\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} \usepackage{cancel} \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, 2016} \rfoot{\footnotesize Last revised: March 18, 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 eqaution, 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 just created.$\vspace{0.05in}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\sqrt{x-1}=-x+3$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square both sides$\vspace{0.05in}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $x-1=\left( -x+3\right) ^{2}$ \ \ \ \ \ \ \ \ \ \ \ \ \ expand complete square$\vspace{0.05in}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $x-1=x^{2}-6x+9$ \ \ \ \ \ \ \ \ \ \ \ \ subtract $x\vspace{% 0.05in}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $-1=x^{2}-7x+9$ \ \ \ \ \ \ \ \ \ \ \ \ add $1\vspace{% 0.05in}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $0=x^{2}-7x+10\vspace{0.05in}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $0=\left( x-2\right) \left( x-5\right) \vspace{% 0.05in}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\ \ \ \ \ \ x_{1}=2\vspace{0.05in}$ \ \ \ $x_{2}=5\vspace{0.05in}$ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion It appears that this equation has two solutions, $2$ and $5$. \ Let us check.% $\vspace{0.05in}$ %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.$\vspace{0.05in}$ %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 this equation. \ Thus this equation has one solution, \fbox{$x=2$}.$\vspace{0.05in}$ \pagebreak %TCIMACRO{% %\TeXButton{\vissza}{\vissza %\setstretch{1.16}}}% %BeginExpansion \vissza \setstretch{1.16}% %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. 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 subtracting $x$.% \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$\vspace{0.05in}$ not a solution. \ The only solution is \fbox{% $x=6$}.$\vspace{0.05in}$ %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} \pagebreak %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{0.2in} \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{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion \begin{enumerate} \item $\sqrt{3x-2}=x\vspace{0.11in}\ $ \item $x+2\sqrt{2x+16}=-2\vspace{0.11in}$ \item $10+\sqrt{4x-7}=7\vspace{0.11in}$ \item $5+\sqrt{x+15}=x\vspace{0.11in}$ \item $2\sqrt{x-1}=x-4\vspace{0.11in}$\ \item $x=2\sqrt{x+8}\vspace{0.11in}$ \item $5\sqrt{x}+1=3\sqrt{x}+17\vspace{0.11in}$ \item $\sqrt{2x+5}+5=x\vspace{0.11in}$ \item $2\sqrt{2x+29}=x+12\vspace{0.11in}$ \item $\sqrt{x+10}+10=x\vspace{0.11in}$ \end{enumerate} %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \bigskip \bigskip \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{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion \begin{enumerate} \item $\sqrt{3x-5}=4\vspace{0.11in}$ \item $2\sqrt{a-1}+7=1\vspace{0.11in}$\ \item $3\sqrt{7x+1}+2=20\vspace{0.11in}$\ \ \item $\sqrt{x+3}=x-9\vspace{0.11in}$\ \item $2\sqrt{x+5}=x-3\vspace{0.11in}$\ \item $p+1=3\sqrt{p+1}\vspace{0.11in}$ \item $2\sqrt{x+5}=x+6\vspace{0.11in}$\ \item $\sqrt{18+x}=x-2\vspace{0.11in}$\ \item $w+13=2\sqrt{3w+30}\vspace{0.11in}$\ \item $3\sqrt{4x-39}=x-3\vspace{0.11in}$ \item $-4a=3\sqrt{2a+22}\vspace{0.11in}$ \item $2y-3=\sqrt{2y+3}\vspace{0.11in}$\ \item $\sqrt{b-2}+b=8\vspace{0.11in}$ \item $\sqrt{3x+1}-x=-1\vspace{0.11in}$ \item $2\sqrt{x+1}=x+2\vspace{0.11in}$ \item $m-4=2\sqrt{2m-8}\vspace{0.11in}$ \item $-k-1=-2\sqrt{k+16}\vspace{0.11in}$ \item $\sqrt{1-x}+1=x+12\vspace{0.11in}$ \item $2=\sqrt{x^{2}+1}-x\vspace{0.11in}$ \item $3\sqrt{4t-35}=t-6\vspace{0.11in}$ \end{enumerate} %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \bigskip \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 \vspace{0.2in} {\Large Sample \ Problems}% %TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}% %BeginExpansion \setcounter{enumi}{0}\RESUME% %EndExpansion \begin{enumerate} \item $1,2$ \ \ \ \ \ 2.\ $-6$\ \ \ \ 3.\ no real solution \ \ \ \ 4.\ $10$ \ \ \ \ 5.\ $10$ \ \ \ \ \ 6.\ $8$ \ \ \ \ \ 7. \ $64$ \ \ \ \ \ 8.\ $10$ \ \ \ 9. $-2$ \ \ \ \ 10.\ $15$\vspace{0.14in} \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.\ $-1,8$ \ \ \ \ \ 7.\ $-4$ \ \ \ \ \ \ 8.\ $7$ \ \ \ \ \ 9.\ $-7$ \item[10.] $12,30$ \ \ \ \ \ 11.\ $-3$ \ \ \ \ \ 12.\ $3$ \ \ \ \ \ 13.\ $6$ \ \ \ \ \ \ 14.\ $5$ \ \ \ \ \ \ 15.\ $0$ \ \ \ \ \ \ 16.\ $4,12$ \ \ \ \ \ \ 17.\ $9$ \ \ \ \ \ 18.\ $-8$ \item[19.] $-\dfrac{3}{4}$ \ \ \ \ \ 20.\ $9,39$\vspace{0.14in} \end{enumerate} {\Large Enrichment}% %TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}% %BeginExpansion \setcounter{enumi}{0}\RESUME% %EndExpansion \begin{enumerate} \item $-4,5$\vspace{0.14in}\pagebreak \end{enumerate} \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) ~~~~\Longrightarrow ~~~~~~x_{1}=2% \text{ \ \ and \ }x_{2}=1 \end{eqnarray*}% We check: if $x=2$, then LHS $=\sqrt{3\left( 2\right) -2}=\sqrt{4}=2=$ RHS $% \checkmark $ and if $x=1$, then LHS$=\sqrt{3\left( 1\right) -2}=\sqrt{1}=1=$ RHS $% \checkmark $ \ \ Therefore, \fbox{$1$ and $2$} are solutions of the equation. \item $x+2\sqrt{2x+16}=-2$ Solution:% \begin{eqnarray*} x+2\sqrt{2x+16} &=&-2\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ add }2\text{ and subtract }2\sqrt{2x+16} \\ x+2 &=&-2\sqrt{2x+16}\text{ \ \ \ \ \ \ \ \ \ \ \ \ square both sides} \\ \left( x+2\right) ^{2} &=&\left( -2\sqrt{2x+16}\right) ^{2} \\ x^{2}+4x+4 &=&4\left( 2x+16\right) \\ x^{2}+4x+4 &=&8x+64\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }\left( 8x+64\right) \\ x^{2}-4x-60 &=&0\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ factor} \\ \left( x-10\right) \left( x+6\right) &=&0 \end{eqnarray*}% \begin{equation*} x_{1}=10\text{ and }x_{2}=-6 \end{equation*}% We check: if $x=10$, then% \begin{equation*} \text{LHS}=10+2\sqrt{2\cdot 10+16}=10+2\sqrt{36}=10+12=22\text{ \ \ \ and \ \ RHS}=-2\text{ \ \ \ \ \ \ LHS}\not=\text{RHS} \end{equation*}% Thus $x=10$ is NOT a solution. \ If $x=-6,$ then% \begin{equation*} \text{LHS}=-6+2\sqrt{2\left( -6\right) +16}=-6+2\sqrt{4}=-6+2\cdot 2=-2=% \text{RHS }\checkmark \end{equation*}% Thus the only solution is \fbox{$-6$}. \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, there is \fbox{no real solution}. \pagebreak \item $5+\sqrt{x+15}=x$ \ %TCIMACRO{% %\TeXButton{heading for subsection}{\thispagestyle{fancy} %\lhead{ \small Solutions for 23.1 -- Radical Equations} %\rhead{\footnotesize page \thepage}}}% %BeginExpansion \thispagestyle{fancy} \lhead{ \small Solutions for 23.1 -- Radical Equations} \rhead{\footnotesize page \thepage}% %EndExpansion 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 \fbox{$x=10$} is the only solution.