Problem 42 - Intermediate Algebra

Question

Solve each equation. See Example 5. $$ (2 b+29)^{1 / 3}=3 $$

Step-by-Step Solution

Answer

The solution is \(b = -1\).

Step 1: Isolate the Power

Given the equation \((2b + 29)^{1/3} = 3\), the cube root is already isolated on one side of the equation. There is no need to do any further isolation before moving to the next step.

Step 2: Remove the Cube Root

To eliminate the cube root, cube both sides of the equation. This gives us: \((2b + 29) = 3^3\).

Step 3: Solve for the Expression

Calculate \(3^3\), which is 27. Substitute this value back into the equation: \(2b + 29 = 27\).

Step 4: Isolate the Variable Term

Subtract 29 from both sides to solve for the variable term: \(2b = 27 - 29\), which simplifies to \(2b = -2\).

Step 5: Solve for the Variable

Finally, divide both sides by 2 to solve for \(b\): \(b = \frac{-2}{2} = -1\).

Key Concepts

Cube Roots

Understanding cube roots is an essential part of intermediate algebra. A cube root of a number is a value that, when multiplied by itself twice, gives the original number. For example, the cube root of 27 is 3 because \(3 \times 3 \times 3 = 27\). Cube roots are denoted using the radical symbol \(^3\sqrt{}\) or exponent notation like \(^{1/3}\). In problems, cube roots help simplify equations by reducing powers. When working with cube roots, remember:

• Cubing both sides of an equation helps eliminate the root.
• This operation maintains equality of both sides, which means any transformations are balanced.

Familiarity with cube roots is a powerful tool, especially when paired with techniques like algebraic manipulation.

Isolate Variable

Isolating a variable is fundamental for solving equations. It refers to the process of rearranging an equation to get the variable by itself on one side. This allows for easily finding its value.In the example provided, the main objective was to isolate the variable \(b\). Initially, the cube root of \(2b + 29\) was isolated, readying it for the next step.To isolate variables effectively:

• Perform inverse operations.
• Maintain balance by performing identical operations on both sides of the equation.

In this context, isolating \(b\) means solving for the variable through staged operations, ultimately achieving a straightforward expression that reveals the variable's value.

Algebraic Manipulation

Algebraic manipulation involves using mathematical operations to transform or solve equations. It’s crucial in simplifying complex expressions and making it easier to isolate variables.For instance, after removing the cube root in the exercise, the equation transformed to \(2b + 29 = 27\). By subtracting 29 from both sides, this maintains equation balance while moving one step closer to solving for \(b\). Key algebraic manipulation strategies include:

• Adding or subtracting terms on both sides.
• Multiplying or dividing, which is crucial when isolating the variable.

Mastering algebraic manipulation enhances problem-solving efficiency and accuracy, mainly when tackling equations with multiple steps.

Intermediate Algebra

Intermediate algebra includes concepts like solving equations involving cube roots, handling polynomials, and performing algebraic manipulations. This level serves as a bridge between basic algebra and advanced courses like calculus. At its core:

• It builds upon foundational algebra.
• It includes methods for systematically solving equations using multiple steps.
• It often involves rearranging complex equations into recognizable forms.

Understanding intermediate algebra equips students with the skills to tackle a range of mathematical challenges effectively, preparing them for more advanced studies in mathematics.