\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=0.6in,right=0.7in,top=0.7in,bottom=0.7in} \pagestyle{fancy} \lhead{\color{blue} \large} \chead{\color{black} \Large Algebraic Transformations} \rhead{\ page \ \thepage} \cfoot{} \lfoot{\small \copyright $\;$ Hidegkuti, Powell, 2009} \textwidth 7.5in \textheight 9.4in \setlength{\headheight}{30pt} \setlength{\parindent}{0in} \begin{document} \begin{enumerate} \item Simplify each of the following expressions. a) \ $\dfrac{a^{2}-9}{a+2}\div \left( 1-\dfrac{5}{a+2}\right) $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ c) \ $\dfrac{1-\dfrac{x^{2}}{x^{2}-1}}{2+\dfrac{3x-1}{1-x}}$ \ \ \ where $% \left\vert x\right\vert \not=1$ b) $\ \dfrac{9a-3b}{9a^{2}-b^{2}}\cdot \dfrac{15a+5b}{3}$\ \ \ where \ $% \left\vert 3a\right\vert \not=\left\vert b\right\vert $ \item Find the exact value of each of the following expressions if $x=2,$ \ $% y=\sqrt{3},$ and $z=0.2009$ a) $\ \dfrac{1}{\left( x-y\right) \left( x-z\right) }+\dfrac{1}{\left( z-x\right) \left( z-y\right) }+\dfrac{1}{\left( y-x\right) \left( y-z\right) }$ \ \ \ \ \ \ d) \ $\dfrac{\left( x+y\right) ^{2}-\left( x-y\right) ^{2}}{% 4xy}$ b) $\ \dfrac{1}{x\left( x+z\right) }+\dfrac{1}{z\left( x+z\right) }+\dfrac{1% }{x\left( x-z\right) }+\dfrac{1}{z\left( z-x\right) }$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e) \ $\dfrac{\left( x^{2}-y^{2}-z^{2}-2yz\right) \left( x+y-z\right) }{% \left( x+y+z\right) \left( x^{2}+z^{2}-2xz-y^{2}\right) }$ c) $\ \left( \dfrac{2x^{2}+x}{x^{3}-1}-\dfrac{x+1}{x^{2}+x+1}\right) \left( 1+\dfrac{x+1}{x}-\dfrac{x^{2}+5x}{x^{2}+x}\right) $ \item Simplify each of the following expressions. a) $\ \dfrac{4-a^{2}-2ab-b^{2}}{2+a+b}$ \ where $a+b\not=-2$ \ \ \ \ \ \ \ \ b) $\ \dfrac{a^{2}+b^{2}-c^{2}+2ab}{a^{2}-b^{2}+c^{2}+2ac}$ \ \ where \ $% \left\vert a+c\right\vert \not=\left\vert b\right\vert $ \item Prove that if $a+b+c=0,$ then $a^{3}+a^{2}c+b^{2}c-abc+b^{3}=0$ \item Simplify each of the following expressions. a) \ $\sqrt{12}+\sqrt{75}-\sqrt{147}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g) \ $\sqrt{7-4\sqrt{3}}-\sqrt{7+\sqrt{48}}$ b) $\ \sqrt{28}+\sqrt{7}-\sqrt{63}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h) \ $\sqrt[3]{7+5\sqrt{2}}$ c) $\ \sqrt{\sqrt{41}+4\sqrt{2}}\cdot \sqrt{\sqrt{41}-\sqrt{32}}$ \ \ \ \ \ \ \ \ \ \ i) \ $\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}$ d) $\ \sqrt{5\sqrt{3}+\sqrt{59}}\cdot \sqrt{\sqrt{75}-\sqrt{59}}$ \ \ \ \ \ \ \ \ \ j) \ $\sqrt[3]{10+6\sqrt{3}}+\sqrt[3]{10-6\sqrt{3}}$ e) \ $\left( \sqrt{6+\sqrt{11}}+\sqrt{6-\sqrt{11}}\right) ^{2}$ \ \ \ \ \ \ \ \ \ \ \ k) \ $\sqrt[4]{7-4\sqrt{5}}$ f) \ $\sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}$ \item Simplify each of the following expressions. a) $\ \dfrac{3-\sqrt{5}}{3+\sqrt{5}}+\dfrac{3+\sqrt{5}}{3-\sqrt{5}}$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\ b) $\ \left( \dfrac{8}{\sqrt{7}+\sqrt{3}}+% \dfrac{12}{\sqrt{7}-\sqrt{3}}\right) \left( 5\sqrt{7}-\sqrt{3}\right) $ \item Rationalize the denominator in each of the following expressions. a) $\ \dfrac{3}{\sqrt{5}-\sqrt{2}}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ b) $\ \dfrac{% a}{\sqrt{a}+\sqrt{b}}$ \ \ where $a,b>0$ \ \ \ \ \ \ \ \ \ \ \ \ \ c) $\ \dfrac{\sqrt{7}-\sqrt{2}}{\sqrt{7}+\sqrt{2}}$ \item Which one is greater? a) $\ 2\sqrt{7}$ \ \ or \ \ $\dfrac{1}{\sqrt{7}-\sqrt{6}}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ c) $\ \dfrac{7}{5-3\sqrt{2}}$ \ \ \ or \ $\sqrt{72}$ b) $\ \sqrt[4]{4}$ \ \ \ or \ \ \ $\sqrt[5]{5}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ d) \ $2\sqrt{3}$ \ \ or \ \ $\dfrac{1}{\sqrt{3}-\sqrt{2}}$ \item For what values of $k$ \ can we factor out $x+3$ \ from the polynomial $2x^{2}+x+k$? \item Find the exact value of the following expression.% \begin{equation*} \dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2% \sqrt{2}} \end{equation*} \item We divided a line segment into two parts so that the ratio between the shorter and longer part is the same as the ratio between the longer part and the entire line segment. \ If $R$ represents this ratio, find the exact value of the following expression.% \begin{equation*} R^{\left( R^{\left( R^{2}+R^{-1}\right) }+R^{-1}\right) }+R^{-1} \end{equation*} \item Find the integer part in $\left( \sqrt{3}+\sqrt{2}\right) ^{6}$. \item If \ $p,~q,$ \ and $r$ are solutions of the equation \ $% x^{3}-x^{2}+x-2=0,$ the find the exact value of $p^{3}+q^{3}+r^{3}$. \item Is the number \ $\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}$ \ a solution of the equation \ $x^{3}-3x-14=0$?\pagebreak \end{enumerate} \begin{center} {\Large Answers\bigskip } \end{center} \begin{enumerate} \item Simplify each of the following expressions. \begin{enumerate} \item $\dfrac{a^{2}-9}{a+2}\div \left( 1-\dfrac{5}{a+2}\right) =a+3$\newline Solution: \ We first perform the subtraction and then divide by multiplying by the reciprocal.