\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=75pt,right=50pt,top=0.5in,bottom=0.5in} \pagestyle{fancy} \lhead{\color{blue} \large Lecture Notes} \chead{\color{black} \Large Radical Equations} \rhead{\small page \ \thepage} \cfoot{} \lfoot{\footnotesize \copyright $\;$ Hidegkuti, Powell, 2008} \rfoot{\footnotesize Last revised: January 13, 2019} \textheight 9.6in \setlength{\headheight}{30pt} \setlength{\parindent}{0in} \newcommand{\defstart}{ \begin{boxedminipage}{7in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt} \setlength{\rightskip}{20pt} \setlength{\leftskip}{20pt} \color{white} . \color{black} } \newcommand{\defcont}{ \setlength{\hangindent}{57pt} } \newcommand{\definside}{ \setlength{\hangindent}{57pt} \setlength{\parindent}{57pt} } \newcommand{\defend}{ \end{boxedminipage} \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} \setlength{\parskip}{0.06in} \setstretch{1.08} } \newcommand{\exstart}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{60pt} \setlength{\parskip}{0.04in} \setstretch{1.08} } \newcommand{\excont}{ \setlength{\parindent}{60pt} \setlength{\hangindent}{75pt} } \newcommand{\solstart}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{63pt} } \newcommand{\solcont}{ \setlength{\parindent}{63pt} \setlength{\hangindent}{63pt} } \newcommand{\solrestart}{ \setlength{\parindent}{48pt} \setlength{\hangindent}{63pt} } \newcommand{\sol}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{50pt} } \newcommand{\solinside}{ \setlength{\parindent}{50pt} \setlength{\hangindent}{50pt} \setstretch{1.08} } \newcommand{\vissza}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} \setlength{\parskip}{0.1in} \setstretch{1.08} \normalsize } \newcommand{\matek}{ \normalsize \setlength{\parskip}{0.04in} \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} \setstretch{1.05} } \setstretch{1.04} \begin{document} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion We will now study radical equations. \ As the name suggests, these are equations with square roots or cubic roots, etc. \ Let us first recall a few facts.\vspace{0.05in} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 1}. \ Simplify each of the given expressions. %TCIMACRO{\TeXButton{\excont}{\excont}}% %BeginExpansion \excont% %EndExpansion a) \ $\left( \sqrt{x-2}\right) ^{2}$ \ \ \ \ \ \ \ \ \ \ b) \ $\left( \sqrt{x% }-2\right) ^{2}$ \ \ \ \ \ \ \ c) \ $\ \left( -3\sqrt{x}\right) ^{2}$ %TCIMACRO{\TeXButton{\solstart}{\solstart}}% %BeginExpansion \solstart% %EndExpansion \textbf{Solution:} \ a) \ Recall that $\sqrt{5}$ is the non-negative number whose square is $5$. \ Similarly, $\sqrt{x-2}$ \ is the non-negative quantity that, when squared, the result is $x-2$. \ That is exactly what happens here and so %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion $\left( \sqrt{x-2}\right) ^{2}=\fbox{$x-2$}\vspace{0.05in}$ %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion b) \ This is a different situation. \ $\sqrt{x}$ is the non-negative number, that, when squared, the result is $x$. \ To square $\sqrt{x}-2$, we need to apply the distributive property.