Problem 77 - Intermediate Algebra

Question

Perform each indicated operation. Write the result in the form \(a+b i\). $$ (\sqrt{6}+i)(\sqrt{6}-i) $$

Step-by-Step Solution

Answer

The result is 7.

Step 1: Apply the Distributive Property

First, we need to distribute each term in the first binomial to each term in the second binomial. This process is often called FOIL (First, Outer, Inner, Last): 1. First: \(\sqrt{6} \cdot \sqrt{6} = 6\)2. Outer: \(\sqrt{6} \cdot (-i) = -\sqrt{6}i\)3. Inner: \(i \cdot \sqrt{6} = \sqrt{6}i\)4. Last: \(i \cdot (-i) = -i^2\)

Step 2: Perform the Multiplication

Now we multiply the terms we found in Step 1:- \(6\) (from \(\sqrt{6} \cdot \sqrt{6}\))- \(-\sqrt{6}i\) and \(+\sqrt{6}i\) (resulting from \(\sqrt{6}(-i)\) and \(i\sqrt{6}\) respectively)- \(-i^2\) (which simplifies as follows)

Step 3: Simplify \(-i^2\)

Recall that \(i^2 = -1\). So, in terms of our calculation:\[-i^2 = -(-1) = 1\].Thus, the expression becomes:\(6 - \sqrt{6}i + \sqrt{6}i + 1\).

Step 4: Combine Like Terms

Now combine the like terms:- The imaginary parts \(-\sqrt{6}i\) and \(+\sqrt{6}i\) cancel each other because they are additive inverses.- Combine the real numbers \(6\) and \(1\):\[6 + 1 = 7\].

Step 5: Final Result

The result of the expression in the form \(a + bi\) is:\[7 + 0i\ = 7\].

Key Concepts

Distributive Property

The distributive property is a fundamental concept in mathematics that allows us to distribute, or "spread out," multiplication over addition or subtraction within parentheses. Essentially, when you are given an expression like \((a + b)(c + d)\), the distributive property lets you multiply each term inside the first pair of parentheses by each term in the second pair of parentheses. It's like ensuring every term in one group gets to interact with every term in the other.

When applied, this becomes:

• First: \(a \cdot c\)
• Outer: \(a \cdot d\)
• Inner: \(b \cdot c\)
• Last: \(b \cdot d\)

Add these products together to simplify or solve the expression fully. This method of breaking down complex multiplication is invaluable, especially when dealing with binomials and polynomial expressions. Notice how in our exercise the same rule was applied with \((\sqrt{6}+i)(\sqrt{6}-i)\). Each term was individually multiplied following the same structured approach.

Binomial Multiplication

Binomial multiplication involves expanding expressions where two terms are multiplied together, typically presented in the form \((a+b)(c+d)\). This is a direct application of the distributive property. When multiplying binomials:

• "First" - multiply the first terms of each binomial.
• "Outer" - multiply the outer terms of the binomials.
• "Inner" - multiply the inner terms.
• "Last" - multiply the last terms of each.

This FOIL method ensures you cover every combination of terms being multiplied between the two sets of parentheses.

In our example, \((\sqrt{6}+i)(\sqrt{6}-i)\), we multiplied as follows:

• First: \(\sqrt{6} \cdot \sqrt{6} = 6\)
• Outer: \(\sqrt{6} \cdot (-i) = -\sqrt{6}i\)
• Inner: \(i \cdot \sqrt{6} = \sqrt{6}i\)
• Last: \(i \cdot (-i) = -i^2\)

Imaginary Unit

The imaginary unit, often represented by \(i\), is a key component in complex numbers. It is defined as the square root of -1. Thus, \[i^2 = -1\]. This property is incredibly important when performing operations with complex numbers because it provides a bridge between real numbers and these new, complex numbers.

In our calculation, when multiplying terms involving \(i\), like \(i \cdot (-i)\), we have to remember \(-i^2\) simplifies to \(1\) because \(-i^2 = -(-1)\). This simplification transforms often complex-looking expressions into something more manageable, as seen by turning \(-i^2\) into \(+1\). This reduction was crucial when completing the operation, allowing us ultimately to reach a simpler final form.

Like Terms

When simplifying expressions, recognizing and combining like terms is essential and can tremendously simplify your calculations. Like terms have identical variables raised to the same power. Therefore, you can combine them into a single term.

For example, consider the expression \(x^2 + 2x^2 = 3x^2\) which has like terms of \(x^2\). In the realm of complex numbers, the concept of like terms extends to matching imaginary units \(i\).

In our exercise, the terms \(-\sqrt{6}i\) and \(+\sqrt{6}i\) are like terms and are additive inverses, which means they cancel each other out to zero. This is why we no longer see any terms involving \(i\) in the final expression of \(7 + 0i = 7\). Recognizing like terms allows for simplification, helping to present expressions in \(a + bi\) form, where \(b\) is the coefficient of \(i\).