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Squaring both sides gives \(\displaystyle x^{2} + 8 x + 16 = 6 x + 64\). The equation is quadratic setting it equal to zero gives \(\displaystyle x^{2} + 2 x - 48 = 0\). Factoring gives \(\displaystyle (x - 6)(x + 8)=0\) so the possible solutions are \(\displaystyle x = 6\) and \(\displaystyle x = -8\). Checking the solution \(\displaystyle x = 6\) in the original equation gives \(\displaystyle 10 = 10\). The solution checks, so \(\displaystyle x = 6\) is a true solution. Checking the solution \(\displaystyle x = -8\) in the original equation gives \(\displaystyle -4 = 4\). The solution does no check, so \(\displaystyle x = -8\) is an extraneous solution.

Download \(\LaTeX\)

\begin{question}Solve $x + 4 = \sqrt{6 x + 64}$.
    \soln{10cm}{Squaring both sides gives $x^{2} + 8 x + 16 = 6 x + 64$. The equation is quadratic setting it equal to zero gives $x^{2} + 2 x - 48 = 0$. Factoring gives $(x - 6)(x + 8)=0$ so the possible solutions are $x = 6$ and $x = -8$. Checking the solution $x = 6$ in the original equation gives $10 = 10$. The solution checks, so $x = 6$ is a true solution. Checking the solution $x = -8$ in the original equation gives $-4 = 4$. The solution does no check, so $x = -8$ is an extraneous solution. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)

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\usepackage{amsmath}
\usepackage[margin=2cm]{geometry}
\usepackage{tcolorbox}

\newcounter{ExamNumber}
\newcounter{questioncount}
\stepcounter{questioncount}

\newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}}
\renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}}

\newif\ifShowSolution
\newcommand{\soln}[2]{%
\ifShowSolution%
\noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else%
\vspace{#1}%
\fi%
}%
\newcommand{\hideifShowSolution}[1]{%
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%
\else%
#1%
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}%
\everymath{\displaystyle}
\ShowSolutiontrue

\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

\end{question}\end{document}

HTML for Canvas

<p> <p>Solve  <img class="equation_image" title=" \displaystyle x + 4 = \sqrt{6 x + 64} " src="/equation_images/%20%5Cdisplaystyle%20x%20%2B%204%20%3D%20%5Csqrt%7B6%20x%20%2B%2064%7D%20" alt="LaTeX:  \displaystyle x + 4 = \sqrt{6 x + 64} " data-equation-content=" \displaystyle x + 4 = \sqrt{6 x + 64} " /> .</p> </p>

HTML for Canvas

<p> <p>Squaring both sides gives  <img class="equation_image" title=" \displaystyle x^{2} + 8 x + 16 = 6 x + 64 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%208%20x%20%2B%2016%20%3D%206%20x%20%2B%2064%20" alt="LaTeX:  \displaystyle x^{2} + 8 x + 16 = 6 x + 64 " data-equation-content=" \displaystyle x^{2} + 8 x + 16 = 6 x + 64 " /> . The equation is quadratic setting it equal to zero gives  <img class="equation_image" title=" \displaystyle x^{2} + 2 x - 48 = 0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%202%20x%20-%2048%20%3D%200%20" alt="LaTeX:  \displaystyle x^{2} + 2 x - 48 = 0 " data-equation-content=" \displaystyle x^{2} + 2 x - 48 = 0 " /> . Factoring gives  <img class="equation_image" title=" \displaystyle (x - 6)(x + 8)=0 " src="/equation_images/%20%5Cdisplaystyle%20%28x%20-%206%29%28x%20%2B%208%29%3D0%20" alt="LaTeX:  \displaystyle (x - 6)(x + 8)=0 " data-equation-content=" \displaystyle (x - 6)(x + 8)=0 " />  so the possible solutions are  <img class="equation_image" title=" \displaystyle x = 6 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%206%20" alt="LaTeX:  \displaystyle x = 6 " data-equation-content=" \displaystyle x = 6 " />  and  <img class="equation_image" title=" \displaystyle x = -8 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-8%20" alt="LaTeX:  \displaystyle x = -8 " data-equation-content=" \displaystyle x = -8 " /> . Checking the solution  <img class="equation_image" title=" \displaystyle x = 6 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%206%20" alt="LaTeX:  \displaystyle x = 6 " data-equation-content=" \displaystyle x = 6 " />  in the original equation gives  <img class="equation_image" title=" \displaystyle 10 = 10 " src="/equation_images/%20%5Cdisplaystyle%2010%20%3D%2010%20" alt="LaTeX:  \displaystyle 10 = 10 " data-equation-content=" \displaystyle 10 = 10 " /> . The solution checks, so  <img class="equation_image" title=" \displaystyle x = 6 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%206%20" alt="LaTeX:  \displaystyle x = 6 " data-equation-content=" \displaystyle x = 6 " />  is a true solution. Checking the solution  <img class="equation_image" title=" \displaystyle x = -8 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-8%20" alt="LaTeX:  \displaystyle x = -8 " data-equation-content=" \displaystyle x = -8 " />  in the original equation gives  <img class="equation_image" title=" \displaystyle -4 = 4 " src="/equation_images/%20%5Cdisplaystyle%20-4%20%3D%204%20" alt="LaTeX:  \displaystyle -4 = 4 " data-equation-content=" \displaystyle -4 = 4 " /> . The solution does no check, so  <img class="equation_image" title=" \displaystyle x = -8 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-8%20" alt="LaTeX:  \displaystyle x = -8 " data-equation-content=" \displaystyle x = -8 " />  is an extraneous solution. </p> </p>