Chapter 7: Problem 78
Perform each indicated operation. Write the result in the form \(a+b i\). $$ (\sqrt{6}+i)(\sqrt{6}-i) $$
Short Answer
Expert verified
The result is \(7 + 0i\).
Step by step solution
01
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 \(b = i\).
02
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:\[ (\sqrt{6})^2 - (i)^2 \]
03
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 \(-1\).
04
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 \(7 + 0i\).
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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)
- \((\sqrt{6} + i)(\sqrt{6} - i)\)
- \(a = \sqrt{6}\)
- \(b = i\)
- \(a^2 - b^2\)
- \((\sqrt{6})^2 - (i)^2\)
Imaginary Unit
The concept of the imaginary unit, denoted by \(i\), is fundamental in dealing with complex numbers. By definition:
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.
- \(i = \sqrt{-1}\)
- \(i^2 = -1\)
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 \((\sqrt{6})^2 - (i)^2\)
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:
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
- \((\sqrt{6} + i)(\sqrt{6} - i)\)
- \(a^2 - b^2\)
Algebraic Expressions
Algebraic expressions are combinations of constants, variables, and operations.
In our problem, we start with
Using algebraic techniques, we transform it to a more manageable form. Understanding such expressions enhances proficiency in manipulating complex algebraic problems.
In our problem, we start with
- the initial expression: \((\sqrt{6} + i)(\sqrt{6} - i)\)
- The expression combines real and imaginary components.
Using algebraic techniques, we transform it to a more manageable form. Understanding such expressions enhances proficiency in manipulating complex algebraic problems.