\vspace{0.12in} \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\left( 1\right) =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{ \ and \ RHS}% =10-4=6=\text{RHS }\checkmark ~~~~~~~~~~~ \end{equation*}% Thus \fbox{$x=10$} is the only solution. \item $x=2\sqrt{x+8}$ %TCIMACRO{\TeXButton{\begin{minipage}{1.1in}}{\begin{minipage}{0.8in}}}% %BeginExpansion \begin{minipage}{0.8in}% %EndExpansion Solution:\vspace{0.1in}\vspace{0.07in}\vspace{0.07in}\vspace{0.07in}\vspace{% 0.07in}\vspace{0.07in}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \ \ %TCIMACRO{\TeXButton{\begin{minipage}{2.5in}}{\begin{minipage}{2.5in}}}% %BeginExpansion \begin{minipage}{2.5in}% %EndExpansion $x=2\sqrt{x+8}$\vspace{0.07in} \ \ \ square both sides $x^{2}=\left( 2\sqrt{x+8}\right) ^{2}$\vspace{0.07in} $x^{2}=4\left( x+8\right) $% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \vspace{0.07in} \ \ \ \ %TCIMACRO{\TeXButton{\begin{minipage}{2.5in}}{\begin{minipage}{2.5in}}}% %BeginExpansion \begin{minipage}{2.5in}% %EndExpansion $x^{2}-4x-32=0$\vspace{0.07in} $\left( x-8\right) \left( x+4\right) =0$\vspace{0.07in} $x_{1}=8$ and $x_{2}=-4$% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion We check: If $x=8,$ then LHS $=8$\ and RHS$=2\sqrt{8+8}=2\sqrt{16}=2\cdot 4=8 $ $\checkmark $. \ \ Thus $x=8$ is indeed a solution. \ If $x=-4,$ then% \begin{equation*} \text{RHS}=2\sqrt{-4+8}=2\sqrt{4}=2\cdot 2=4=\text{LHS \ \ \ \ \ \ RHS }\not=% \text{ LHS} \end{equation*}% Thus $x=-4$ is NOT a solution. The only solution is \fbox{$x=8$}. \item $5\sqrt{x}+1=3\sqrt{x}+17$ 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\text{ }\checkmark \end{equation*}% Thus \fbox{$64$} is indeed a solution.\vspace{0.07in}\vspace{0.07in} \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{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. \ 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. \pagebreak \item $2\sqrt{2x+29}=x+12$ Solution:% \begin{eqnarray*} 2\sqrt{2x+29} &=&x+12\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ square both sides} \\ \left( 2\sqrt{2x+29}\right) ^{2} &=&\left( x+12\right) ^{2}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 4\left( 2x+29\right) &=&\left( x+12\right) ^{2} \\ 8x+116 &=&x^{2}+24x+144\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ reduce one side to zero} \\ 0 &=&x^{2}+16x+28 \\ 0 &=&\left( x+14\right) \left( x+2\right) ~~\Longrightarrow ~~x_{1}=-14\text{ \ \ and \ \ \ }x_{2}=-2 \end{eqnarray*}% We check: if $x=-14$, then \begin{equation*} \text{LHS}=2\sqrt{2\left( -14\right) +29}=2\sqrt{1}=2\text{ \ \ and \ RHS}% =-14+12=-2\text{ \ \ \ LHS }\not=\text{ RHS} \end{equation*}% Since the two sides are not equal, $x=-14$ is NOT a solution. \ If $x=-2$, then \begin{equation*} \text{LHS}=2\sqrt{2\left( -2\right) +29}=2\sqrt{25}=2\cdot 5=10\text{ \ \ and \ RHS}=-12+12=10\text{ }\checkmark \end{equation*}% Thus \fbox{$-2$} is the only solution. \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*}% 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{$15$}$.$% %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion \vspace{0.22in}\vspace{0.2in}\vspace{1in} \end{enumerate} \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}