$\vspace{0.05in}$ %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion $\left( \sqrt{x}-2\right) ^{2}=\left( \sqrt{x}-2\right) \left( \sqrt{x}% -2\right) =\sqrt{x}\sqrt{x}-2\sqrt{x}-2\sqrt{x}+4=\left( \sqrt{x}\right) ^{2}-4\sqrt{x}+4=$ \fbox{$x-4\sqrt{x}+4$}$\vspace{0.05in}$ %TCIMACRO{\TeXButton{\solcont}{\solcont}}% %BeginExpansion \solcont% %EndExpansion This computation shows that the radical is not always eliminated just becuse we square the expression.$\vspace{0.05in}$ %TCIMACRO{\TeXButton{\solrestart}{\solrestart}}% %BeginExpansion \solrestart% %EndExpansion b) \ $\left( -3\sqrt{x}\right) ^{2}=\left( -3\sqrt{x}\right) \left( -3\sqrt{x% }\right) =9\left( \sqrt{x}\right) ^{2}=$ \fbox{$9x$}$\vspace{0.05in}$ %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion When we are dealing with radical expressions and want to get rifd of radicals, squaring is useful for expressions such as $\sqrt{x-2}$ or $-3% \sqrt{x}$, but not for expressions such as $\sqrt{x}-2$. \ This will be important to keep in mind.$\vspace{0.05in}$ Also recall that when solving an equations, an \textbf{equivalent step} is one that does not change the solution set. $\vspace{0.05in}$ %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 2}. \ Find all real solutions of the equation \ $\sqrt{x-1}% =-x+3\vspace{0.05in}$ %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution:} \ If we square both sides of the equation, the radical on the left will disappear, and the expression on the right becomes quadratic, resulting in a quadratic equation. \ So it looks like it is a good idea to square both sides first and then solve the quadratic equation we obtained.% \begin{eqnarray*} \sqrt{x-1} &=&-x+3\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square both sides} \\ x-1 &=&\left( -x+3\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ expand complete square} \\ x-1 &=&x^{2}-6x+9\text{ \ \ \ \ \ \ \ \ \ \ subtract }x \\ -1 &=&x^{2}-7x+9\text{ \ \ \ \ \ \ \ \ \ \ add }1 \\ 0 &=&x^{2}-7x+10 \\ 0 &=&\left( x-2\right) \left( x-5\right) \end{eqnarray*}% Therefore, it appears that this eqaution has two solutions, $2$ and $5$. \ Let us check. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion If $x=2$, then LHS $=\sqrt{2-1}=\sqrt{1}$ $=1$ \ and RHS $=-2+3=1$ and so $% x=2$ works. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion If $x=5$, then LHS $=\sqrt{5-1}=\sqrt{4}=2$ and RHS $=-5+3=-2,$ and $% 2\not=-2,$ so $x=5$ is not a solution of theis equation. \ Thus thiis equation has one solution, \fbox{$x=2$}. %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion Until now, checking a solution was just a matter of making certain that we did not make a mistake. \ This is a different situation: our computation was correct, and yet we have a number that is a solution of the last equation, but not of the first. \ This is because squaring both sides of an equation is a \textbf{non-equaivalent step} that increases the solution set. \pagebreak When we substituted $x=5$ into the original equation, we had $2=-2,$ which is a false statement. \ But next we squared both sides, and the false statement $2=-2$ became $4=4$, which \ is true. \ This is how $x=5$ works after we squared both sides, but not before. If we square both sides of an equation $L=R$, and solve the equation $% L^{2}=R^{2}$, then some of the solutions of $L^{2}=R^{2}$ might be numbers for which $L=-R$. \ Such a number is called an \textbf{extreneous solution}. \ When we square both sides of an equation, we \textit{must }check our solution(s) because squaring both sides of an equation is a non-equivalent step. \ In our previous example, $x=5$ was an extreneous solution. \vspace{% 0.05in} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 3}. \ Find all real solutions of the equation \ $11=\sqrt{% 4x+1}+x\vspace{0.05in}$ %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution:} \ If\ we squared both sides as they are, we would \ not eliminate \ the radical. \ Recall that sums involving radicals do not respond well to squaring both sides. \ Before we square, we need to isolate the radical expression on one side. \ Therefore, we will start by sutracting $x$.% \begin{eqnarray*} 11 &=&\sqrt{4x+1}+x\text{ \ \ \ \ \ \ \ \ \ \ \ subtract }x \\ 11-x &=&\sqrt{4x+1}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square both sides} \\ \left( 11-x\right) ^{2} &=&4x+1 \\ x^{2}-22x+121 &=&4x+1\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }x\text{, subtract }1 \\ x^{2}-26x+120 &=&0 \\ \left( x-6\right) \left( x-20\right) &=&0\text{ \ \ \ }\Longrightarrow \text{ \ \ }x_{1}=6\text{, \ \ \ }x_{2}=20 \end{eqnarray*}% We check: \ If $x=6,$ then RHS $=\sqrt{4\cdot 6+1}+6=\sqrt{25}+6=5+6=11=$ LHS $\checkmark $ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion If $x=20$, then RHS $=\sqrt{4\cdot 20+1}+20=\sqrt{81}+20=9+20=29\not=11$. \ Thus $x=20$ is not a solution. \ The only solution is \fbox{$x=6$}. %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion Why would $x=20$ show up as a solution? \ Let us substitute $20$ into the equation we squared. \ That is the second line, $11-x=\sqrt{4x+1}$. \ If we substitute $20$ into $x$, we get $-9=9.$ \ That is a false statement, but we see that it will become true after squaring both sides. We have only seen equations so far that had one solution and one extreneous solution. \ This is not the only possibility: some radical equations have two solution, some have none.\vspace{0.05in} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 4}. \ Find all real solutions of the equation \ $\sqrt{3x+1}-% \sqrt{x-1}=2\vspace{0.05in}$ %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution:} \ If\ we squared both sides as they are, we would be left with expressions such as $2\sqrt{3x+1}\sqrt{x-1}$. \ To avoid that,we should isolate the radical expression. \ In this equation, however, there are two radical expressions, and we cannot isolate both. \ We should select the more complicated one, isolate that and square. \ Then we will only left with one radical expression, so we repeat the process of isolating it and then sqjaring both sides.% \begin{eqnarray*} \sqrt{3x+1}-\sqrt{x-1} &=&2\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ add }\sqrt{x-1} \\ \sqrt{3x+1} &=&\sqrt{x-1}+2\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square} \\ 3x+1 &=&\left( \sqrt{x-1}+2\right) ^{2} \end{eqnarray*}% To expand $\left( \sqrt{x-1}+2\right) ^{2}$, we apply the distributive law:% \begin{eqnarray*} \left( \sqrt{x-1}+2\right) ^{2} &=&\left( \sqrt{x-1}+2\right) \left( \sqrt{% x-1}+2\right) =\sqrt{x-1}\sqrt{x-1}+2\sqrt{x-1}+2\sqrt{x-1}+4 \\ &=&\left( \sqrt{x-1}\right) ^{2}+4\sqrt{x-1}+4=x-1+4\sqrt{x-1}+4=x+4\sqrt{x-1% }+3 \end{eqnarray*}% and so our equation is \begin{eqnarray*} 3x+1 &=&\left( \sqrt{x-1}+2\right) ^{2} \\ 3x+1 &=&x+4\sqrt{x-1}+3\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }x \\ 2x+1 &=&4\sqrt{x-1}+3\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }3 \\ 2x-2 &=&4\sqrt{x-1} \end{eqnarray*}% Notice that all coefficients are even. \ Therefore, we may dividde both sides by $2$. \ This is a step that is not necessary, but it saves us work as the numbers don't get as large.% \begin{eqnarray*} 2x-2 &=&4\sqrt{x-1}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }2 \\ x-1 &=&2\sqrt{x-1}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square \ both sides} \\ \left( x-1\right) ^{2} &=&\left( 2\sqrt{x-1}\right) ^{2} \\ x^{2}-2x+1 &=&4\left( x-1\right) \\ x^{2}-2x+1 &=&4x-4\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }4x \\ x^{2}-6x+1 &=&-4\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ add }4 \\ x^{2}-6x+5 &=&0 \\ \left( x-1\right) \left( x-5\right) &=&0\text{ \ \ \ \ \ \ \ \ \ \ }% \Longrightarrow \text{ \ \ \ \ }x_{1}=1\text{, \ }x_{2}=5 \end{eqnarray*}% We check both candidates. \ If $x=1,$ then \begin{equation*} \text{LHS}=\sqrt{3\cdot 1+1}-\sqrt{1-1}=\sqrt{4}-\sqrt{0}=2-0=2=\text{RHS }% \checkmark \end{equation*}% and if $x=5$, then \begin{equation*} \text{LHS}=\sqrt{3\cdot 5+1}-\sqrt{5-1}=\sqrt{16}-\sqrt{4}=4-2=2=\text{RHS }% \checkmark \end{equation*}% In case of this equation, \ both \fbox{$1$ and $5$} are solutions.\vspace{% 0.15in} \FRAME{itbpF}{0.8631in}{0.6832in}{0.2603in}{}{}{tree.jpg}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.8631in;height 0.6832in;depth 0.2603in;original-width 4.4685in;original-height 3.531in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../../book99/weeks1to8/tree.jpg';file-properties "XNPEU";}}\ \ \ %TCIMACRO{\TeXButton{\sffamily}{\sffamily}}% %BeginExpansion \sffamily% %EndExpansion {\LARGE Enrichment} %TCIMACRO{% %\TeXButton{font family sans sherif}{\sffamily %\small}}% %BeginExpansion \sffamily \small% %EndExpansion \begin{enumerate} \item Consider the equation $x^{2}-x-10=8+\sqrt{x^{2}-x-16}$ The problem here is that if we isolate the radical expression an square, we will end up with an equation of degree $4$. \ It is worth a try, as we can solve some degree 4 equations, but chances are that this one would be tougher. Here is another method. Let us introduce a new variable, $a=\sqrt{x^{2}-x-16}$. \ Then $x^{2}-x-10$ on the left-hand side can be written as \ $x^{2}-x-10=x^{2}-x-16+6=a^{2}+6$. Substituting $a$ on both sides, our equation becomes $a^{2}+6=8+a$. \ Solve for $a$. \ Once you have $a$, solve for $x$. \end{enumerate} %TCIMACRO{% %\TeXButton{\rmfamily}{\rmfamily %\normalsize}}% %BeginExpansion \rmfamily \normalsize% %EndExpansion \vspace{0.15in} \pagebreak \FRAME{itbpF}{0.7922in}{0.6452in}{0.2811in}{}{}{sample.jpg}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.7922in;height 0.6452in;depth 0.2811in;original-width 8.4267in;original-height 6.8441in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../../book99/oldBook99/week6/exponents2/sample.jpg';file-properties "XNPEU";}% }\ \ {\LARGE Sample Problems}% %TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}% %BeginExpansion \setcounter{enumi}{0}\RESUME% %EndExpansion %TCIMACRO{\TeXButton{\begin{minipage}{1.7in}}{\begin{minipage}{1.7in}}}% %BeginExpansion \begin{minipage}{1.7in}% %EndExpansion \begin{enumerate} \item $\sqrt{3x-2}=x\bigskip \ $ \item $\sqrt{3x+4}+2=x$ \bigskip \item $10+\sqrt{4x-7}=7\bigskip ~$ \item $5+\sqrt{x+15}=x~$\bigskip\ \item $2\sqrt{x-1}=x-4~~$ \end{enumerate} %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{\TeXButton{\begin{minipage}{2.3in}}{\begin{minipage}{2.3in}}}% %BeginExpansion \begin{minipage}{2.3in}% %EndExpansion \begin{enumerate} \item[6.] $2\sqrt{x+4}=1+\sqrt{2x+9}~$\bigskip \item[7.] $5\sqrt{x}+1=3\sqrt{x}+17~~$\bigskip \item[8.] $\sqrt{2x+5}+5=x~$\bigskip \item[9.] $\sqrt{2x+5}-\sqrt{x-1}=\sqrt{x+2}$\bigskip \item[10.] $\sqrt{10x-1}+2=-x$ \end{enumerate} %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{\TeXButton{\begin{minipage}{2.4in}}{\begin{minipage}{2.4in}}}% %BeginExpansion \begin{minipage}{2.4in}% %EndExpansion \begin{enumerate} \item[11.] $\sqrt{3x+1}-\sqrt{x-4}=3$\bigskip \item[12.] $\sqrt{x+10}+10=x$\bigskip \item[13.] $\sqrt{4x-11}=\sqrt{x-1}+\sqrt{x-4}$\bigskip \item[14.] $\sqrt{4x+6}=\sqrt{x+1}-\sqrt{x+5}$\bigskip \bigskip \end{enumerate} %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \bigskip \bigskip \bigskip $\FRAME{itbpF}{0.7031in}{0.729in}{0.2413in}{}{}{work.jpg}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.7031in;height 0.729in;depth 0.2413in;original-width 4.6977in;original-height 4.875in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../../book99/oldBook99/week2/square_root/work.jpg';file-properties "XNPEU";}}$ \ \ {\LARGE Practice Problems}{\Large \ }% %TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}% %BeginExpansion \setcounter{enumi}{0}\RESUME% %EndExpansion %TCIMACRO{\TeXButton{\begin{minipage}{1.9in}}{\begin{minipage}{1.9in}}}% %BeginExpansion \begin{minipage}{1.9in}% %EndExpansion \begin{enumerate} \item $\sqrt{3x-5}=4$\vspace{0.13in} \item $2\sqrt{a-1}+7=1$\vspace{0.13in}\ \item $3\sqrt{7x+1}+2=20$\vspace{0.13in}\ \ \item $\sqrt{x+3}=x-9$\vspace{0.13in} \item $2\sqrt{x+5}=x-3$\vspace{0.13in}\ \item $\sqrt{2p+4}-\sqrt{p+3}=1$\vspace{0.13in} \item $\sqrt[3]{x^{3}+16}=x+4$\vspace{0.13in} \item $\sqrt{18+x}=x-2$ \end{enumerate} %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{\TeXButton{\begin{minipage}{2.3in}}{\begin{minipage}{2.3in}}}% %BeginExpansion \begin{minipage}{2.3in}% %EndExpansion \begin{enumerate} \item[9.] $\sqrt{w}+\sqrt{w+3}=3$\vspace{0.13in}\ \item[10.] $\sqrt{2x+1}+\sqrt{5-x}=4$\vspace{0.13in}\ \item[11.] $2\sqrt{x}-\sqrt{x-3}=\sqrt{x+7}$\vspace{0.13in} \item[12.] $2\sqrt{y+4}+\sqrt{y-5}=\sqrt{9y+7}$\vspace{0.13in}\ \item[13.] $\sqrt{b-2}+b=8$\vspace{0.13in} \item[14.] $\sqrt{3x+1}-\sqrt{x+4}=1$\vspace{0.13in} \item[15.] $\sqrt[3]{x-8}-\sqrt[3]{4x+1}=0$\vspace{0.13in} \item[16.] $\sqrt{5m-9}=\sqrt{5m}-3$ \end{enumerate} %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{\TeXButton{\begin{minipage}{2.4in}}{\begin{minipage}{2.4in}}}% %BeginExpansion \begin{minipage}{2.4in}% %EndExpansion \begin{enumerate} \item[17] $\sqrt{3k-5}-\sqrt{3k}=-1$\vspace{0.13in} \item[18.] $\sqrt{1-x}+1=x+12$\vspace{0.13in} \item[19.] $2=\sqrt{x^{2}+1}-x$\vspace{0.13in} \item[20.] $\sqrt{3x+1}-4=8-2\sqrt{3x+1}$\vspace{0.13in} \item[21.] $\sqrt{x}=\sqrt{10+3\sqrt{x}}$\vspace{0.13in} \item[22.] $\sqrt[3]{x^{3}+7}=x+1$\vspace{0.13in} \item[23.] $\sqrt{x-6}=\sqrt{x+2}-4$\vspace{0.13in} \item[24.] $\sqrt{6x+7}-\sqrt{3x+3}=1$\vspace{0.13in} \end{enumerate} %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \pagebreak \FRAME{itbpF}{0.8311in}{0.8311in}{0.2006in}{}{}{answers.jpg}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.8311in;height 0.8311in;depth 0.2006in;original-width 8.6455in;original-height 8.6455in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../2_algebra1/simplifying_algebraic_expressions/answers.jpg';file-properties "XNPEU";}% } \ \ \ {\LARGE Answers}% %TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}% %BeginExpansion \setcounter{enumi}{0}\RESUME% %EndExpansion \bigskip {\Large Sample \ Problems} \begin{enumerate} \item[1.] $1,2$ \ \ \ \ \ 2. \ $7$\ \ \ \ \ \ \ 3. \ no real solution \ \ \ \ 4. \ $10$ \ \ \ \ \ \ 5. \ $10$ \ \ \ \ \ \ 6. \ $0$ \ \ \ \ \ \ \ 7. \ $% 64 $ \ \ \ \ \ 8. \ $10$ \item[9.] $2$ \ \ \ \ \ 10. no solution\ \ \ \ 11. \ $5,8$ \ \ \ \ \ \ 12. \ $15$ \ \ \ \ \ \ \ 13. \ $5$ \ \ \ \ \ 14. \ no real solution\bigskip \end{enumerate} {\Large Practice \ Problems}% %TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}% %BeginExpansion \setcounter{enumi}{0}\RESUME% %EndExpansion \begin{enumerate} \item $7$ \ \ \ \ 2. \ no real solution\ \ \ \ \ \ 3. \ $5$\ \ \ \ \ \ \ 4. \ $13$ \ \ \ \ \ \ 5. \ $11$ \ \ \ \ \ 6. \ $6$ \ \ \ \ \ 7. \ $-2$ \ \ \ \ \ \ 8. \ $7$ \ \ \ \ \ 9. \ $1$ \item[10.] $4,\dfrac{20}{9}$ \ \ \ \ \ 11. \ $\dfrac{25}{8}$ \ \ \ \ \ 12. \ $21$ \ \ \ \ \ 13. \ $6$ \ \ \ \ \ \ 14. \ $5$ \ \ \ \ \ \ 15. \ $-3$ \ \ \ \ \ \ 16. \ $\dfrac{9}{5}$ \ \ \ \ \ \ 17. \ $3$ \item[18.] $-8$ \ \ \ \ \ \ 19. \ $-\dfrac{3}{4}$ \ \ \ \ \ 20. \ $5$ \ \ \ \ \ 21. \ $25$ \ \ \ \ \ 22. \ $-2,1$ \ \ \ \ \ 23. \ no real solution \ \ \ \ \ \ 24. \ $\dfrac{1}{3},-1$ \end{enumerate} \bigskip {\Large Enrichment}% %TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}% %BeginExpansion \setcounter{enumi}{0}\RESUME% %EndExpansion \begin{enumerate} \item $-4,5$ \end{enumerate} \pagebreak \begin{center} {\Large Sample Problems \FRAME{itbpF}{1.0957in}{0.6434in}{0.1712in}{}{}{% pencil.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 1.0957in;height 0.6434in;depth 0.1712in;original-width 1.9735in;original-height 1.1467in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../2_algebra1/simplifying_algebraic_expressions/pencil.bmp';file-properties "XNPEU";}% } Solutions}% %TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}% %BeginExpansion \setcounter{enumi}{0}\RESUME% %EndExpansion \end{center} \begin{enumerate} \item $\sqrt{3x-2}=x$ 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 \ \ \ \ LHS $=\sqrt{3\cdot 2-2}=\sqrt{4}=2=$ RHS \ $% \checkmark \vspace{0.05in}$ and if $x=1$, then \ \ \ \ \ LHS $=\sqrt{3\cdot 1-2}=\sqrt{1}=1=$ RHS \ $% \checkmark $ Thus the solutions are \fbox{$1$ and $2$} \vspace{0.05in} \item $\sqrt{3x+4}+2=x$ Solution:% \begin{eqnarray*} \sqrt{3x+4}+2 &=&x\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }2 \\ \sqrt{3x+4} &=&x-2\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square} \\ 3x+4 &=&\left( x-2\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }3 \\ 3x+4 &=&x^{2}-4x+4\text{ \ \ \ \ \ \ \ \ \ \ subtract }3x \\ 4 &=&x^{2}-7x+4\text{ \ \ \ \ \ \ \ \ \ \ subtract }4 \\ 0 &=&x^{2}-7x \\ 0 &=&x\left( x-7\right) \text{~~~}\Longrightarrow ~~x_{1}=2\text{, }x_{2}=1 \end{eqnarray*}% We check \ both candidates: if $x=0$, then% \begin{equation*} \text{LHS}=\sqrt{3\cdot 0+4}+2=\sqrt{4}+2=2+2=4\text{ \ \ \ \ and \ \ \ \ RHS }=2\text{ \ \ \ \ \ \ \ \ \ LHS}\not=\text{RHS} \end{equation*}% so $0$ is not a solution. \ If $x=7$, then% \begin{equation*} \text{LHS}=\sqrt{3\cdot 7+4}+2=\sqrt{25}+2=5+2=7=\text{RHS }\checkmark \end{equation*}% Therefore, \fbox{$7$} is the only solution.\vspace{0.05in} \item $10+\sqrt{4x-7}=7$ 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, we can already see there is \ \fbox{ no real solution }. \pagebreak \item $5+\sqrt{x+15}=x$ 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{equation*} \text{LHS}=5+\sqrt{1+15}=5+\sqrt{16}=5+4=9\text{ \ \ \ \ and \ \ \ RHS}=1% \text{ \ \ \ \ \ \ \ \ \ RHS }\not=\text{ LHS} \end{equation*} Thus $x=1$ is NOT a solution. \ \ If $x=10,$ then% \begin{equation*} \text{LHS}=5+\sqrt{10+15}=5+\sqrt{25}=5+5=10=\text{RHS }\checkmark \end{equation*}% Thus $x=$ \fbox{$10$} is the only solution. \item $2\sqrt{x-1}=x-4$ 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{equation*} \text{LHS}=2\sqrt{2-1}=2\sqrt{1}=2\cdot 1=2\text{ \ and \ RHS}=2-4=-2\text{ \ \ \ \ \ RHS }\not=\text{ LHS} \end{equation*}% Thus $x=2$ is NOT a solution. If $x=10,$ then% \begin{equation*} \text{LHS}=2\sqrt{10-1}=2\sqrt{9}=2\left( 3\right) =6=\text{RHS }\checkmark \end{equation*}% Thus \fbox{$x=10$} is the only solution. \pagebreak \item $2\sqrt{x+4}=1+\sqrt{2x+9}$ 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\text{ }\checkmark \end{equation*}% Thus $x=0$ is indeed a solution. \ 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 \fbox{$x=0$}. \item $5\sqrt{x}+1=3\sqrt{x}+17$ Solution: \ This equation is a bit different. \ Although there are two radical expressions, they are identical. \ This equation is linear in $\sqrt{% x}$. \ We can solve for $\sqrt{x}$ and then for $x$.% \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\text{ \ }\checkmark \end{equation*}% Thus $x=$\fbox{$64$} is indeed a solution. \item $\sqrt{2x+5}+5=x$ 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{equation*} \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{equation*}% Thus $x=2$ is NOT a solution. \ If $x=10,$ then% \begin{equation*} \text{LHS}=\sqrt{2\left( 10\right) +5}+5=\sqrt{20+5}+5=\sqrt{25}+5=5+5=10=% \text{RHS \ }\checkmark \end{equation*}% Thus \fbox{ $x=10$} is the only solution. \item $\sqrt{2x+5}-\sqrt{x-1}=\sqrt{x+2}$ Solution: \ This equation contains three different radical expressions. \ Our \ method still works. \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 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ LHS $=\sqrt{% 2\left( -3\right) +5}-\sqrt{\left( -3\right) -1}=\sqrt{-1}-\sqrt{-4}=$ undefined Since the left hand side is undefined, $x=-3$ is NOT a solution. \ \ If $% x=2, $ then \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ LHS $=\sqrt{% 2\left( 2\right) +5}-\sqrt{2-1}=\sqrt{9}-\sqrt{1}=3-1=2=$RHS $\checkmark $ Thus \fbox{$x=2$} is the only solution. \item $\sqrt{10x-1}+2=-x$ Solution: \begin{eqnarray*} \sqrt{10x-1}+2 &=&-x\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }2 \\ \sqrt{10x-1} &=&-2-x\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square} \\ 10x-1 &=&x^{2}+4x+4\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }10x\text{, add }1 \\ 0 &=&x^{2}-6x+5 \\ 0 &=&\left( x-1\right) \left( x-5\right) \text{~~~}\Longrightarrow ~~x_{1}=1% \text{ \ and\ \ }x_{2}=1 \end{eqnarray*}% We check both answers: if $x=1$, then \begin{equation*} \text{LHS}=\sqrt{10\cdot 1-1}=\sqrt{9}=3\text{ \ \ \ and \ \ RHS}=-1\text{ \ \ \ \ \ \ \ LHS}\not=\text{RHS} \end{equation*}% thus $1$ does not work. \ If $x=5$, then \begin{equation*} \text{LHS}=\sqrt{10\cdot 5-1}=\sqrt{49}=7\text{ \ \ \ and \ \ RHS}=-5\text{ \ \ \ \ \ \ \ LHS}\not=\text{RHS} \end{equation*}% thus $5$ does not work either. \ This equation has \fbox{no real solution}. \item $\sqrt{3x+1}-\sqrt{x-4}=3$ 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 }\checkmark \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 }\checkmark \end{equation*}% The solutions are \fbox{$5$ and $8$}. \item $\sqrt{x+10}+10=x~$ 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{equation*} \text{LHS}=\sqrt{6+10}+10=\sqrt{16}+10=4+10=14\text{ \ and \ RHS}=6\text{ \ \ \ \ \ \ \ \ LHS }\not=\text{ RHS} \end{equation*}% and if $x=15$, then \begin{equation*} \text{LHS}=\sqrt{15+10}+10=\sqrt{25}+10=5+10=15=\text{RHS }\checkmark \end{equation*}% since $x=6$ doesn't work, the only solution is \fbox{$15$}$.$ \item $\sqrt{4x-11}=\sqrt{x-1}+\sqrt{x-4}$ Solution:% \begin{eqnarray*} \sqrt{4x-11} &=&\sqrt{x-1}+\sqrt{x-4}\text{ \ \ \ \ \ \ \ \ \ square} \\ \left( \sqrt{4x-11}\right) ^{2} &=&\left( \sqrt{x-1}+\sqrt{x-4}\right) ^{2} \\ 4x-11 &=&\left( \sqrt{x-1}+\sqrt{x-4}\right) \left( \sqrt{x-1}+\sqrt{x-4}% \right) \text{ \ \ \ \ \ \ \ \ \ \ FOIL} \\ 4x-11 &=&\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\text{ \ \ \ \ \ \ combine like terms} \\ 4x-11 &=&2x-5+2\sqrt{x-1}\sqrt{x-4}\text{ \ \ \ \ \ \ subtract \ }2x \\ 2x-11 &=&-5+2\sqrt{x-1}\sqrt{x-4}\text{ \ \ \ \ \ \ add }5 \\ 2x-6 &=&2\sqrt{x-1}\sqrt{x-4} \\ 2\left( x-3\right) &=&2\sqrt{x-1}\sqrt{x-4}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }2 \\ x-3 &=&\sqrt{\left( x-1\right) \left( x-4\right) }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square} \\ \left( x-3\right) ^{2} &=&\left( x-1\right) \left( x-4\right) \text{ \ \ \ \ \ \ \ \ \ \ \ \ \ FOIL} \\ x^{2}-6x+9 &=&x^{2}-5x+4\text{ \ \ \ \ \ \ \ \ subtract }x^{2} \\ -6x+9 &=&-5x+4\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ add }6x \\ 9 &=&x+4\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }4 \\ 5 &=&x \end{eqnarray*}% We check:% \begin{equation*} \text{LHS}=\sqrt{5-1}+\sqrt{5-4}=\sqrt{4}+\sqrt{1}=2+1=3\text{ \ \ and \ \ RHS}=\sqrt{4x-11}=\sqrt{4\left( 5\right) -11}=\sqrt{9}=3 \end{equation*}% And so the solution is $\ $\fbox{$5$}. \item $\sqrt{4x+6}=\sqrt{x+1}-\sqrt{x+5}$ 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 \fbox{no real solution}% . \end{enumerate} \bigskip \vspace{1in}\vspace{0.5in}\bigskip \bigskip %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion \href{http://www.teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html% }{For more documents like this, visit our page at\ http://www.teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}