Question

question 3
solve the radical equation, and check all proposed solutions.
\\(\sqrt{3x - 2} = 2\\)
\\(\boldsymbol{\left\\{\frac{2}{3}\
ight\\}}\\)
\\(\boldsymbol{\emptyset}\\)
\\(\boldsymbol{\\{4\\}}\\)
\\(\boldsymbol{\\{2\\}}\\)

Explanation:

Step1: Square both sides to eliminate the square root

To solve the radical equation \(\sqrt{3x - 2} = 2\), we first square both sides of the equation. Squaring the left side \(\sqrt{3x - 2}\) will give us \(3x - 2\) (since \((\sqrt{a})^2=a\) for \(a\geq0\)), and squaring the right side \(2\) gives \(2^2 = 4\). So we have the equation:
\[
(\sqrt{3x - 2})^2=2^2
\]
\[
3x - 2 = 4
\]

Step2: Solve for \(x\)

Now we solve the linear equation \(3x - 2 = 4\) for \(x\). First, we add \(2\) to both sides of the equation:
\[
3x-2 + 2=4 + 2
\]
\[
3x=6
\]
Then, we divide both sides by \(3\):
\[
x=\frac{6}{3}=2
\]

Step3: Check the solution

We need to check if \(x = 2\) is a valid solution by substituting it back into the original equation \(\sqrt{3x - 2}\).
Substitute \(x = 2\) into the left - hand side: \(\sqrt{3(2)-2}=\sqrt{6 - 2}=\sqrt{4}=2\), which is equal to the right - hand side of the original equation.

Answer:

The solution to the equation \(\sqrt{3x - 2}=2\) is \(x = 2\), so the answer is \(\{2\}\) (the option with \(\{2\}\)).