\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.5in,right=0.5in,top=0.5in,bottom=0.5in} \pagestyle{fancy} \lhead{\color{blue} \Large Lecture Notes} \chead{\color{black} \LARGE Radical Equations} \rhead{\large page \ \thepage} \cfoot{} \lfoot{\small \copyright $\;$ copyright Hidegkuti, Powell, 2008} \rfoot{\small Last revised: October 2, 2009} \textwidth 7.5in \textheight 9.6in \setlength{\headheight}{30pt} \setlength{\parindent}{0in} \begin{document} \begin{center} {\Large Sample Problems }\bigskip \bigskip \bigskip \end{center} %TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion \begin{enumerate} \item $\sqrt{3x-2}=x\bigskip \ $ \item $\sqrt[3]{3x-6}=3$ \bigskip \item $10+\sqrt{4x-7}=7\bigskip ~$ \item $5+\sqrt{x+15}=x~$\bigskip\ \item $2\sqrt{x-1}=x-4~~$\bigskip\ \item $2\sqrt{x+4}=1+\sqrt{2x+9}~$\bigskip \item $5\sqrt{x}+1=3\sqrt{x}+17~~$\bigskip \item $\sqrt{2x+5}+5=x~$\bigskip \item $\sqrt{2x+5}-\sqrt{x-1}=\sqrt{x+2}$\bigskip \item $\sqrt[3]{x^{3}+26}=x+2$\bigskip \item $\sqrt{3x+1}-\sqrt{x-4}=3$\bigskip \item $\sqrt{x+10}+10=x$\bigskip \item $\sqrt[3]{x^{3}+208}=x+4$\bigskip \item $\sqrt{x-1}+\sqrt{x-4}=\sqrt{4x-11}$\bigskip \item $\sqrt{4x+6}=\sqrt{x+1}-\sqrt{x+5}$\bigskip \end{enumerate} %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \bigskip \bigskip \bigskip \bigskip \bigskip \bigskip \begin{center} {\Large Practice Problems}\bigskip \bigskip \end{center} %TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion \begin{enumerate} \item $\sqrt{3x-5}=4$\bigskip \item $2\sqrt{a-1}+7=1$\bigskip\ \item $3\sqrt{7x+1}+2=20$\bigskip\ \ \item $\sqrt{x+3}=x-9$\bigskip\ \ \ \ \item $2\sqrt{x+5}=x-3$\bigskip\ \item $\sqrt{2p+4}-\sqrt{p+3}=1$\bigskip \item $\sqrt[3]{x^{3}+16}=x+4$\bigskip\ \item $\sqrt{18+x}=x-2$\bigskip\ \item $\sqrt{w}+\sqrt{w+3}=3$\bigskip\ \item $\sqrt{2x+1}+\sqrt{5-x}=4$\bigskip\ \item $2\sqrt{x}-\sqrt{x-3}=\sqrt{x+7}$\bigskip \item $2\sqrt{y+4}+\sqrt{y-5}=\sqrt{9y+7}$\bigskip\ \item $\sqrt{b-2}+b=8$\bigskip \item $\sqrt{3x+1}-\sqrt{x+4}=1$\bigskip \item $\sqrt[3]{x-8}-\sqrt[3]{4x+1}=0$\bigskip \item $\sqrt{5m-9}=\sqrt{5m}-3$\bigskip \item $\sqrt{3k-5}-\sqrt{3k}=-1$\bigskip \item $\sqrt{1-x}+1=x+12$\bigskip \item $2=\sqrt{x^{2}+1}-x$\bigskip \item $\sqrt{3x+1}-4=8-2\sqrt{3x+1}$\bigskip \item $\sqrt{x}=\sqrt{10+3\sqrt{x}}$\bigskip \item $\sqrt[3]{x^{3}+7}=x+1$\bigskip \item $\sqrt{x-6}=\sqrt{x+2}-4$\bigskip \item $\sqrt{6x+7}-\sqrt{3x+3}=1$\bigskip \end{enumerate} %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \bigskip \pagebreak \begin{center} {\Large Sample Problems -- Answers}\bigskip \end{center} \bigskip 1.) \ $1,2$ \ \ \ \ \ 2.) \ $11$ \ \ \ \ \ \ \ 3.) \ no real solution \ \ \ \ 4.) \ $10$ \ \ \ \ \ \ \ 5.) \ $10$ \ \ \ \ \ 6.) \ $0$ \ \ \ \ \ \ \ 7.) \ $64$ \ \ \ \ \ 8.) \ $10$\bigskip 9.) \ $2$ \ \ \ \ \ 10.) \ $-3,1$ \ \ \ \ \ 11.) \ $5,8$ \ \ \ \ \ \ 12.) \ $% 15$ \ \ \ \ \ \ \ 13.) \ $-6,2$ \ \ \ \ \ 14.) \ $5$ \ \ \ \ \ 15.) \ no real solution\bigskip \bigskip \bigskip \begin{center} {\Large Practice Problems -- Answers}\bigskip \bigskip \end{center} 1.) \ $7$ \ \ \ \ 2.) \ no real solution\bigskip\ \ \ \ \ \ 3.) \ $5$\ \ \ \ \ \ \ 4.) \ $13$ \ \ \ \ \ \ 5.) \ $11$ \ \ \ \ \ 