Answer
The solution set is $\left\{8\right\}$.
Work Step by Step
Subtract 3 on both sides of the equation to obtain:
$\sqrt{3x+1}=x-3$
Square both sides to obtain:
$3x+1=(x-3)^2$
Use the rule $(a-b)^2 = a^2-2ab+b^2$ to obtain:
$3x+1=x^2-2(x)(3) + 3^2
\\3x+1=x^2-6x+9$
Subtract $3x$ and $1$ on both sides of the equation to obtain:
$\begin{array}{ccc}
&3x+1-3x-1 &= &x^2-6x+9-3x-1
\\&0 &= &x^2-9x+8
\end{array}$
Factor the trinomial to obtain:
$0=(x-8)(x-1)$
Equate each factor to zero, and then solve each equation to obtain:
$\begin{array}{ccc}
&x-8=0 &\text{ or } &x-1=0
\\&x=8 &\text{ or } &x=1
\end{array}$
$1$ is an extraneous solution since it does not satisfy the original equation.
Thus, the solution set is $\left\{8\right\}$.
