Problem 43 - Intermediate Algebra

Question

See Example 6. Let \(f(x)=\sqrt[4]{3 x+1}\). For what value(s) of \(x\) is \(f(x)=4 ?\)

Step-by-Step Solution

Answer

The value of \(x\) is 85.

Step 1: Set up the Equation

We are given the function \(f(x) = \sqrt[4]{3x + 1}\). To find the values of \(x\) where \(f(x) = 4\), we set up the equation: \(\sqrt[4]{3x + 1} = 4\).

Step 2: Remove the Fourth Root

To eliminate the fourth root, raise both sides of the equation to the power of 4. This gives us \((\sqrt[4]{3x + 1})^4 = 4^4\), simplifying to \(3x + 1 = 256\).

Step 3: Solve for x

Subtract 1 from both sides of the equation: \(3x = 256 - 1\), which simplifies to \(3x = 255\). Then, divide both sides by 3 to solve for \(x\): \(x = \frac{255}{3}\), simplifying to \(x = 85\).

Step 4: Verify the Solution

Substitute \(x = 85\) back into the original function to verify: \(f(85) = \sqrt[4]{3(85) + 1} = \sqrt[4]{255 + 1} = \sqrt[4]{256} = 4\), which matches \(f(x) = 4\). Therefore, the solution is correct.

Key Concepts

Fourth Roots

When solving equations involving fourth roots, understanding the concept of fourth roots is key. A fourth root of a number is a value that, when raised to the power of four, gives the original number back. For instance, if you take the number 16, the fourth root is 2 because

• Raising 2 to the power of 4 (2 \(^4\)) gives you 16.
• Mathematically, if you have a number 'a', the fourth root is represented as \( \sqrt[4]{a} \).

To find the fourth root of an expression such as \(\sqrt[4]{3x + 1}\), you need to solve for 'x' in such a way that when the expression is raised to the fourth power, it equals another number. In our example, raising both sides of the equation to the fourth power was necessary to eliminate the fourth root. This essential step allows us to work with a polynomial equation instead.

Verifying Solutions

Verification is like double-checking your work in math. After finding a solution, it's important to plug it back into your original equation to ensure it truly satisfies the given conditions. For instance, in our solution process with the equation \(\sqrt[4]{3x + 1} = 4\):

• We found that \(x = 85\) was the solution.
• To verify, substitute \(x = 85\) back into the function to see if it holds: \(f(85) = \sqrt[4]{3(85) + 1} = \sqrt[4]{256}\).
• Since \(\sqrt[4]{256} = 4\), it matches the right side of the original equation \(f(x) = 4\).

Verification not only ensures accuracy, but also boosts your confidence in the solution. This process reinforces the correctness of how you solved the problem, ensuring that all manipulations and assumptions were correct.

Algebraic Manipulation

Algebraic manipulation is a powerful tool that enables us to change the form of an equation in order to solve it. Here, we used it to solve for 'x' in the equation \(\sqrt[4]{3x + 1} = 4\). Let me walk you through the main steps:

• Raising each side of the equation to the fourth power eliminated the radical (fourth root) and simplified our task.
• We ended up with the equation \(3x + 1 = 256\) after raising the fourth root to the power of 4.
• Next, we performed standard algebraic steps by subtracting 1 from both sides to isolate the term with 'x'. This is a common technique to simplify equations.
• Lastly, we divided by the coefficient of 'x', which was 3, to finally solve for 'x'. Hence, \(x = 85\) was our solution.

Algebraic manipulation allows us to re-arrange terms, clear fractions, combine like terms, and ultimately make the steps straightforward to find our solutions in an organized and efficient way.