6.) \ $6$ \ \ \ \ \ 7.) \ $% -2$ \ \ \ \ \ \ 8.) \ $7$ \ \ \ \ \ 9.) \ $1$ 10.) \ $4,\dfrac{20}{9}$ \ \ \ \ \ 11.) \ $\dfrac{25}{8}$ \ \ \ \ \ 12.) \ $% 21$ \ \ \ \ \ 13.) \ $6$ \ \ \ \ \ \ 14.) \ $5$ \ \ \ \ \ \ 15.) \ $-3$ \ \ \ \ \ \ 16.) \ $\dfrac{9}{5}$ \ \ \ \ \ \ 17.) \ $3$\bigskip 18.) \ $-8$ \ \ \ \ \ \ 19.) \ $-\dfrac{3}{4}$ \ \ \ \ \ 20.) \ $5$ \ \ \ \ \ 21.) \ $25$ \ \ \ \ \ 22.) \ $-2,1$ \ \ \ \ \ 23.) \ no real solution \ \ \ \ \ \ 24.) \ $\dfrac{1}{3},-1$\pagebreak \begin{center} {\Large Sample Problems -- Solutions}\bigskip \end{center} \begin{enumerate} \item $\sqrt{3x-2}=x~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 1,2$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution:% \begin{eqnarray*} \sqrt{3x-2} &=&x\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square} \\ 3x-2 &=&x^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ reduce one side to zero} \\ 0 &=&x^{2}-3x+2\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ factor} \\ 0 &=&\left( x-2\right) \left( x-1\right) \text{~~~}\Longrightarrow ~~x_{1}=2% \text{ \ \ and \ }x_{2}=1 \end{eqnarray*}% We check: if $x=2$, then% \begin{eqnarray*} \text{LHS} &=&\sqrt{3\left( 2\right) -2}=\sqrt{4}=2 \\ \text{RHS} &=&2 \end{eqnarray*}% and if $x=1$, then \begin{eqnarray*} \text{LHS} &=&\sqrt{3\left( 1\right) -2}=\sqrt{1}=1 \\ \text{RHS} &=&1 \end{eqnarray*}% Thus the solution set is: $\left\{ 1,2\right\} $ \item $\sqrt[3]{3x-6}=3~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 11$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution:% \begin{eqnarray*} \sqrt[3]{3x-6} &=&3\text{ \ \ \ \ \ \ \ \ \ \ \ raise both sides to the third power} \\ 3x-6 &=&27\text{ \ \ \ \ \ \ \ \ \ add }6 \\ 3x &=&33\text{ \ \ \ \ \ \ \ \ \ divide by }3 \\ x &=&11 \end{eqnarray*}% We check: if $x=11$, then% \begin{equation*} \text{LHS}=\sqrt[3]{3\left( 11\right) -6}=\sqrt[3]{27}=3=\text{RHS} \end{equation*} \item $10+\sqrt{4x-7}=7~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion no real solution% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: \begin{eqnarray*} 10+\sqrt{4x-7} &=&7\text{ \ \ \ \ \ subtract }10 \\ \sqrt{4x-7} &=&-3 \end{eqnarray*}% Since the square root of no real number is negative, there is no real solution. \item $5+\sqrt{x+15}=x~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 10$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution:% \begin{eqnarray*} 5+\sqrt{x+15} &=&x\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }5 \\ \sqrt{x+15} &=&x-5\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square both sides} \\ x+15 &=&\left( x-5\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ FOIL right hand side} \\ x+15 &=&x^{2}-10x+25\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ reduce one side to zero} \\ 0 &=&x^{2}-11x+10\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ factor} \\ 0 &=&\left( x-1\right) \left( x-10\right) \text{~~~}\Longrightarrow ~~x_{1}=1% \text{ \ \ and }x_{2}=10 \end{eqnarray*}% We check: If $x=1,$ then% \begin{eqnarray*} \text{LHS} &=&5+\sqrt{1+15}=5+\sqrt{16}=5+4=9 \\ \text{RHS} &=&1 \\ \text{RHS } &\not=&\text{ LHS} \end{eqnarray*} Thus $x=1$ is NOT a solution. If $x=10,$ then% \begin{eqnarray*} \text{LHS} &=&5+\sqrt{10+15}=5+\sqrt{25}=5+5=10 \\ \text{RHS} &=&10 \\ \text{RHS } &=&\text{ LHS} \end{eqnarray*}% Thus $x=10$ is the only solution. \item $2\sqrt{x-1}=x-4~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 10$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution:% \begin{eqnarray*} 2\sqrt{x-1} &=&x-4\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square both sides} \\ 4\left( x-1\right) &=&\left( x-4\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ FOIL, distribute} \\ 4x-4 &=&x^{2}-8x+16\text{ \ \ \ \ \ \ \ reduce one side to zero} \\ 0 &=&x^{2}-12x+20\text{ \ \ \ \ \ factor} \\ 0 &=&\left( x-2\right) \left( x-10\right) \text{~~~}\Longrightarrow ~~x_{1}=2% \text{ \ \ and \ }x_{2}=10 \end{eqnarray*}% We check: If $x=2,$ then% \begin{eqnarray*} \text{LHS} &=&2\sqrt{2-1}=2\sqrt{1}=2\left( 1\right) =2 \\ \text{RHS} &=&2-4=-2 \\ \text{RHS } &\not=&\text{ LHS} \end{eqnarray*}% Thus $x=2$ is NOT a solution.\newline If $x=10,$ then% \begin{eqnarray*} \text{LHS} &=&2\sqrt{10-1}=2\sqrt{9}=2\left( 3\right) =6\text{ \ and \ RHS}% =10-4=6 \\ \text{RHS} &=&\text{LHS} \end{eqnarray*}% Thus $x=10$ is the only solution. \item $2\sqrt{x+4}=1+\sqrt{2x+9}~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 0$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution:% \begin{eqnarray*} 2\sqrt{x+4} &=&1+\sqrt{2x+9}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square both sides} \\ \left( 2\sqrt{x+4}\right) ^{2} &=&\left( 1+\sqrt{2x+9}\right) ^{2}\text{ } \\ 2^{2}\left( \sqrt{x+4}\right) ^{2} &=&\left( 1+\sqrt{2x+9}\right) \left( 1+% \sqrt{2x+9}\right) \\ 4\left( x+4\right) &=&1+\sqrt{2x+9}+\sqrt{2x+9}+2x+9\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ combine like terms} \\ 4x+16 &=&2x+10+2\sqrt{2x+9}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }2x \\ 2x+16 &=&10+2\sqrt{2x+9}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }10 \\ 2x+6 &=&2\sqrt{2x+9}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ factor out }2 \end{eqnarray*}% \begin{eqnarray*} 2\left( x+3\right) &=&2\sqrt{2x+9}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }2 \\ x+3 &=&\sqrt{2x+9}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square both sides} \\ \left( x+3\right) ^{2} &=&2x+9 \\ x^{2}+6x+9 &=&2x+9\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ reduce one side to zero} \\ x^{2}+4x &=&0\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ factor} \\ x\left( x+4\right) &=&0\text{~~~}\Longrightarrow ~~x_{1}=0\text{ \ \ \ \ and \ \ \ }x_{2}=-4\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \end{eqnarray*}% We check: If $x=0,$ then% \begin{equation*} \text{LHS}=2\sqrt{0+4}=2\sqrt{4}=2\cdot 2=4\text{ \ and \ RHS}=1+\sqrt{% 2\left( 0\right) +9}=1+3=4 \end{equation*}% Thus $x=0$ is indeed a solution.