require extra attention to detail.

Imaginary Numbers

Imaginary numbers might sound strange at first, but they serve a critical role in mathematics. Key to understanding them is the unit imaginary number, denoted as \(i\). It is defined so that \(i^2 = -1\).

Because of this unique property, calculations involving imaginary numbers often involve managing \(i^2\) terms. For example, when you multiply two imaginary numbers together, like \(2i \cdot (-2i)\), you actually face multiplying \(2 \cdot -2 \cdot i^2\), which simplifies to \(4\) because \(i^2 = -1\).
This manipulation of \(i\) is pivotal in working through problems and ensures expressions remain in the friendly form \(a + bi\), where \(a\) and \(b\) are real numbers.

Multiplication of Binomials

Multiplying binomials is a fundamental skill in algebra that becomes increasingly important when complex numbers are involved. When you encounter a problem, such as multiplying \((\sqrt{3}+2i)(\sqrt{3}-2i)\), you are essentially performing the multiplication by combining like terms.

Remember that the structure of binomial multiplication allows each term from the first binomial to be distributed across each term in the second. This ensures that each part contributes to the final product. Importantly, when dealing with complex numbers, simplification often involves:

• Recognizing \(i^2\) terms and replacing them with \(-1\).
• Combining real number terms separately from imaginary number terms.
• Observing that certain terms cancel out, reducing the overall complexity of the expression.

By following these steps, you can ensure that the product of the binomials is computed effectively and that the result is expressed in the standard form \(a + bi\). This helps in both clarity and in comparing or solving complex equations.