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Chapter 10: Problem 18

Solve each radical equation. $$\sqrt[3]{6 x-3}=3$$

Short Answer

Expert verified
The solution to the equation is \( x = 5 \)

Step by step solution

01

Isolate the cube root

First, we will need to isolate the cube root on one side of the equation. In this case, it is already isolated, so we can proceed directly to the next step.
02

Eliminate the cube root

To eliminate the cube root from \(\sqrt[3]{6 x-3}\), we must cube both sides of the equation. This gives \( (\sqrt[3]{6 x-3})^3 = 3^3\). Which simplifies to \( 6 x-3 = 27\).
03

Solve for x

Next, solve for x by transposing the equation and dividing by 6. Adding 3 to both sides of the equation gives \( 6x = 27 + 3 = 30\). Dividing both sides by 6 results in \( x = \frac{30}{6}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cube Root
Understanding cube roots is essential for working with radical equations. A cube root is the number that, when multiplied by itself three times, gives the original number. For example, if \(x^3 = a\), then \(\sqrt[3]{a} = x\).
Cube roots can simplify equations involving radicals, like the equation given, \(\sqrt[3]{6x-3} = 3\).
To solve such equations, our goal is to remove the cube root so that the variable (\(x\)) can be isolated.
Isolation of Variables
Isolating variables is a crucial step in solving equations. Here, the cube root term, \(\sqrt[3]{6x-3}\), is already on one side of the equation. This means we've isolated the term from any other numbers or variables.
Isolation involves rearranging the equation to place the variable term separately. In equations where the cube root is not isolated initially, you may need to add, subtract, multiply, or divide both sides of the equation to achieve isolation.
This simplification helps in performing other operations, such as removing radicals or solving for variables.
Equation Solving
Solving equations involves methodically working through a series of steps:
  • Cubing both sides to eliminate the cube root: \(\bigg(\sqrt[3]{6x-3}\bigg)^3 = 3^3\).
  • This simplifies to \(6x-3 = 27\).
By performing these steps correctly, the equation becomes simpler to work with, allowing you to solve for the unknown variable.
Once the equation is free of radicals, you can use arithmetic operations to isolate the variable further, as seen in the final solution of this exercise.
Algebraic Manipulation
Algebraic manipulation is the art of reshaping equations. It allows you to isolate variables and solve for them. In our equation, once \(6x - 3 = 27\) is attained, additional steps are used to solve for \(x\):
  • Add 3 to both sides to get \(6x = 30\).
  • Divide both sides by 6 to solve for \(x\), resulting in \(x = 5\).
This process demonstrates how algebraic manipulation can simplify complex expressions and find solutions effectively. It underscores the importance of structure and sequence in solving algebraic equations.

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Most popular questions from this chapter

In Exercises \(93-104\), rationalize each numerator. Simplify, if possible. $$\frac{\sqrt{x}+\sqrt{y}}{x^{2}-y^{2}}$$

Use a graphing utility to solve each radical equation. Graph each side of the equation in the given viewing rectangle. The equation's solution set is given by the \(x\) -coordinate(s) of the point (s) of intersection. Check by substitution. $$\begin{aligned} &\sqrt{2 x+2}=\sqrt{3 x-5}\\\ &[-1,10,1] \text { by }|-1,5,1| \end{aligned}$$

What is an extraneous solution to a radical equation?

Solve: \(7[(2 x-5)-(x+1)]=(\sqrt{7}+2)(\sqrt{7}-2)\)

Determine whether each statement "makes sense" or "does not make sense" and explain your reasoning. Now that I know how to solve radical equations, I can use models that are radical functions to determine the value of the independent variable when a function value is known.
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