\newline If $x=-4,$ then% \begin{eqnarray*} \text{LHS} &=&2\sqrt{\left( -4\right) +4}=2\sqrt{0}=2\left( 0\right) =0 \\ \text{RHS} &=&1+\sqrt{2\left( -4\right) +9}=1+\sqrt{-8+9}=1+\sqrt{1}=1+1=2 \\ \text{RHS } &\not=&\text{ LHS} \end{eqnarray*}% Thus $x=-4$ is NOT a solution. The only solution is $x=0$. \item $5\sqrt{x}+1=3\sqrt{x}+17~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 64$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution:% \begin{eqnarray*} 5\sqrt{x}+1 &=&3\sqrt{x}+17\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }3\sqrt{x} \\ 2\sqrt{x}+1 &=&17\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }1 \\ 2\sqrt{x} &=&16\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }2 \\ \sqrt{x} &=&8\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square both sides} \\ x &=&64 \end{eqnarray*}% We check: If $x=64,$ then% \begin{equation*} \text{LHS}=5\sqrt{64}+1=5\left( 8\right) +1=41\text{ \ and \ RHS}=3\sqrt{64}% +17=3\left( 8\right) +17=24+17=41 \end{equation*}% Thus $x=64$ is indeed a solution. \item $\sqrt{2x+5}+5=x~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 10$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution:% \begin{eqnarray*} \sqrt{2x+5}+5 &=&x\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }5 \\ \sqrt{2x+5} &=&x-5\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square} \\ 2x+5 &=&x^{2}-10x+25\text{ \ \ \ \ \ \ \ \ reduce one side to zero} \\ 0 &=&x^{2}-12x+20\text{ \ \ \ \ \ \ \ \ factor} \\ 0 &=&\left( x-2\right) \left( x-10\right) \text{~~~}\Longrightarrow ~~x_{1}=2% \text{ \ \ and \ }x_{2}=10 \end{eqnarray*}% We check: If $x=2,$ then% \begin{eqnarray*} \text{LHS} &=&\sqrt{2\left( 2\right) +5}+5=\sqrt{4+5}+5=\sqrt{9}+5=3+5=8% \text{ \ and \ RHS}=2 \\ \text{RHS } &\not=&\text{ LHS} \end{eqnarray*}% Thus $x=2$ is NOT a solution.\newline If $x=10,$ then% \begin{eqnarray*} \text{LHS} &=&\sqrt{2\left( 10\right) +5}+5=\sqrt{20+5}+5=\sqrt{25}+5=5+5=10 \\ \text{RHS} &=&10 \\ \text{RHS } &=&\text{ LHS} \end{eqnarray*}% Thus $x=10$ is the only solution. \item $\sqrt{2x+5}-\sqrt{x-1}=\sqrt{x+2}~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 2$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution:% \begin{eqnarray*} \sqrt{2x+5}-\sqrt{x-1} &=&\sqrt{x+2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ add \ }\sqrt{x-1} \\ \sqrt{2x+5} &=&\sqrt{x+2}+\sqrt{x-1}\text{ \ \ \ \ \ \ \ \ \ \ \ square} \\ \left( \sqrt{2x+5}\right) ^{2} &=&\left( \sqrt{x+2}+\sqrt{x-1}\right) ^{2} \\ 2x+5 &=&\left( \sqrt{x+2}+\sqrt{x-1}\right) \left( \sqrt{x+2}+\sqrt{x-1}% \right) \\ 2x+5 &=&\underset{\text{{\LARGE F}}}{\underbrace{\sqrt{x+2}\sqrt{x+2}}}+% \underset{\text{{\LARGE O}}}{\underbrace{\sqrt{x+2}\sqrt{x-1}}}+\underset{% \text{{\LARGE I}}}{\underbrace{\sqrt{x-1}\sqrt{x+2}}}+\underset{\text{% {\LARGE L}}}{\underbrace{\sqrt{x-1}\sqrt{x-1}}} \\ 2x+5 &=&x+2+2\sqrt{x-1}\sqrt{x+2}+x-1 \\ 2x+5 &=&2x+1+2\sqrt{x-1}\sqrt{x+2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }% 2x \\ 5 &=&1+2\sqrt{\left( x-1\right) \left( x+2\right) }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }1 \\ 4 &=&2\sqrt{\left( x-1\right) \left( x+2\right) }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }2 \\ 2 &=&\sqrt{\left( x-1\right) \left( x+2\right) }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square} \\ 4 &=&\left( x-1\right) \left( x+2\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ FOIL} \\ 4 &=&x^{2}+x-2\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ reduce one seide to zero} \\ 0 &=&x^{2}+x-6\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ factor} \\ 0 &=&\left( x+3\right) \left( x-2\right) \text{~~~}\Longrightarrow ~~x_{1}=-3% \text{ \ \ and \ \ }x_{2}=2 \end{eqnarray*}% We check: If $x=-3,$ then% \begin{equation*} \text{LHS}=\sqrt{2\left( -3\right) +5}-\sqrt{\left( -3\right) -1}=\sqrt{-1}-% \sqrt{-4}=\text{undefined} \end{equation*}% Since the left hand side is undefined, $x=-3$ is NOT a solution.