% \begin{eqnarray*} \dfrac{a^{2}-9}{a+2}\div \left( 1-\dfrac{5}{a+2}\right) &=&\dfrac{a^{2}-9}{% a+2}\div \left( \dfrac{a+2}{a+2}-\dfrac{5}{a+2}\right) =\dfrac{a^{2}-9}{a+2}% \div \dfrac{a+2-5}{a+2}=\dfrac{a^{2}-9}{a+2}\div \dfrac{a-3}{a+2} \\ &=&\dfrac{a^{2}-9}{a+2}\cdot \dfrac{a+2}{a-3}=\dfrac{a^{2}-9}{a-3}=\dfrac{% \left( a+3\right) \left( a-3\right) }{a-3}=a+3 \end{eqnarray*} \item $\dfrac{9a-3b}{9a^{2}-b^{2}}\cdot \dfrac{15a+5b}{3}=5$ \ \ \ where \ $% \left\vert 3a\right\vert \not=\left\vert b\right\vert $\newline Solution: \item $\dfrac{1-\dfrac{x^{2}}{x^{2}-1}}{2+\dfrac{3x-1}{1-x}}=\dfrac{1}{% \left( x+1\right) ^{2}}$ \ \ \ where $\left\vert x\right\vert \not=1$ \end{enumerate} \item Find the exact value of each of the following expressions if $x=2,$ \ $% y=\sqrt{3},$ and $z=0.2009$ \begin{enumerate} \item $\dfrac{1}{\left( x-y\right) \left( x-z\right) }+\dfrac{1}{\left( z-x\right) \left( z-y\right) }+\dfrac{1}{\left( y-x\right) \left( y-z\right) }=0$\newline Solution: \ \begin{equation*} \dfrac{1}{\left( x-y\right) \left( x-z\right) }+\dfrac{1}{\left( z-x\right) \left( z-y\right) }+\dfrac{1}{\left( y-x\right) \left( y-z\right) }= \end{equation*}% \begin{eqnarray*} \dfrac{1}{\left( x-y\right) \left( x-z\right) }+\dfrac{1}{\left( z-x\right) \left( z-y\right) }+\dfrac{1}{\left( y-x\right) \left( y-z\right) } &=& \\ \dfrac{y-z}{\left( x-y\right) \left( x-z\right) \left( y-z\right) }+\dfrac{% x-y}{\left( -1\right) \left( x-z\right) \left( -1\right) \left( y-z\right) \left( x-y\right) }+\dfrac{x-z}{\left( -1\right) \left( x-y\right) \left( y-z\right) \left( x-z\right) } &=& \\ \dfrac{\left( y-z\right) +\left( x-y\right) -\left( x-z\right) }{\left( x-y\right) \left( x-z\right) \left( y-z\right) } &=& \\ \dfrac{y-z+x-y-x+z}{\left( x-y\right) \left( x-z\right) \left( y-z\right) } &=&0 \end{eqnarray*} \item $\dfrac{1}{x\left( x+z\right) }+\dfrac{1}{z\left( x+z\right) }+\dfrac{1% }{x\left( x-z\right) }+\dfrac{1}{z\left( z-x\right) }=$\newline Solution:% \begin{eqnarray*} \dfrac{1}{x\left( x+z\right) }+\dfrac{1}{z\left( x+z\right) }+\dfrac{1}{% x\left( x-z\right) }+\dfrac{1}{z\left( z-x\right) } &=& \\ \dfrac{z\left( x-z\right) }{xz\left( x+z\right) \left( x-z\right) }+\dfrac{% x\left( x-z\right) }{xz\left( x+z\right) \left( x-z\right) }+\dfrac{z\left( x+z\right) }{xz\left( x+z\right) \left( x-z\right) }+\dfrac{x\left( x+z\right) }{zx\left( -1\right) \left( x-z\right) \left( x+z\right) } &=& \\ \dfrac{z\left( x-z\right) +x\left( x-z\right) +z\left( x+z\right) -x\left( x+z\right) }{xz\left( x+z\right) \left( x-z\right) } &=& \\ \dfrac{xz-z^{2}+x^{2}-xz+xz+z^{2}-x^{2}-xz}{xz\left( x+z\right) \left( x-z\right) } &=&0 \end{eqnarray*} \item $\left( \dfrac{2x^{2}+x}{x^{3}-1}-\dfrac{x+1}{x^{2}+x+1}\right) \left( 1+\dfrac{x+1}{x}-\dfrac{x^{2}+5x}{x^{2}+x}\right) =$% \begin{eqnarray*} \left( \dfrac{2x^{2}+x}{x^{3}-1}-\dfrac{x+1}{x^{2}+x+1}\right) \left( 1+% \dfrac{x+1}{x}-\dfrac{x^{2}+5x}{x^{2}+x}\right) &=& \\ \left( \dfrac{2x^{2}+x}{\left( x-1\right) \left( x^{2}+x+1\right) }-\dfrac{% \left( x-1\right) \left( x+1\right) }{\left( x-1\right) \left( x^{2}+x+1\right) }\right) \left( 1+\dfrac{x+1}{x}-\dfrac{x\left( x+5\right) }{x\left( x+1\right) }\right) &=& \\ \dfrac{2x^{2}+x-\left( x^{2}-1\right) }{\left( x-1\right) \left( x^{2}+x+1\right) }\left( \dfrac{x\left( x+1\right) }{x\left( x+1\right) }+% \dfrac{\left( x+1\right) ^{2}}{x\left( x+1\right) }-\dfrac{x\left( x+5\right) }{x\left( x+1\right) }\right) &=& \\ \dfrac{x^{2}+x+1}{\left( x-1\right) \left( x^{2}+x+1\right) }\cdot \dfrac{% x^{2}-2x+1}{x\left( x+1\right) } &=& \\ \dfrac{1}{\left( x-1\right) }\cdot \dfrac{\left( x-1\right) ^{2}}{x\left( x+1\right) } &=& \\ \dfrac{\left( x-1\right) }{x\left( x+1\right) } &=&\dfrac{1}{2\cdot 3}=% \dfrac{1}{6} \end{eqnarray*} \item $\dfrac{\left( x+y\right) ^{2}-\left( x-y\right) ^{2}}{4xy}=1$\newline Solution:% \begin{eqnarray*} \dfrac{\left( x+y\right) ^{2}-\left( x-y\right) ^{2}}{4xy} &=&\dfrac{% x^{2}+2xy+y^{2}-\left( x^{2}-2xy+y^{2}\right) }{4xy} \\ &=&\dfrac{x^{2}+2xy+y^{2}-x^{2}+2xy-y^{2}}{4xy}=\dfrac{4xy}{4xy}=1 \end{eqnarray*} \item $\dfrac{\left( x^{2}-y^{2}-z^{2}-2yz\right) \left( x+y-z\right) }{% \left( x+y+z\right) \left( x^{2}+z^{2}-2xz-y^{2}\right) }=1$\newline Solution:% \begin{eqnarray*} \dfrac{\left( x^{2}-y^{2}-z^{2}-2yz\right) \left( x+y-z\right) }{\left( x+y+z\right) \left( x^{2}+z^{2}-2xz-y^{2}\right) } &=&\dfrac{\left( x^{2}-\left( y+z\right) ^{2}\right) \left( x+y-z\right) }{\left( x+y+z\right) \left( \left( x-z\right) ^{2}-y^{2}\right) } \\ &=&\dfrac{\left( x+y+z\right) \left( x-y-z\right) \left( x+y-z\right) }{% \left( x+y+z\right) \left( x-z+y\right) \left( x-z-y\right) } \\ &=&\dfrac{\left( x-y-z\right) \left( x+y-z\right) }{\left( x-z+y\right) \left( x-z-y\right) } \\ &=&\dfrac{\left( x-y-z\right) \left( x+y-z\right) }{\left( x-y-z\right) \left( x+y-z\right) }=1 \end{eqnarray*} \end{enumerate} \item Simplify each of the following expressions. \begin{enumerate} \item $\dfrac{4-a^{2}-2ab-b^{2}}{2+a+b}$ \ where $a+b\not=-2$\newline Solution:% \begin{equation*} \dfrac{4-a^{2}-2ab-b^{2}}{2+a+b}=\dfrac{4-\left( a+b\right) ^{2}}{2+a+b}=% \dfrac{\left( 2-\left( a+b\right) \right) \left( 2+a+b\right) }{2+a+b}=2-a-b \end{equation*} \item $\dfrac{a^{2}+b^{2}-c^{2}+2ab}{a^{2}-b^{2}+c^{2}+2ac}$ \ \ where \ $% \left\vert a+c\right\vert \not=\left\vert b\right\vert $\newline Solution:% \begin{equation*} \dfrac{a^{2}+b^{2}-c^{2}+2ab}{a^{2}-b^{2}+c^{2}+2ac}=\dfrac{\left( a+b\right) ^{2}-c^{2}}{\left( a+c\right) ^{2}-b^{2}}=\dfrac{\left( a+b+c\right) \left( a+b-c\right) }{\left( a+c+b\right) \left( a+c-b\right) }=% \dfrac{a+b-c}{a-b+c} \end{equation*} \end{enumerate} \item Prove that if $a+b+c=0,$ then $a^{3}+a^{2}c+b^{2}c-abc+b^{3}=0$\newline Solution:% \begin{eqnarray*} a^{3}+a^{2}c+b^{2}c-abc+b^{3} &=&a^{3}+b^{3}+a^{2}c+b^{2}c-abc= \\ &=&\left( a+b\right) \left( a^{2}-ab+b^{2}\right) +c\left( a^{2}+b^{2}-ab\right) \\ &=&\left( a+b+c\right) \left( a^{2}-ab+b^{2}\right) \\ &=&0\left( a^{2}-ab+b^{2}\right) =0 \end{eqnarray*} \item Simplify each of the following expressions. \begin{enumerate} \item $\sqrt{12}+\sqrt{75}-\sqrt{147}=0$ \item $\sqrt{28}+\sqrt{7}-\sqrt{63}=0$ \item $\sqrt{\sqrt{41}+4\sqrt{2}}\cdot \sqrt{\sqrt{41}-\sqrt{32}}=3$\newline Solution:% \begin{eqnarray*} \sqrt{\sqrt{41}+4\sqrt{2}}\cdot \sqrt{\sqrt{41}-\sqrt{32}} &=&\sqrt{\sqrt{41}% +\sqrt{32}}\cdot \sqrt{\sqrt{41}-\sqrt{32}}=\sqrt{\left( \sqrt{41}+\sqrt{32}% \right) \left( \sqrt{41}-\sqrt{32}\right) } \\ &=&\sqrt{\left( \left( \sqrt{41}\right) ^{2}-\left( \sqrt{32}\right) ^{2}\right) }=\sqrt{41-32}=\sqrt{9}=3 \end{eqnarray*} \item $\sqrt{5\sqrt{3}+\sqrt{59}}\cdot \sqrt{\sqrt{75}-\sqrt{59}}=4$\newline Solution:% \begin{eqnarray*} \sqrt{5\sqrt{3}+\sqrt{59}}\cdot \sqrt{\sqrt{75}-\sqrt{59}} &=&\sqrt{\sqrt{75}% +\sqrt{59}}\cdot \sqrt{\sqrt{75}-\sqrt{59}}=\sqrt{\left( \sqrt{75}+\sqrt{59}% \right) \left( \sqrt{75}-\sqrt{59}\right) } \\ &=&\sqrt{\left( \left( \sqrt{75}\right) ^{2}-\left( \sqrt{59}\right) ^{2}\right) }=\sqrt{75-59}=\sqrt{16}=4 \end{eqnarray*} \item $\left( \sqrt{6+\sqrt{11}}+\sqrt{6-\sqrt{11}}\right) ^{2}=22$\newline