Question

solve.
\\(\sqrt4{3x + 2} - 3 = 0\\)

select the correct choice below and fill in any answer boxes present in your choice.

\\(\bigcirc\\) a. the solution(s) is(are) \\(x = \square\\).
(type an integer or a simplified fraction. use a comma to separate answers as needed.)
\\(\bigcirc\\) b. there is no solution.

Explanation:

Step1: Isolate the radical term

Add 3 to both sides of the equation \(\sqrt[4]{3x + 2}-3 = 0\) to get \(\sqrt[4]{3x + 2}=3\).

Step2: Raise both sides to the 4th power

Raise each side of the equation \(\sqrt[4]{3x + 2}=3\) to the 4th power. Using the property \((\sqrt[n]{a})^n=a\) (for \(n\) even, we assume \(a\geq0\) which we will check later), we have \((\sqrt[4]{3x + 2})^4 = 3^4\), so \(3x+2 = 81\).

Step3: Solve for x

Subtract 2 from both sides of \(3x + 2=81\): \(3x=81 - 2=79\). Then divide both sides by 3: \(x=\frac{79}{3}\).

Step4: Check the solution

Substitute \(x = \frac{79}{3}\) into the original equation. First, calculate \(3x+2\): \(3\times\frac{79}{3}+2=79 + 2=81\). Then \(\sqrt[4]{81}-3=\sqrt[4]{3^4}-3 = 3-3 = 0\), which satisfies the original equation.

Answer:

A. The solution(s) is(are) \(x=\frac{79}{3}\)