Problem 24 Exer. 23-30: Find the solutions ... [FREE SOLUTION]

91影视

Algebra and Trigonometry with Analytic Geometry

Earl Swokowski, Jeffery A. Cole

$Math Studyset 91影视 Explanations$ Math

12 Edition

Chapter 8: Problem 24

Exer. 23-30: Find the solutions of the equation. $$ x^{6}-64=0 $$

Short Answer

Expert verified

The real solutions are $ x = 2 $ and $ x = -2 $.

Step by step solution

Identify the equation type

The given equation is $ x^6 - 64 = 0 $. This is a difference of two perfect squares since $ 64 $ is $ 8^2 $. Therefore, this can be rewritten using the difference of squares formula: $ a^2 - b^2 = (a + b)(a - b) $.

Express given equation as difference of squares

Recognize that $ x^6 - 64 $ can be expressed as $ (x^3)^2 - 8^2 $. Thus, the equation can be rewritten as $ (x^3 - 8)(x^3 + 8) = 0 $.

Solve each factor for its roots

Set each factor equal to zero and solve:1. $ x^3 - 8 = 0 $ gives $ x^3 = 8 $, so $ x = root {3} 8 = 2 $.2. $ x^3 + 8 = 0 $ gives $ x^3 = -8 $, so $ x = root {3} -8 = -2 $.

Consider all real roots

The real solutions from each factor are $ x = 2 $ and $ x = -2 $. These are the real roots of the given equation.

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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 and solving equations. It applies when you have two squared terms separated by a subtraction sign. The formula is:

$ a^2 - b^2 = (a + b)(a - b) $

This formula allows you to break down the equation into simpler multiplicative factors.

Consider the equation $ x^6 - 64 = 0 $. This can be viewed as a difference of squares, where $ a = x^3 $ and $ b = 8 $. Thus, it can be expressed as:

$ (x^3)^2 - 8^2 $

Recognizing this form helps us apply the formula and rewrite it as:

$ (x^3 - 8)(x^3 + 8) = 0 $

By factoring this way, we simplify the task of finding the roots of the equation as solving two simpler equations instead of one complex polynomial.

Polynomial Roots

Finding the roots of a polynomial is equivalent to finding where the polynomial equals zero. For our equation, once we employ the difference of squares, we find:

First Factor: $ x^3 - 8 = 0 $
Second Factor: $ x^3 + 8 = 0 $

To find the roots, or solutions, set each factor to zero and solve individually. Solving $ x^3 - 8 = 0 $ involves finding the cube root of 8:

$ x = \sqrt[3]{8} = 2 $

Similarly, for $ x^3 + 8 = 0 $, taking the cube root of $-8$ yields:

$ x = \sqrt[3]{-8} = -2 $

By solving these simpler cubic equations, we determine that the polynomial has real roots of 2 and -2. These are the solutions to the original polynomial equation.

Cubic Equations

Cubic equations are polynomial equations of degree three, with the general form $ ax^3 + bx^2 + cx + d = 0 $. In our exercise, both factors $ x^3 - 8 $ and $ x^3 + 8 $ are cubic equations:

$ x^3 = 8 $ reduces to $ x = \sqrt[3]{8} $
$ x^3 = -8 $ reduces to $ x = \sqrt[3]{-8} $

The solutions to these equations come from applying basic algebraic operations like taking the cube root. Cubic equations can often yield real and/or complex roots, but in this case, we have two real solutions: 2 and -2.

These solutions indicate where the original polynomial crosses the x-axis, a clear visual representation of its roots. Understanding the nature of cubic equations allows us to solve higher-degree polynomials by reducing them to simpler, solvable components.

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91影视

Exer. 23-30: Find the solutions of the equation. $$ x^{6}-64=0 $$

Short Answer

Step by step solution

Identify the equation type

Express given equation as difference of squares

Solve each factor for its roots

Consider all real roots

Key Concepts

Difference of Squares

Polynomial Roots

Cubic Equations

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