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Questions: Algebra BusinessCalculus
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Solve. \(\displaystyle \sqrt{x + 30}=\sqrt{x + 6} + 4\)
Squaring both sides gives \(\displaystyle x + 30=x + 8 \sqrt{x + 6} + 22\). Isolating the radical gives \(\displaystyle 1=\sqrt{x + 6}\) Squaring again gives \(\displaystyle 1=x + 6\). Solving for \(\displaystyle x\) gives \(\displaystyle x=-5\). Checking the solution, \(\displaystyle x = -5\), in the original equation gives \(\displaystyle 5 = 5\) which is true so the solution checks.
\begin{question}Solve. $\sqrt{x + 30}=\sqrt{x + 6} + 4$
\soln{10cm}{Squaring both sides gives $x + 30=x + 8 \sqrt{x + 6} + 22$. Isolating the radical gives $1=\sqrt{x + 6}$ Squaring again gives $1=x + 6$. Solving for $x$ gives $x=-5$. Checking the solution, $x = -5$, in the original equation gives $5 = 5$ 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 + 30}=\sqrt{x + 6} + 4 " src="/equation_images/%20%5Cdisplaystyle%20%5Csqrt%7Bx%20%2B%2030%7D%3D%5Csqrt%7Bx%20%2B%206%7D%20%2B%204%20" alt="LaTeX: \displaystyle \sqrt{x + 30}=\sqrt{x + 6} + 4 " data-equation-content=" \displaystyle \sqrt{x + 30}=\sqrt{x + 6} + 4 " /> </p> </p><p> <p>Squaring both sides gives <img class="equation_image" title=" \displaystyle x + 30=x + 8 \sqrt{x + 6} + 22 " src="/equation_images/%20%5Cdisplaystyle%20x%20%2B%2030%3Dx%20%2B%208%20%5Csqrt%7Bx%20%2B%206%7D%20%2B%2022%20" alt="LaTeX: \displaystyle x + 30=x + 8 \sqrt{x + 6} + 22 " data-equation-content=" \displaystyle x + 30=x + 8 \sqrt{x + 6} + 22 " /> . Isolating the radical gives <img class="equation_image" title=" \displaystyle 1=\sqrt{x + 6} " src="/equation_images/%20%5Cdisplaystyle%201%3D%5Csqrt%7Bx%20%2B%206%7D%20" alt="LaTeX: \displaystyle 1=\sqrt{x + 6} " data-equation-content=" \displaystyle 1=\sqrt{x + 6} " /> Squaring again gives <img class="equation_image" title=" \displaystyle 1=x + 6 " src="/equation_images/%20%5Cdisplaystyle%201%3Dx%20%2B%206%20" alt="LaTeX: \displaystyle 1=x + 6 " data-equation-content=" \displaystyle 1=x + 6 " /> . 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=-5 " src="/equation_images/%20%5Cdisplaystyle%20x%3D-5%20" alt="LaTeX: \displaystyle x=-5 " data-equation-content=" \displaystyle x=-5 " /> . Checking the solution, <img class="equation_image" title=" \displaystyle x = -5 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-5%20" alt="LaTeX: \displaystyle x = -5 " data-equation-content=" \displaystyle x = -5 " /> , 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 " /> which is true so the solution checks.</p> </p>