Solve the following radical equation \$ 2\sqrt(3x +1) + 4 = 8\$.

The given radical equation is

\$ 2\sqrt(3x +1) + 4 = 8\$

Move constant tersm to right side

\$2\sqrt(3x + 1) = 8 - 4\$

\$2\sqrt(3x + 1) = 4\$

Divide both sides by 2 to isolate the square root:

\$2/2 \sqrt(3x +1) = \cancel(4)^2 /\cancel(2)\$

\$ \sqrt(3x +1) = 2\$

Square both sides to eliminate the square root:

\$(\sqrt(3x +1))^2 = 2^2\$

3x +1 = 4

Solve for x:

3x = 4 - 1

\$ x = 3/3\$

x = 1

Therefore, the solution to the equation \$ 2\sqrt(3x +1) + 4 = 8\$ is 1.

Since \$\sqrt(7)\$ is not an integer, \$2\sqrt(7) + 4\$ is not equal to 8. Therefore, x = 2 does not satisfy the original equation, making option A incorrect

x = 2

The left side \$2\sqrt(10) + 4\$ does not equal the right side 8, so x = 3 is not a solution to the equation

x = 3

x = 1 is correct because it satisfies the original equation

x = 1

Since \$\sqrt(13)\$ is not a rational number and certainly not equal to 2, it is clear that \$2\sqrt(13) +4 ≠ 8\$. Thus, x = 4 does not satisfy the original equation, confirming that it it wrong

x = 4

Option

C

Can you summarize what you’ve understood in the above steps?