Problem 79 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt[3]{x+8}=-2 $$

Step-by-Step Solution

Answer

The solution is \( x = -16 \).

Step 1: Understanding the Equation

We start with the equation \( \sqrt[3]{x+8} = -2 \). This equation involves a cube root and our goal is to solve for \( x \). We need to find values of \( x \) that satisfy this equation.

Step 2: Isolating the Cube Root

The cube root \( \sqrt[3]{x+8} \) is already isolated on the left side. So, we can move on to eliminating the cube root to solve for \( x \).

Step 3: Eliminating the Cube Root

To eliminate the cube root, raise both sides of the equation to the power of 3. This means we will have: \[ (\sqrt[3]{x+8})^3 = (-2)^3 \].

Step 4: Simplifying Both Sides

Simplify both sides of the equation: - The left side becomes \( x+8 \).- The right side, \( (-2)^3 \), simplifies to \( -8 \). Thus, the equation becomes \( x + 8 = -8 \).

Step 5: Solving for x

Subtract 8 from both sides of the equation to solve for \( x \): \[ x + 8 - 8 = -8 - 8 \] This simplifies to \( x = -16 \).

Step 6: Checking for Extraneous Solutions

Substitute \( x = -16 \) back into the original equation to check: \( \sqrt[3]{-16 + 8} = \sqrt[3]{-8} = -2 \). The original equation is satisfied, so \( x = -16 \) is not an extraneous solution. It is the correct solution.

Key Concepts

Extraneous Solutions

In mathematics, an extraneous solution is a solution derived from the process of solving an equation that does not satisfy the original equation. This often occurs when dealing with equations involving radicals or rational expressions.
When solving equations, especially those involving cube roots, it is crucial to verify each solution by substituting it back into the original equation.
Extraneous solutions can sometimes appear during the algebraic manipulations needed to isolate variables or eliminate complex expressions. This is why the final step in solving equations, as shown in our solution with the equation \( \sqrt[3]{x+8} = -2 \), involves checking potential solutions to confirm their validity.In our example, the solution \( x = -16 \) satisfies the original equation, meaning it is not extraneous. Always make it a habit to check solutions, as this ensures accuracy and helps avoid rushed mistakes.

Solving Equations

Solving equations involves finding the values of variables that make the equation true. It often requires several steps of manipulation using algebraic principles.In our given exercise with \( \sqrt[3]{x+8} = -2 \), solving the equation primarily involved removing the cube root to make it simpler:

• First, realize that the cube root was already isolated.
• Then, eliminate the cube root by raising both sides to the power of 3.

This simplification transforms the equation into a more straightforward linear form \( x + 8 = -8 \). Solving equations like this often starts with isolating expressions and progressively working to simplify until you can find the variable's value.Remember that the approach to solving equations can vary depending on the complexity and style of the equation, but the fundamental steps—isolating, simplifying, and solving—remain consistent.

Isolating Variables

Isolating the variable, typically \( x \), is a central step in solving equations. It involves making the variable the subject of the formula so it stands alone on one side of the equation.In our cube root equation \( \sqrt[3]{x+8} = -2 \), the expression \( x+8 \) is nested within a cube root. By raising both sides to the power of 3, the cube root is "canceled out," thereby isolating \( x + 8 \) directly.The strategy for isolating a variable varies depending on the equation's components:

• With linear equations, this might involve simple addition or subtraction.
• With roots, powers, or fractions, neutralizing them using inverse operations is necessary.

After isolating \( x \), solving becomes straightforward, often just requiring one more arithmetic step. This critical maneuver simplifies solving complex-looking equations by focusing on the variable of interest and the operations needed to isolate it.