Problem 13
Question
Solve. $$ \sqrt[3]{x-2}-3=0 $$
Step-by-Step Solution
Verified Answer
The solution is \(x = 29\).
1Step 1: Isolate the Cube Root
Start by isolating the cube root on the left side of the equation. The given equation is: \[ \sqrt[3]{x-2} - 3 = 0 \]Add 3 to both sides to isolate the cube root:\[ \sqrt[3]{x-2} = 3 \]
2Step 2: Eliminate the Cube Root
To eliminate the cube root, cube both sides of the equation. Cubing both sides gives:\[ (\sqrt[3]{x-2})^3 = 3^3 \]Since cubing a cube root cancels each other out, we have:\[ x - 2 = 27 \]
3Step 3: Solve for x
Now solve for \(x\) by adding 2 to both sides of the equation:\[ x - 2 + 2 = 27 + 2 \]which simplifies to:\[ x = 29 \]
4Step 4: Verify the Solution
Substitute \(x = 29\) back into the original equation to check your work:\[ \sqrt[3]{29-2} - 3 = 0 \]Simplifying inside the cube root:\[ \sqrt[3]{27} - 3 = 0 \]Since \(\sqrt[3]{27} = 3\), the equation holds:\[ 3 - 3 = 0 \].Thus, the solution is correct.
Key Concepts
Cube RootIsolate TermsVerify Solution
Cube Root
When solving equations involving cube roots, it's essential to understand what a cube root is. The cube root of a number is a value that, when multiplied by itself three times, gives the original number. In mathematical notation, it is represented as \( \sqrt[3]{x} \). For example, if \( \sqrt[3]{x} = 3 \), then \( x = 3 \times 3 \times 3 = 27 \).
In the context of solving equations, cube roots can often be found in problems where a term is raised to the power of \/3. To solve such problems, you'll need to "undo" the cubing by taking the cube root. This is akin to how you would take a square root to undo squaring in quadratic equations. Remember, if you have \( \sqrt[3]{a} = b \), then \( a = b^3 \).
Understanding this concept is crucial for isolating and solving equations where terms are raised to any power, especially \/3, where the cube root becomes involved.
In the context of solving equations, cube roots can often be found in problems where a term is raised to the power of \/3. To solve such problems, you'll need to "undo" the cubing by taking the cube root. This is akin to how you would take a square root to undo squaring in quadratic equations. Remember, if you have \( \sqrt[3]{a} = b \), then \( a = b^3 \).
Understanding this concept is crucial for isolating and solving equations where terms are raised to any power, especially \/3, where the cube root becomes involved.
Isolate Terms
Isolating terms in an equation is a foundational step in solving the equation. The goal is to get the variable of interest by itself on one side of the equation. In our example, we started with the equation \( \sqrt[3]{x-2} - 3 = 0 \).
To isolate the cube root term \( \sqrt[3]{x-2} \), we added 3 to both sides of the equation, resulting in \( \sqrt[3]{x-2} = 3 \).
This step is significant as it sets up the equation for solving or simplification in the subsequent steps.
When isolating terms, always perform the same mathematical operation on both sides of the equation. This maintains the equation's balance, similar to keeping both sides of a scale even. Whether you're dealing with addition, subtraction, multiplication, or division, ensuring that operations are consistently applied to both sides is key to solving any equation correctly.
To isolate the cube root term \( \sqrt[3]{x-2} \), we added 3 to both sides of the equation, resulting in \( \sqrt[3]{x-2} = 3 \).
This step is significant as it sets up the equation for solving or simplification in the subsequent steps.
When isolating terms, always perform the same mathematical operation on both sides of the equation. This maintains the equation's balance, similar to keeping both sides of a scale even. Whether you're dealing with addition, subtraction, multiplication, or division, ensuring that operations are consistently applied to both sides is key to solving any equation correctly.
Verify Solution
Verifying a solution is an important step in solving any equation. Once you have found a solution, substitute it back into the original equation to ensure it satisfies all the conditions given. For our exercise, we found that \( x = 29 \) was the solution.
To verify, substitute \( x=29 \) into the original equation: \( \sqrt[3]{29-2} - 3 = 0 \). Simplifying further, we find \( \sqrt[3]{27} - 3 = 0 \). Since \( \sqrt[3]{27} = 3 \), it holds that \( 3 - 3 = 0 \), confirming that our solution is correct.
Verification is crucial because it helps catch any possible errors in calculations or logic that may have occurred. It ensures the solution truly works with the original conditions of the problem. Always remember that a solution isn't truly complete until it has been verified for accuracy.
To verify, substitute \( x=29 \) into the original equation: \( \sqrt[3]{29-2} - 3 = 0 \). Simplifying further, we find \( \sqrt[3]{27} - 3 = 0 \). Since \( \sqrt[3]{27} = 3 \), it holds that \( 3 - 3 = 0 \), confirming that our solution is correct.
Verification is crucial because it helps catch any possible errors in calculations or logic that may have occurred. It ensures the solution truly works with the original conditions of the problem. Always remember that a solution isn't truly complete until it has been verified for accuracy.
Other exercises in this chapter
Problem 12
Use the product rule to multiply. Assume that all variables represent positive real numbers. $$ \sqrt[4]{a b^{2}} \cdot \sqrt[4]{27 a b} $$
View solution Problem 12
Multiply or divide as indicated. $$ \sqrt{-2} \cdot \sqrt{-6} $$
View solution Problem 13
Rationalize each denominator. Assume that all variables represent positive real numbers. \(\frac{3}{\sqrt[3]{2}}\)
View solution Problem 13
Find each square root. Assume that all variables represent nonnegative real numbers. $$ -\sqrt{36} $$
View solution