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Squaring both sides gives \(\displaystyle x^{2} + 2 x + 1 = 4 x + 9\). The equation is quadratic setting it equal to zero gives \(\displaystyle x^{2} - 2 x - 8 = 0\). Factoring gives \(\displaystyle (x - 4)(x + 2)=0\) so the possible solutions are \(\displaystyle x = 4\) and \(\displaystyle x = -2\). Checking the solution \(\displaystyle x = 4\) in the original equation gives \(\displaystyle 5 = 5\). The solution checks, so \(\displaystyle x = 4\) is a true solution. Checking the solution \(\displaystyle x = -2\) in the original equation gives \(\displaystyle -1 = 1\). The solution does no check, so \(\displaystyle x = -2\) is an extraneous solution.

Download \(\LaTeX\)

\begin{question}Solve $x + 1 = \sqrt{4 x + 9}$.
    \soln{10cm}{Squaring both sides gives $x^{2} + 2 x + 1 = 4 x + 9$. The equation is quadratic setting it equal to zero gives $x^{2} - 2 x - 8 = 0$. Factoring gives $(x - 4)(x + 2)=0$ so the possible solutions are $x = 4$ and $x = -2$. Checking the solution $x = 4$ in the original equation gives $5 = 5$. The solution checks, so $x = 4$ is a true solution. Checking the solution $x = -2$ in the original equation gives $-1 = 1$. The solution does no check, so $x = -2$ 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}%
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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 + 1 = \sqrt{4 x + 9} " src="/equation_images/%20%5Cdisplaystyle%20x%20%2B%201%20%3D%20%5Csqrt%7B4%20x%20%2B%209%7D%20" alt="LaTeX:  \displaystyle x + 1 = \sqrt{4 x + 9} " data-equation-content=" \displaystyle x + 1 = \sqrt{4 x + 9} " /> .</p> </p>

HTML for Canvas

<p> <p>Squaring both sides gives  <img class="equation_image" title=" \displaystyle x^{2} + 2 x + 1 = 4 x + 9 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%202%20x%20%2B%201%20%3D%204%20x%20%2B%209%20" alt="LaTeX:  \displaystyle x^{2} + 2 x + 1 = 4 x + 9 " data-equation-content=" \displaystyle x^{2} + 2 x + 1 = 4 x + 9 " /> . The equation is quadratic setting it equal to zero gives  <img class="equation_image" title=" \displaystyle x^{2} - 2 x - 8 = 0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20-%202%20x%20-%208%20%3D%200%20" alt="LaTeX:  \displaystyle x^{2} - 2 x - 8 = 0 " data-equation-content=" \displaystyle x^{2} - 2 x - 8 = 0 " /> . Factoring gives  <img class="equation_image" title=" \displaystyle (x - 4)(x + 2)=0 " src="/equation_images/%20%5Cdisplaystyle%20%28x%20-%204%29%28x%20%2B%202%29%3D0%20" alt="LaTeX:  \displaystyle (x - 4)(x + 2)=0 " data-equation-content=" \displaystyle (x - 4)(x + 2)=0 " />  so the possible solutions are  <img class="equation_image" title=" \displaystyle x = 4 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%204%20" alt="LaTeX:  \displaystyle x = 4 " data-equation-content=" \displaystyle x = 4 " />  and  <img class="equation_image" title=" \displaystyle x = -2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-2%20" alt="LaTeX:  \displaystyle x = -2 " data-equation-content=" \displaystyle x = -2 " /> . Checking the solution  <img class="equation_image" title=" \displaystyle x = 4 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%204%20" alt="LaTeX:  \displaystyle x = 4 " data-equation-content=" \displaystyle x = 4 " />  in the original equation gives  <img class="equation_image" title=" \displaystyle 5 = 5 " src="/equation_images/%20%5Cdisplaystyle%205%20%3D%205%20" alt="LaTeX:  \displaystyle 5 = 5 " data-equation-content=" \displaystyle 5 = 5 " /> . The solution checks, so  <img class="equation_image" title=" \displaystyle x = 4 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%204%20" alt="LaTeX:  \displaystyle x = 4 " data-equation-content=" \displaystyle x = 4 " />  is a true solution. Checking the solution  <img class="equation_image" title=" \displaystyle x = -2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-2%20" alt="LaTeX:  \displaystyle x = -2 " data-equation-content=" \displaystyle x = -2 " />  in the original equation gives  <img class="equation_image" title=" \displaystyle -1 = 1 " src="/equation_images/%20%5Cdisplaystyle%20-1%20%3D%201%20" alt="LaTeX:  \displaystyle -1 = 1 " data-equation-content=" \displaystyle -1 = 1 " /> . The solution does no check, so  <img class="equation_image" title=" \displaystyle x = -2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-2%20" alt="LaTeX:  \displaystyle x = -2 " data-equation-content=" \displaystyle x = -2 " />  is an extraneous solution. </p> </p>