Problem 78 - 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 \(7 + 0i\).

Step 1: Recognize the Expression as a Difference of Squares

The given expression \((\sqrt{6} + i)(\sqrt{6} - i)\) is a difference of squares. This is because it follows the form \((a + b)(a - b)\) where \(a = \sqrt{6}\) and \(b = i\).

Step 2: Apply the Difference of Squares Formula

The formula for the difference of squares is \((a + b)(a - b) = a^2 - b^2\). Applying this to our numbers: \(a = \sqrt{6}\) and \(b = i\), we get:\[ (\sqrt{6})^2 - (i)^2 \]

Step 3: Calculate the Individual Squares

First, calculate each square:- \((\sqrt{6})^2 = 6\) because squaring a square root gives the original number.- \((i)^2 = -1\) because the square of the imaginary unit \(i\) is \(-1\).

Step 4: Substitute and Simplify

Substitute these values back into the expression from Step 2:\[ (\sqrt{6})^2 - (i)^2 = 6 - (-1) \]Simplifying further, this becomes:\[ 6 + 1 = 7 \]Since no imaginary part remains, it is equivalent to \(7 + 0i\).

Step 5: Present the Result in the Form \(a+bi\)

The result of the expression, written in the form \(a+bi\), is \(7 + 0i\). In this form, \(a = 7\) and \(b = 0\), indicating a purely real result with no imaginary part.

Key Concepts

Difference of Squares

In algebra, the difference of squares is a powerful tool for simplifying certain expressions. It refers to the product of a binomial and its conjugate, which follows the pattern

• (a + b)(a - b)

In our exercise, the expression

• \((\sqrt{6} + i)(\sqrt{6} - i)\)

represents the difference of squares because it fits the form with

• \(a = \sqrt{6}\)
• \(b = i\)

The formula for the difference of squares is

• \(a^2 - b^2\)

Applying this,

• \((\sqrt{6})^2 - (i)^2\)

gives us the simplified real number result by effectively removing the imaginary component.

Imaginary Unit

The concept of the imaginary unit, denoted by \(i\), is fundamental in dealing with complex numbers. By definition:

• \(i = \sqrt{-1}\)
• \(i^2 = -1\)

In the realm of complex numbers, the imaginary unit allows for solutions that real numbers alone cannot provide.
In our exercise, \(i\) plays a crucial role in the simplification process. It is part of the difference of squares whereby one of the components \((b)\) is the imaginary unit.

• This gives \((\sqrt{6})^2 - (i)^2\)

The negative sign from \(i^2\) effectively adds to the result, simplifying the operation to be purely real with no imaginary part left.

Complex Number Operations

Operations with complex numbers often involve addition, subtraction, multiplication, and division.
In our given task, we are multiplying two complex numbers. This involves:

• Using the form \((a+b)(a-b)\)
• Applying algebraic rules to simplify the expression

Multiplying

• \((\sqrt{6} + i)(\sqrt{6} - i)\)

requires recognizing the relationship as a difference of squares. By applying

• \(a^2 - b^2\)

we simplify the operation to achieve real results.

Algebraic Expressions

Algebraic expressions are combinations of constants, variables, and operations.
In our problem, we start with

• the initial expression: \((\sqrt{6} + i)(\sqrt{6} - i)\)

These expressions often undergo several transformation steps, requiring the application of different algebraic rules.

• The expression combines real and imaginary components.

Simplifying the expression involves recognizing patterns like the difference of squares.
Using algebraic techniques, we transform it to a more manageable form. Understanding such expressions enhances proficiency in manipulating complex algebraic problems.