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Algebra
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Solve. \(\displaystyle \sqrt{x + 12}=\sqrt{x - 2} + 4\)

Squaring both sides gives \(\displaystyle x + 12=x + 8 \sqrt{x - 2} + 14\). Isolating the radical gives \(\displaystyle - \frac{1}{4}=\sqrt{x - 2}\) Squaring again gives \(\displaystyle \frac{1}{16}=x - 2\). Solving for \(\displaystyle x\) gives \(\displaystyle x=\frac{33}{16}\). Checking the solution, \(\displaystyle x = \frac{33}{16}\), in the original equation gives \(\displaystyle \frac{15}{4} = \frac{17}{4}\) which is false so it is an extraneous solution and there is no solution.

Download \(\LaTeX\) 

\begin{question}Solve. $\sqrt{x + 12}=\sqrt{x - 2} + 4$
    \soln{10cm}{Squaring both sides gives $x + 12=x + 8 \sqrt{x - 2} + 14$. Isolating the radical gives $- \frac{1}{4}=\sqrt{x - 2}$ Squaring again gives $\frac{1}{16}=x - 2$.  Solving for $x$ gives $x=\frac{33}{16}$. Checking the solution, $x = \frac{33}{16}$, in the original equation gives $\frac{15}{4} = \frac{17}{4}$ which is false so it is an extraneous solution and there is no solution.}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>Solve.  <img class="equation_image" title=" \displaystyle \sqrt{x + 12}=\sqrt{x - 2} + 4 " src="/equation_images/%20%5Cdisplaystyle%20%5Csqrt%7Bx%20%2B%2012%7D%3D%5Csqrt%7Bx%20-%202%7D%20%2B%204%20" alt="LaTeX:  \displaystyle \sqrt{x + 12}=\sqrt{x - 2} + 4 " data-equation-content=" \displaystyle \sqrt{x + 12}=\sqrt{x - 2} + 4 " /> </p> </p>
HTML for Canvas
<p> <p>Squaring both sides gives  <img class="equation_image" title=" \displaystyle x + 12=x + 8 \sqrt{x - 2} + 14 " src="/equation_images/%20%5Cdisplaystyle%20x%20%2B%2012%3Dx%20%2B%208%20%5Csqrt%7Bx%20-%202%7D%20%2B%2014%20" alt="LaTeX:  \displaystyle x + 12=x + 8 \sqrt{x - 2} + 14 " data-equation-content=" \displaystyle x + 12=x + 8 \sqrt{x - 2} + 14 " /> . Isolating the radical gives  <img class="equation_image" title=" \displaystyle - \frac{1}{4}=\sqrt{x - 2} " src="/equation_images/%20%5Cdisplaystyle%20-%20%5Cfrac%7B1%7D%7B4%7D%3D%5Csqrt%7Bx%20-%202%7D%20" alt="LaTeX:  \displaystyle - \frac{1}{4}=\sqrt{x - 2} " data-equation-content=" \displaystyle - \frac{1}{4}=\sqrt{x - 2} " />  Squaring again gives  <img class="equation_image" title=" \displaystyle \frac{1}{16}=x - 2 " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B1%7D%7B16%7D%3Dx%20-%202%20" alt="LaTeX:  \displaystyle \frac{1}{16}=x - 2 " data-equation-content=" \displaystyle \frac{1}{16}=x - 2 " /> .  Solving for  <img class="equation_image" title=" \displaystyle x " src="/equation_images/%20%5Cdisplaystyle%20x%20" alt="LaTeX:  \displaystyle x " data-equation-content=" \displaystyle x " />  gives  <img class="equation_image" title=" \displaystyle x=\frac{33}{16} " src="/equation_images/%20%5Cdisplaystyle%20x%3D%5Cfrac%7B33%7D%7B16%7D%20" alt="LaTeX:  \displaystyle x=\frac{33}{16} " data-equation-content=" \displaystyle x=\frac{33}{16} " /> . Checking the solution,  <img class="equation_image" title=" \displaystyle x = \frac{33}{16} " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20%5Cfrac%7B33%7D%7B16%7D%20" alt="LaTeX:  \displaystyle x = \frac{33}{16} " data-equation-content=" \displaystyle x = \frac{33}{16} " /> , in the original equation gives  <img class="equation_image" title=" \displaystyle \frac{15}{4} = \frac{17}{4} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B15%7D%7B4%7D%20%3D%20%5Cfrac%7B17%7D%7B4%7D%20" alt="LaTeX:  \displaystyle \frac{15}{4} = \frac{17}{4} " data-equation-content=" \displaystyle \frac{15}{4} = \frac{17}{4} " />  which is false so it is an extraneous solution and there is no solution.</p> </p>