Problem 24 - Algebra and Trigonometry with Analytic Geometry

Question

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

Step-by-Step Solution

Answer

The real solutions are \( x = 2 \) and \( x = -2 \).

Step 1: 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: \( a^2 - b^2 = (a + b)(a - b) \).

Step 2: 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 \( (x^3 - 8)(x^3 + 8) = 0 \).

Step 3: 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 \( x = root {3} -8 = -2 \).

Step 4: 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.

Key Concepts

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.