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Squaring both sides gives \(\displaystyle x^{2} + 6 x + 9 = 5 x + 39\). The equation is quadratic setting it equal to zero gives \(\displaystyle x^{2} + x - 30 = 0\). Factoring gives \(\displaystyle (x - 5)(x + 6)=0\) so the possible solutions are \(\displaystyle x = 5\) and \(\displaystyle x = -6\). Checking the solution \(\displaystyle x = 5\) in the original equation gives \(\displaystyle 8 = 8\). The solution checks, so \(\displaystyle x = 5\) is a true solution. Checking the solution \(\displaystyle x = -6\) in the original equation gives \(\displaystyle -3 = 3\). The solution does no check, so \(\displaystyle x = -6\) is an extraneous solution.

Download \(\LaTeX\)

\begin{question}Solve $x + 3 = \sqrt{5 x + 39}$.
    \soln{10cm}{Squaring both sides gives $x^{2} + 6 x + 9 = 5 x + 39$. The equation is quadratic setting it equal to zero gives $x^{2} + x - 30 = 0$. Factoring gives $(x - 5)(x + 6)=0$ so the possible solutions are $x = 5$ and $x = -6$. Checking the solution $x = 5$ in the original equation gives $8 = 8$. The solution checks, so $x = 5$ is a true solution. Checking the solution $x = -6$ in the original equation gives $-3 = 3$. The solution does no check, so $x = -6$ 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%
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\newcommand{\hideifShowSolution}[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 + 3 = \sqrt{5 x + 39} " src="/equation_images/%20%5Cdisplaystyle%20x%20%2B%203%20%3D%20%5Csqrt%7B5%20x%20%2B%2039%7D%20" alt="LaTeX:  \displaystyle x + 3 = \sqrt{5 x + 39} " data-equation-content=" \displaystyle x + 3 = \sqrt{5 x + 39} " /> .</p> </p>

HTML for Canvas

<p> <p>Squaring both sides gives  <img class="equation_image" title=" \displaystyle x^{2} + 6 x + 9 = 5 x + 39 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%206%20x%20%2B%209%20%3D%205%20x%20%2B%2039%20" alt="LaTeX:  \displaystyle x^{2} + 6 x + 9 = 5 x + 39 " data-equation-content=" \displaystyle x^{2} + 6 x + 9 = 5 x + 39 " /> . The equation is quadratic setting it equal to zero gives  <img class="equation_image" title=" \displaystyle x^{2} + x - 30 = 0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%20x%20-%2030%20%3D%200%20" alt="LaTeX:  \displaystyle x^{2} + x - 30 = 0 " data-equation-content=" \displaystyle x^{2} + x - 30 = 0 " /> . Factoring gives  <img class="equation_image" title=" \displaystyle (x - 5)(x + 6)=0 " src="/equation_images/%20%5Cdisplaystyle%20%28x%20-%205%29%28x%20%2B%206%29%3D0%20" alt="LaTeX:  \displaystyle (x - 5)(x + 6)=0 " data-equation-content=" \displaystyle (x - 5)(x + 6)=0 " />  so the possible solutions are  <img class="equation_image" title=" \displaystyle x = 5 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%205%20" alt="LaTeX:  \displaystyle x = 5 " data-equation-content=" \displaystyle x = 5 " />  and  <img class="equation_image" title=" \displaystyle x = -6 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-6%20" alt="LaTeX:  \displaystyle x = -6 " data-equation-content=" \displaystyle x = -6 " /> . Checking the solution  <img class="equation_image" title=" \displaystyle x = 5 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%205%20" alt="LaTeX:  \displaystyle x = 5 " data-equation-content=" \displaystyle x = 5 " />  in the original equation gives  <img class="equation_image" title=" \displaystyle 8 = 8 " src="/equation_images/%20%5Cdisplaystyle%208%20%3D%208%20" alt="LaTeX:  \displaystyle 8 = 8 " data-equation-content=" \displaystyle 8 = 8 " /> . The solution checks, so  <img class="equation_image" title=" \displaystyle x = 5 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%205%20" alt="LaTeX:  \displaystyle x = 5 " data-equation-content=" \displaystyle x = 5 " />  is a true solution. Checking the solution  <img class="equation_image" title=" \displaystyle x = -6 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-6%20" alt="LaTeX:  \displaystyle x = -6 " data-equation-content=" \displaystyle x = -6 " />  in the original equation gives  <img class="equation_image" title=" \displaystyle -3 = 3 " src="/equation_images/%20%5Cdisplaystyle%20-3%20%3D%203%20" alt="LaTeX:  \displaystyle -3 = 3 " data-equation-content=" \displaystyle -3 = 3 " /> . The solution does no check, so  <img class="equation_image" title=" \displaystyle x = -6 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-6%20" alt="LaTeX:  \displaystyle x = -6 " data-equation-content=" \displaystyle x = -6 " />  is an extraneous solution. </p> </p>