Answer
The solution set of the equation is\[\left\{ 6 \right\}\].
Work Step by Step
Consider the provided radical equation:
$\sqrt{3x+7}+1=x$.
Re-write the above expression as,
$\sqrt{3x+7}=x-1$
Now take the square root of both sides,
$\begin{align}
& {{\left( \sqrt{3x+7} \right)}^{2}}={{\left( x-1 \right)}^{2}} \\
& 3x+7={{x}^{2}}-2x+1 \\
& {{x}^{2}}-2x-3x+1-7=0 \\
& {{x}^{2}}-5x-6=0
\end{align}$
Now factor:
$\begin{align}
& {{x}^{2}}-6x+x-6=0 \\
& x\left( x-6 \right)+1\left( x-6 \right)=0 \\
& \left( x+1 \right)\left( x-6 \right)=0
\end{align}$
Then equate each term to zero,
$\begin{align}
& x+1=0 \\
& x=-1
\end{align}$
And,
$\begin{align}
& x-6=0 \\
& x=6
\end{align}$
Therefore, the possible set of solutions is $\left\{ -1,6 \right\}$.
Check:
Now, put $x=-1$ in the provided equation and verify,
$\begin{align}
\sqrt{3\left( -1 \right)+7}+1\overset{?}{\mathop{=}}\,-1 & \\
\sqrt{-3+7}+1\overset{?}{\mathop{=}}\,-1 & \\
\sqrt{4}+1\overset{?}{\mathop{=}}\,-1 & \\
2+1\overset{?}{\mathop{=}}\,-1 & \\
\end{align}$
Further, simplify
$\begin{align}
2+1\overset{?}{\mathop{=}}\,-1 & \\
3\ne -1 & \\
\end{align}$
Which is not true.
So, this is not a solution.
Now, put $x=6$ in the provided equation and verify,
$\begin{align}
\sqrt{3\left( 6 \right)+7}+1\overset{?}{\mathop{=}}\,6 & \\
\sqrt{18+7}+1\overset{?}{\mathop{=}}\,6 & \\
\sqrt{25}+1\overset{?}{\mathop{=}}\,6 & \\
5+1\overset{?}{\mathop{=}}\,6 & \\
\end{align}$
Further, simplify
$\begin{align}
5+1\overset{?}{\mathop{=}}\,6 & \\
6=6 & \\
\end{align}$
Which is satisfied in the given equation. So, this is a solution.
Hence, the solution set of the provided equation is $\left\{ 6 \right\}$.