\newline If $x=2,$ then% \begin{eqnarray*} \text{LHS} &=&\sqrt{2\left( 2\right) +5}-\sqrt{2-1}=\sqrt{9}-\sqrt{1}=3-1=2 \\ \text{RHS} &=&\sqrt{2+2}=\sqrt{4}=2 \\ \text{RHS } &=&\text{ LHS} \end{eqnarray*}% Thus $x=2$ is the only solution. \item $\sqrt[3]{x^{3}+26}=x+2~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion -3,1$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: \begin{eqnarray*} \left( \sqrt[3]{x^{3}+26}\right) ^{3} &=&\left( x+2\right) ^{3} \\ x^{3}+26 &=&\left( x+2\right) ^{3} \\ x^{3}+26 &=&\allowbreak x^{3}+6x^{2}+12x+8 \\ 0 &=&\allowbreak x^{3}+6x^{2}+12x+8-x^{3}-26 \\ 0 &=&\allowbreak 6x^{2}+12x-18 \\ 0 &=&\allowbreak 6\left( x^{2}+2x-3\right) \\ 0 &=&\allowbreak 6\left( x+3\right) \left( x-1\right) \text{~~~}% \Longrightarrow ~~x_{1}=-3\text{ \ and\ \ }x_{2}=1 \end{eqnarray*}% We check both answers: if $x=-3$, then \begin{eqnarray*} \text{LHS} &=&\sqrt[3]{\left( -3\right) ^{3}+26}=\sqrt[3]{-27+26}=\sqrt[3]{-1% }=-1 \\ \text{RHS} &=&\left( -3\right) +2=-1 \end{eqnarray*}% thus $-3$ does work. \ If $x=1$, then \begin{eqnarray*} \text{LHS} &=&\sqrt[3]{1^{3}+26}=\sqrt[3]{1+26}=\sqrt[3]{27}=3 \\ \text{RHS} &=&1+2=3 \end{eqnarray*}% The solution set is $\left\{ -3,1\right\} $. \item $\sqrt{3x+1}-\sqrt{x-4}=3~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 5,8$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution:% \begin{eqnarray*} \sqrt{3x+1}-\sqrt{x-4} &=&3\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ add \ \ }\sqrt{x-4}\text{ \ to both sides} \\ \sqrt{3x+1} &=&3+\sqrt{x-4}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square both sides} \\ 3x+1 &=&\left( 3+\sqrt{x-4}\right) ^{2} \\ 3x+1 &=&\left( 3+\sqrt{x-4}\right) \left( 3+\sqrt{x-4}\right) \\ 3x+1 &=&9+3\sqrt{x-4}+3\sqrt{x-4}+x-4 \\ 3x+1 &=&x+5+6\sqrt{x-4}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }x \\ 2x+1 &=&5+6\sqrt{x-4}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }5 \\ 2x-4 &=&6\sqrt{x-4} \\ 2\left( x-2\right) &=&6\sqrt{x-4}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }2 \\ x-2 &=&3\sqrt{x-4}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square both sides} \\ \left( x-2\right) ^{2} &=&9\left( x-4\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ FOIL, distribute \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x^{2}-4x+4 &=&9x-36\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ reduce one side to zero} \\ x^{2}-13x+40 &=&0\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ factor} \\ \left( x-5\right) \left( x-8\right) &=&0\text{~~~}\Longrightarrow ~~x_{1}=5% \text{ \ \ and \ \ \ }x_{2}=8 \end{eqnarray*}% We check: if $x=5$, then \begin{equation*} \text{LHS}=\sqrt{3\left( 5\right) +1}-\sqrt{5-4}=\sqrt{16}-\sqrt{1}=4-1=3=% \text{RHS} \end{equation*}% If $x=8$, then \begin{equation*} \text{LHS}=\sqrt{3\left( 