/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 91 Simplify each expression and wri... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Simplify each expression and write it in the standard form \(a+b i\). \((6-i)(6+i)\)

Short Answer

Expert verified
The expression simplifies to 37.

Step by step solution

01

Identify and apply the formula for difference of squares

Recognize that the expression (6-i)(6+i) is a difference of squares: (a-b)(a+b) = a^2 - b^2 where a=6 and b=i.
02

Simplify using the difference of squares

Apply the formula:\[(6-i)(6+i) = 6^2 - (i)^2\]Calculate 6^2 = 36 and (i)^2 = i^2 = -1.
03

Substitute and simplify

Substitute back into the equation:\[36 - (-1) = 36 + 1 = 37\] Hence, the expression simplifies to 37 + 0i.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference of Squares
The difference of squares is a special algebraic identity that helps simplify expressions of the form \((a-b)(a+b)\). Using this identity means you don't have to multiply terms individually; you can directly calculate:\[a^2 - b^2\]In the expression given in the exercise, \((6-i)(6+i)\), the structure perfectly matches the difference of squares formula, where \(a = 6\) and \(b = i\). By applying the formula, you simplify calculations considerably. Instead of performing tedious distribution, just square the first and second terms separately and subtract.Here’s a quick recap:- For \(a = 6\), \(a^2 = 6^2 = 36\).- For \(b = i\), \(b^2 = i^2\), and we know \(i^2 = -1\) (this is crucial!).Putting it all together:- The difference of squares becomes \(36 - (-1) = 37\). It's a powerful technique that saves time and reduces the risk of error.
Standard Form
In mathematics, especially when dealing with complex numbers, writing expressions in standard form is a must-do practice. The standard form is usually expressed as \(a + bi\), where \(a\) is the real part and \(bi\) is the imaginary part of the complex number.In our exercise, after simplifying \((6-i)(6+i)\), we reached the expression \(37 + 0i\). This fits the standard form perfectly, where \(a = 37\) and \(b = 0\). Here's a quick breakdown of why standard form matters:- It clearly separates the real and imaginary parts, making it easier to add, subtract, and compare complex numbers.- It is the universally accepted format in mathematics, so using standard form ensures consistency and understanding.Writing answers in standard form provides clarity, especially in contexts where further mathematical operations might be needed.Even though the imaginary part is zero here, it is still usually written as \(0i\) to indicate completeness.
Imaginary Unit
The imaginary unit, denoted by \(i\), is a mathematical concept specifically used with complex numbers. Its defining characteristic is that \(i^2 = -1\). Understanding this is vital when working with expressions involving \(i\).In the difference of squares problem, the presence of \(i\) allows us to explore numbers that go beyond the real number line. \(i\) is not a real number but it behaves in a manner that expands our system of numbers to include complex numbers.Some quick facts about the imaginary unit:
  • Multiplying by \(i\) rotates a number on the complex plane by 90 degrees.
  • The powers of \(i\) cycle every four intervals: \(i, -1, -i, 1\).
  • Using \(i\) simplifies the equation \((6-i)(6+i)\) since \(i^2=-1\), streamlining calculations.
Understanding \(i\) transforms how you approach problems and provides an entirely new dimension to solving equations and expressions.

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Most popular questions from this chapter

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