\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
Solve. \(\displaystyle \sqrt{x + 20}=\sqrt{x - 8} - 2\)
Squaring both sides gives \(\displaystyle x + 20=x - 4 \sqrt{x - 8} - 4\). Isolating the radical gives \(\displaystyle -6=\sqrt{x - 8}\) Squaring again gives \(\displaystyle 36=x - 8\). Solving for \(\displaystyle x\) gives \(\displaystyle x=44\). Checking the solution, \(\displaystyle x = 44\), in the original equation gives \(\displaystyle 8 = 4\) which is false so it is an extraneous solution and there is no solution.
\begin{question}Solve. $\sqrt{x + 20}=\sqrt{x - 8} - 2$
\soln{10cm}{Squaring both sides gives $x + 20=x - 4 \sqrt{x - 8} - 4$. Isolating the radical gives $-6=\sqrt{x - 8}$ Squaring again gives $36=x - 8$. Solving for $x$ gives $x=44$. Checking the solution, $x = 44$, in the original equation gives $8 = 4$ which is false so it is an extraneous solution and there is no solution.}
\end{question}
\documentclass{article}
\usepackage{tikz}
\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%
}%
\newcommand{\hideifShowSolution}[1]{%
\ifShowSolution%
%
\else%
#1%
\fi%
}%
\everymath{\displaystyle}
\ShowSolutiontrue
\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 + 20}=\sqrt{x - 8} - 2 " src="/equation_images/%20%5Cdisplaystyle%20%5Csqrt%7Bx%20%2B%2020%7D%3D%5Csqrt%7Bx%20-%208%7D%20-%202%20" alt="LaTeX: \displaystyle \sqrt{x + 20}=\sqrt{x - 8} - 2 " data-equation-content=" \displaystyle \sqrt{x + 20}=\sqrt{x - 8} - 2 " /> </p> </p><p> <p>Squaring both sides gives <img class="equation_image" title=" \displaystyle x + 20=x - 4 \sqrt{x - 8} - 4 " src="/equation_images/%20%5Cdisplaystyle%20x%20%2B%2020%3Dx%20-%204%20%5Csqrt%7Bx%20-%208%7D%20-%204%20" alt="LaTeX: \displaystyle x + 20=x - 4 \sqrt{x - 8} - 4 " data-equation-content=" \displaystyle x + 20=x - 4 \sqrt{x - 8} - 4 " /> . Isolating the radical gives <img class="equation_image" title=" \displaystyle -6=\sqrt{x - 8} " src="/equation_images/%20%5Cdisplaystyle%20-6%3D%5Csqrt%7Bx%20-%208%7D%20" alt="LaTeX: \displaystyle -6=\sqrt{x - 8} " data-equation-content=" \displaystyle -6=\sqrt{x - 8} " /> Squaring again gives <img class="equation_image" title=" \displaystyle 36=x - 8 " src="/equation_images/%20%5Cdisplaystyle%2036%3Dx%20-%208%20" alt="LaTeX: \displaystyle 36=x - 8 " data-equation-content=" \displaystyle 36=x - 8 " /> . 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=44 " src="/equation_images/%20%5Cdisplaystyle%20x%3D44%20" alt="LaTeX: \displaystyle x=44 " data-equation-content=" \displaystyle x=44 " /> . Checking the solution, <img class="equation_image" title=" \displaystyle x = 44 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%2044%20" alt="LaTeX: \displaystyle x = 44 " data-equation-content=" \displaystyle x = 44 " /> , in the original equation gives <img class="equation_image" title=" \displaystyle 8 = 4 " src="/equation_images/%20%5Cdisplaystyle%208%20%3D%204%20" alt="LaTeX: \displaystyle 8 = 4 " data-equation-content=" \displaystyle 8 = 4 " /> which is false so it is an extraneous solution and there is no solution.</p> </p>