Solution:% \begin{eqnarray*} \left( \sqrt{6+\sqrt{11}}+\sqrt{6-\sqrt{11}}\right) ^{2} &=&\left( \sqrt{6+% \sqrt{11}}\right) ^{2}+\left( \sqrt{6-\sqrt{11}}\right) ^{2}+2\sqrt{6+\sqrt{% 11}}\sqrt{6-\sqrt{11}} \\ &=&6+\sqrt{11}+6-\sqrt{11}+2\sqrt{\left( 6+\sqrt{11}\right) \left( 6-\sqrt{11% }\right) } \\ &=&12+2\sqrt{36-11}=12+2\sqrt{25}=12+2\cdot 5=22 \end{eqnarray*} \item $\sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}}=2$\newline Solution: \begin{eqnarray*} \sqrt{7+2\sqrt{6}}-\sqrt{7-2\sqrt{6}} &=&\sqrt{\left( \sqrt{6}+1\right) ^{2}}% -\sqrt{\left( \sqrt{6}-1\right) ^{2}}= \\ &=&\sqrt{6}+1-\left( \sqrt{6}-1\right) =\sqrt{6}+1-\sqrt{6}+1=2 \end{eqnarray*} \item $\sqrt{7-4\sqrt{3}}-\sqrt{7+\sqrt{48}}=-2\sqrt{3}$\newline Solution:% \begin{eqnarray*} \sqrt{7-4\sqrt{3}}-\sqrt{7+\sqrt{48}} &=&\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}% } \\ &=&\sqrt{\left( 2-\sqrt{3}\right) ^{2}}-\sqrt{\left( 2+\sqrt{3}\right) ^{2}} \\ &=&2-\sqrt{3}-\left( 2+\sqrt{3}\right) =2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3} \end{eqnarray*} \item $\sqrt[3]{7+5\sqrt{2}}=\sqrt{2}+1$\newline Solution:% \begin{equation*} \left( \sqrt{2}+1\right) ^{3}=\left( \sqrt{2}\right) ^{3}+3\left( \sqrt{2}% \right) ^{2}+3\left( \sqrt{2}\right) +1=2\sqrt{2}+6+3\sqrt{2}+1=7+5\sqrt{2} \end{equation*}% \begin{equation*} \sqrt[3]{7+5\sqrt{2}}=\sqrt{2}+1 \end{equation*} \item $\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=4$\newline Solution: \ $\left( 2+\sqrt{2}\right) ^{3}=20+14\sqrt{2}$ \ and \ $\left( 2-% \sqrt{2}\right) ^{3}=20-14\sqrt{2}$% \begin{equation*} \sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}=2+\sqrt{2}+2-\sqrt{2}=4 \end{equation*} \item $\sqrt[3]{10+6\sqrt{3}}+\sqrt[3]{10-6\sqrt{3}}=2$\newline Solution: \ $\left( 1+\sqrt{3}\right) ^{3}=10+6\sqrt{3}$ \ and \ $\left( 1-% \sqrt{3}\right) ^{3}=10-6\sqrt{3}$% \begin{equation*} \sqrt[3]{10+6\sqrt{3}}+\sqrt[3]{10-6\sqrt{3}}=1+\sqrt{3}+1-\sqrt{3}=2 \end{equation*} \item $\sqrt[4]{7-4\sqrt{5}}=\func{undefined}$% \begin{eqnarray*} \sqrt{5} &>&2 \\ -4\sqrt{5} &<&-8 \\ 7-4\sqrt{5} &<&-1 \end{eqnarray*} \end{enumerate} \item Simplify each of the following expressions. \begin{enumerate} \item $\dfrac{3-\sqrt{5}}{3+\sqrt{5}}+\dfrac{3+\sqrt{5}}{3-\sqrt{5}}=7$% \newline Solution: \begin{equation*} \dfrac{3-\sqrt{5}}{3+\sqrt{5}}+\dfrac{3+\sqrt{5}}{3-\sqrt{5}}=\dfrac{\left( 