8\right) +1}-\sqrt{8-4}=\sqrt{25}-\sqrt{4}=5-2=3=% \text{RHS} \end{equation*}% The solution set is $\left\{ 5,8\right\} $. \item $\sqrt{x+10}+10=x~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 15$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution:% \begin{eqnarray*} \sqrt{x+10}+10 &=&x\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract \ }10 \\ \sqrt{x+10} &=&x-10\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square} \\ x+10 &=&\left( x-10\right) ^{2} \\ x+10 &=&x^{2}-20x+100\text{ \ \ \ \ \ \ \ \ \ reduce one side to zero} \\ 0 &=&x^{2}-21x+90\text{ \ \ \ \ \ \ \ \ \ \ \ factor} \\ 0 &=&\left( x-6\right) \left( x-15\right) \text{~~~}\Longrightarrow ~~x_{1}=6% \text{ \ \ and \ \ }x_{2}=15 \end{eqnarray*}% We check: if $x=6$, then \begin{eqnarray*} \text{LHS} &=&\sqrt{6+10}+10=\sqrt{16}+10=4+10=14 \\ \text{RHS} &=&6 \\ \text{LHS } &\not=&\text{ RHS} \end{eqnarray*}% If $x=15$, then \begin{equation*} \text{LHS}=\sqrt{15+10}+10=\sqrt{25}+10=5+10=15=\text{RHS} \end{equation*}% since $x=6$ doesn't work, the only solution is $15.$ In set notation: the solution set is $\left\{ 15\right\} $. \item $\sqrt[3]{x^{3}+208}=x+4~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion -6,2$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: \begin{eqnarray*} \sqrt[3]{x^{3}+208} &=&x+4\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ raise both sides to the third power} \\ \left( \sqrt[3]{x^{3}+208}\right) ^{3} &=&\left( x+4\right) ^{3} \\ x^{3}+208 &=&x^{3}+12x^{2}+48x+64\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract \ }x^{3}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 208 &=&\allowbreak 12x^{2}+48x+64\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ reduce one side to zero} \\ 0 &=&\allowbreak 12x^{2}+48x-144\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ factor out }12 \\ 0 &=&\allowbreak 12\left( x^{2}+4x-12\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ factor} \\ 0 &=&\allowbreak 12\left( x+6\right) \left( x-2\right) \text{~~~}% \Longrightarrow ~~x_{1}=-6\text{ \ \ \ and \ \ \ }x=2 \end{eqnarray*}% Solution: $-6$ and $2$. \ We check. They both work. \item $\sqrt{x-1}+\sqrt{x-4}=\sqrt{4x-11}~~~% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion 5$% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution:% \begin{eqnarray*} \sqrt{x-1}+\sqrt{x-4} &=&\sqrt{4x-11}\text{ \ \ \ \ \ \ \ \ \ square} \\ \left( \sqrt{x-1}+\sqrt{x-4}\right) ^{2} &=&\left( \sqrt{4x-11}\right) ^{2} \\ \left( \sqrt{x-1}+\sqrt{x-4}\right) \left( \sqrt{x-1}+\sqrt{x-4}\right) &=&4x-11\text{ \ \ \ \ \ \ \ \ \ \ FOIL} \\ \underset{\text{{\LARGE F}}}{\underbrace{\sqrt{x-1}\sqrt{x-1}}}+\underset{% \text{{\LARGE O}}}{\underbrace{\sqrt{x-1}\sqrt{x-4}}}+\underset{\text{% {\LARGE I}}}{\underbrace{\sqrt{x-4}\sqrt{x-1}}}+\underset{\text{{\LARGE L}}}{% \underbrace{\sqrt{x-4}\sqrt{x-4}}} &=&4x-11 \\ x-1+2\sqrt{x-1}\sqrt{x-4}+x-4 &=&4x-11\text{ \ \ \ \ \ \ combine like terms} \\ 2x-5+2\sqrt{x-1}\sqrt{x-4} &=&4x-11\text{ \ \ \ \ \ \ subtract \ }2x \\ -5+2\sqrt{x-1}\sqrt{x-4} &=&2x-11\text{ \ \ \ \ \ \ add }5 \\ 2\sqrt{x-1}\sqrt{x-4} &=&2x-6 \\ 2\sqrt{x-1}\sqrt{x-4} &=&2\left( x-3\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }2 \\ \sqrt{\left( x-1\right) \left( x-4\right) } &=&x-3\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square} \\ \left( x-1\right) \left( x-4\right) &=&\left( x-3\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ FOIL} \\ x^{2}-5x+4 &=&x^{2}-6x+9\text{ \ \ \ \ \ \ \ \ subtract }x^{2} \\ -5x+4 &=&-6x+9\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ add }6x \\ x+4 &=&9\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }4 \\ x &=&5 \end{eqnarray*}% We check:% \begin{eqnarray*} \text{LHS} &=&\sqrt{5-1}+\sqrt{5-4}=\sqrt{4}+\sqrt{1}=2+1=3 \\ \text{RHS} &=&\sqrt{4x-11}=\sqrt{4\left( 5\right) -11}=\sqrt{9}=3 \end{eqnarray*}% And so the solution set is $\left\{ 5\right\} $.\pagebreak \item $\sqrt{4x+6}=\sqrt{x+1}-\sqrt{x+5}~~~$% %TCIMACRO{\TeXButton{red}{\color{red}}}% %BeginExpansion \color{red}% %EndExpansion no real solution% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \newline Solution: \begin{eqnarray*} \sqrt{4x+6} &=&\sqrt{x+1}-\sqrt{x+5}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square} \\ \left( \sqrt{4x+6}\right) ^{2} &=&\left( \sqrt{x+1}-\sqrt{x+5}\right) ^{2} \\ 4x+6 &=&\underset{\text{{\LARGE F}}}{\underbrace{\sqrt{x+1}\sqrt{x+1}}}-% \underset{\text{{\LARGE O}}}{\underbrace{\sqrt{x+1}\sqrt{x+5}}}-\underset{% \text{{\LARGE I}}}{\underbrace{\sqrt{x+5}\sqrt{x+1}}}+\underset{\text{% {\LARGE L}}}{\underbrace{\sqrt{x+5}\sqrt{x+5}}} \\ 4x+6 &=&x+1-2\sqrt{\left( x+1\right) \left( x+5\right) }+x+5 \\ 4x+6 &=&2x+6-2\sqrt{\left( x+1\right) \left( x+5\right) }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }2x \\ 2x+6 &=&6-2\sqrt{\left( x+1\right) \left( x+5\right) }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }6 \\ \text{\ \ \ \ }2x &=&-2\sqrt{\left( x+1\right) \left( x+5\right) }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }2 \\ x &=&-\sqrt{\left( x+1\right) \left( x+5\right) }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square} \\ x^{2} &=&\left( -\sqrt{\left( x+1\right) \left( x+5\right) }\right) ^{2} \\ x^{2} &=&\left( x+1\right) \left( x+5\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ FOIL right hand side} \\ x^{2} &=&x^{2}+6x+5\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }x^{2} \\ 0 &=&6x+5\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }5 \\ -5 &=&6x\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }6 \\ -\dfrac{5}{6} &=&x \end{eqnarray*}% We check: if $x=-\dfrac{5}{6}$, then% \begin{eqnarray*} \text{LHS} &=&\sqrt{4\left( -\dfrac{5}{6}\right) +6}=\sqrt{-\dfrac{10}{3}+6}=% \sqrt{-\dfrac{10}{3}+\dfrac{18}{3}}=\sqrt{\dfrac{8}{3}} \\ \text{RHS} &=&\sqrt{-\dfrac{5}{6}+1}-\sqrt{-\dfrac{5}{6}+5}=\sqrt{\dfrac{1}{6% }}-\sqrt{\left( -\dfrac{5}{6}\right) +\dfrac{30}{6}}=\sqrt{\dfrac{1}{6}}-% \sqrt{\dfrac{25}{6}}=\dfrac{\sqrt{1}}{\sqrt{6}}-\dfrac{\sqrt{25}}{\sqrt{6}} \\ &=&\dfrac{1}{\sqrt{6}}-\dfrac{5}{\sqrt{6}}=\dfrac{1-5}{\sqrt{6}}=-\dfrac{4}{% \sqrt{6}} \end{eqnarray*}% Since the left hand side is positive, and the right hand side is negative, these two numbers can not be equal. This equation has no real solution. \end{enumerate} \bigskip \bigskip \bigskip \bigskip \bigskip \vspace{1.6in} %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}