Problem 71 - Algebra for College Students

Question

Set up an equation and solve each of the following problems. The cube of a number equals nine times the same number. Find the number.

Step-by-Step Solution

Answer

The possible numbers are 0, 3, and -3.

Step 1: Define the Variable

Let the unknown number be represented by the variable \( x \). Our goal is to find the value of \( x \) that satisfies the given condition.

Step 2: Translate the Problem into an Equation

The problem states, "The cube of a number equals nine times the same number." We can translate this into the equation: \( x^3 = 9x \).

Step 3: Simplify the Equation

Rearrange the equation \( x^3 = 9x \) to bring all terms to one side, resulting in: \( x^3 - 9x = 0 \).

Step 4: Factor the Equation

Factor the equation \( x^3 - 9x = 0 \) to find the solutions. Take \( x \) as a common factor: \( x(x^2 - 9) = 0 \).

Step 5: Solve for Zero Products

To solve \( x(x^2 - 9) = 0 \), set each factor equal to zero: \( x = 0 \) or \( x^2 - 9 = 0 \).

Step 6: Solve the Quadratic Equation

The equation \( x^2 - 9 = 0 \) can be written as \( (x - 3)(x + 3) = 0 \). Solving for \( x \), we find: \( x = 3 \) or \( x = -3 \).

Step 7: Write the Final Solutions

The possible values of \( x \) are \( 0 \,\), \( 3 \,\) and \( -3 \). Therefore, the solutions to the problem are: \( x = 0, x = 3, \text{ or } x = -3 \).

Key Concepts

Factoring Equations

Factoring equations is a pivotal skill in algebra that simplifies expressions and helps solve equations. The essence of factoring is breaking down a complex expression into simpler, multiply-able components. These components are called 'factors'.

In terms of solving equations, factoring is used to replace a cumbersome expression with a product of simpler terms. This strategy is beneficial when solving polynomial equations. For example, in the given problem, we encounter the equation:

• \( x^3 - 9x = 0 \).

The first step in factoring this equation is to extract a common factor from the terms. Here, \( x \) is a common factor, resulting in:

• \( x(x^2 - 9) = 0 \).

Further factoring \( x^2 - 9 \) can be achieved by recognizing it as a difference of squares, leading to:

• \( (x - 3)(x + 3) = 0 \).

By breaking down the equation into these factors, we can employ the zero product property and solve for the variable by setting each factor to zero, determining all possible solutions.

Quadratic Equations

Quadratic equations are polynomial equations of the second degree. They take the standard form:

• \( ax^2 + bx + c = 0 \),

where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). A unique characteristic of quadratic equations is that they can have two, one, or no real solutions.

To solve them, one can often employ methods such as factoring, completing the square, or using the quadratic formula. In our exercise, we reduced the problem to solving a related quadratic equation:

• \( x^2 - 9 = 0 \).

Recognizing this equation as a difference of squares, it factors into \( (x - 3)(x + 3) = 0 \). This technique uncovers the values of \( x \) that satisfy the original polynomial equation. Solutions are then found by setting each factor equal to zero:

• \( x - 3 = 0 \), giving \( x = 3 \),
• \( x + 3 = 0 \), resulting in \( x = -3 \).

Utilizing these straightforward solutions provides insight and control over quadratic equations.

Problem Solving in Algebra

Problem solving in algebra is like piecing together the clues to find a solution, much like solving a puzzle. The approach often includes defining a variable to represent the unknown, translating a word problem into an equation, and strategically using algebraic techniques to solve that equation.

In the problem statement given, our objective was to determine the unknown number whose cube equals nine times the number itself. By setting up the equation \( x^3 = 9x \), we reformulated the word statement into an algebraic form. This crucial step is all about changing words into a mathematical expression that can be manipulated.

Next, using algebraic techniques like rearranging terms and factoring, we simplify the problem to make it manageable. These steps are followed by solving the resulting simpler equations:

• First, factor and reduce the problem: \( x(x^2 - 9) = 0 \).
• Then solve each factor using zero product property: \( x = 0 \), \( x = 3 \), or \( x = -3 \).

Ultimately, problem solving in algebra requires a blend of creativity, logic, and proficiency in algebraic concepts, ensuring that solutions are not only found but understood.