Answer
$x=-\sqrt{99} $, $x=\sqrt{99} $
Work Step by Step
Given\begin{equation}
\sqrt{x^2+1}=10.
\end{equation} Let's take a look at the student's solution and see where she went wrong and make the necessary corrections.
\begin{equation}
\begin{aligned}
\sqrt{x^2+1} & =10 \\
x+1 & =10 \\
x & =9.
\end{aligned}
\end{equation} The first mistake is that the student did not square both sides to get rid of the radical sign on the left. Second, he simply wrote down the wrong radicand, $x+1$ which should be $x^2+1$ if he had squared both sides as required.
Here is how the radical equation should be solved.
\begin{equation}
\begin{aligned}
\sqrt{x^2+1} & =10 \\\
\left( \sqrt{x^2+1} \right)^2 & =\left(10 \right) \\
x^2+1& =100 \\
x^2& =100-1 \\
x^2& =99\\
x& = \pm\sqrt{99}\\
\implies x&= -\sqrt{99}\\
x&= \sqrt{99}.
\end{aligned}
\end{equation} Check.
\begin{equation}
\begin{aligned}
\sqrt{\left( -\sqrt{99} \right)^2+1}& \stackrel{?}{=}10 \\
10& =10\quad \textbf{True}\\
\sqrt{\left( \sqrt{99} \right)^2+1}& \stackrel{?}{=}10 \\
10& =10\quad \textbf{True}
\end{aligned}
\end{equation} The solution is $x=-\sqrt{99} $, $x=\sqrt{99} $.
