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Questions: Algebra BusinessCalculus
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Solve. \(\displaystyle \sqrt{x + 18}=\sqrt{x + 2} + 4\)
Squaring both sides gives \(\displaystyle x + 18=x + 8 \sqrt{x + 2} + 18\). Isolating the radical gives \(\displaystyle 0=\sqrt{x + 2}\) Squaring again gives \(\displaystyle 0=x + 2\). Solving for \(\displaystyle x\) gives \(\displaystyle x=-2\). Checking the solution, \(\displaystyle x = -2\), in the original equation gives \(\displaystyle 4 = 4\) which is true so the solution checks.
\begin{question}Solve. $\sqrt{x + 18}=\sqrt{x + 2} + 4$
\soln{10cm}{Squaring both sides gives $x + 18=x + 8 \sqrt{x + 2} + 18$. Isolating the radical gives $0=\sqrt{x + 2}$ Squaring again gives $0=x + 2$. Solving for $x$ gives $x=-2$. Checking the solution, $x = -2$, in the original equation gives $4 = 4$ which is true so the solution checks.}
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Solve. <img class="equation_image" title=" \displaystyle \sqrt{x + 18}=\sqrt{x + 2} + 4 " src="/equation_images/%20%5Cdisplaystyle%20%5Csqrt%7Bx%20%2B%2018%7D%3D%5Csqrt%7Bx%20%2B%202%7D%20%2B%204%20" alt="LaTeX: \displaystyle \sqrt{x + 18}=\sqrt{x + 2} + 4 " data-equation-content=" \displaystyle \sqrt{x + 18}=\sqrt{x + 2} + 4 " /> </p> </p><p> <p>Squaring both sides gives <img class="equation_image" title=" \displaystyle x + 18=x + 8 \sqrt{x + 2} + 18 " src="/equation_images/%20%5Cdisplaystyle%20x%20%2B%2018%3Dx%20%2B%208%20%5Csqrt%7Bx%20%2B%202%7D%20%2B%2018%20" alt="LaTeX: \displaystyle x + 18=x + 8 \sqrt{x + 2} + 18 " data-equation-content=" \displaystyle x + 18=x + 8 \sqrt{x + 2} + 18 " /> . Isolating the radical gives <img class="equation_image" title=" \displaystyle 0=\sqrt{x + 2} " src="/equation_images/%20%5Cdisplaystyle%200%3D%5Csqrt%7Bx%20%2B%202%7D%20" alt="LaTeX: \displaystyle 0=\sqrt{x + 2} " data-equation-content=" \displaystyle 0=\sqrt{x + 2} " /> Squaring again gives <img class="equation_image" title=" \displaystyle 0=x + 2 " src="/equation_images/%20%5Cdisplaystyle%200%3Dx%20%2B%202%20" alt="LaTeX: \displaystyle 0=x + 2 " data-equation-content=" \displaystyle 0=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=-2 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-2%20" alt="LaTeX: \displaystyle x=-2 " data-equation-content=" \displaystyle x=-2 " /> . 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 4 = 4 " src="/equation_images/%20%5Cdisplaystyle%204%20%3D%204%20" alt="LaTeX: \displaystyle 4 = 4 " data-equation-content=" \displaystyle 4 = 4 " /> which is true so the solution checks.</p> </p>