Answer
$x=2$
Work Step by Step
The provided radical equation is $\sqrt{x+7}-1=x$.
Isolate the term $\sqrt{x+7}$ by adding $1$ to both sides of the equation. This gives
$\begin{align}
& \sqrt{x+7}-1+1=x+1 \\
& \sqrt{x+7}=x+1
\end{align}$
Square both sides of the equation.
${{\left( \sqrt{x+7} \right)}^{2}}={{\left( x+1 \right)}^{2}}$
Use the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ to solve the above equation further as,
$\begin{align}
& x+7={{x}^{2}}+2\times x\times 1+{{\left( 1 \right)}^{2}} \\
& x+7={{x}^{2}}+2x+1
\end{align}$
Combine all the like terms and solve as,
$\begin{align}
& x+7={{x}^{2}}+2x+1 \\
& {{x}^{2}}+2x+1-x-7=0 \\
& {{x}^{2}}+x-6=0
\end{align}$
Then writing the factors as,
$\begin{align}
& {{x}^{2}}+3x-2x-6=0 \\
& \left( x+3 \right)\left( x-2 \right)=0
\end{align}$
Equating each factor to 0 and solving further,
$x=2\text{ or -3}$
To check the solutions:
Substituting $x=2$ in the equation $\sqrt{x+7}-1=x$
$\begin{matrix}
\sqrt{x+7}-1=x \\
\sqrt{2+7}-1\overset{?}{\mathop{=}}\,2 \\
3-1\overset{?}{\mathop{=}}\,2 \\
2=2,\text{ true} \\
\end{matrix}$
Now, substituting $x=-3$ in the equation $\sqrt{x+7}-1=x$
$\begin{matrix}
\sqrt{x+7}-1=x \\
\sqrt{-3+7}-1\overset{?}{\mathop{=}}\,-3 \\
2-1\overset{?}{\mathop{=}}\,-3 \\
1\ne -3,\text{ False} \\
\end{matrix}$
Thus, $x=2$ is the right solution for the equation $\sqrt{x+7}-1=x$.