3-\sqrt{5}\right) ^{2}+\left( 3+\sqrt{5}\right) ^{2}}{\left( 3+\sqrt{5}% \right) \left( 3-\sqrt{5}\right) }=\dfrac{14-6\sqrt{5}+14+6\sqrt{5}}{9-5}=% \dfrac{28}{4}=7 \end{equation*} \item $\left( \dfrac{8}{\sqrt{7}+\sqrt{3}}+\dfrac{12}{\sqrt{7}-\sqrt{3}}% \right) \left( 5\sqrt{7}-\sqrt{3}\right) =172$\newline Solution: \begin{equation*} \left( \dfrac{8}{\sqrt{7}+\sqrt{3}}+\dfrac{12}{\sqrt{7}-\sqrt{3}}\right) \left( 5\sqrt{7}-\sqrt{3}\right) =\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \end{equation*}% \begin{eqnarray*} &=&\dfrac{8\left( \sqrt{7}-\sqrt{3}\right) +12\left( \sqrt{7}+\sqrt{3}% \right) }{\left( \sqrt{7}+\sqrt{3}\right) \left( \sqrt{7}-\sqrt{3}\right) }% \left( 5\sqrt{7}-\sqrt{3}\right) \\ &=&\dfrac{8\sqrt{7}-8\sqrt{3}+12\sqrt{7}+12\sqrt{3}}{7-3}\left( 5\sqrt{7}-% \sqrt{3}\right) \\ &=&\dfrac{20\sqrt{7}+4\sqrt{3}}{4}\left( 5\sqrt{7}-\sqrt{3}\right) =\left( 5% \sqrt{7}+\sqrt{3}\right) \left( 5\sqrt{7}-\sqrt{3}\right) \\ &=&25\left( 7\right) -3=175-3=172 \end{eqnarray*} \end{enumerate} \pagebreak \item Rationalize the denominator in each of the following expressions. \begin{enumerate} \item $\dfrac{3}{\sqrt{5}-\sqrt{2}}=\sqrt{5}+\sqrt{2}$ \item $\dfrac{a}{\sqrt{a}+\sqrt{b}}=\dfrac{a\left( \sqrt{a}-\sqrt{b}\right) }{a-b}$ \ \ where $a,b>0$ \item $\dfrac{\sqrt{7}-\sqrt{2}}{\sqrt{7}+\sqrt{2}}=\dfrac{9-2\sqrt{14}}{5}$ \end{enumerate} \item Which one is greater? \begin{enumerate} \item $2\sqrt{7}$ \ \ or \ \ $\dfrac{1}{\sqrt{7}-\sqrt{6}}$ \ \ \ \ \ \ \ \ \ \ $2\sqrt{7}$ \item $\sqrt[4]{4}$ \ \ \ or \ \ \ $\sqrt[5]{5}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\sqrt[4]{4}$ \item $\dfrac{7}{5-3\sqrt{2}}$ \ \ \ or \ $\sqrt{72}$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\dfrac{7}{5-3\sqrt{2}}$\ \newline Solution:% \begin{eqnarray*} \dfrac{7}{5-3\sqrt{2}} &=&\dfrac{7\left( 5+3\sqrt{2}\right) }{\left( 5-3% \sqrt{2}\right) \left( 5+3\sqrt{2}\right) }=\dfrac{7\left( 5+3\sqrt{2}% \right) }{5^{2}-\left( 3\sqrt{2}\right) ^{2}}=\dfrac{7\left( 5+3\sqrt{2}% \right) }{25-18}=5+3\sqrt{2} \\ \sqrt{72} &=&6\sqrt{2} \end{eqnarray*}% \begin{eqnarray*} 5+3\sqrt{2} &>&6\sqrt{2} \\ 5 &>&3\sqrt{2} \\ \sqrt{25} &>&\sqrt{18} \end{eqnarray*} \item $2\sqrt{3}$ \ \ or \ \ $\dfrac{1}{\sqrt{3}-\sqrt{2}}$ \ \ \ \ \ \ \ \ \ \ $2\sqrt{3}$ \end{enumerate} \item For what values of $k$ \ can we factor out $x+3$ \ from the polynomial $2x^{2}+x+k$? \ \ \ \ $-15$ \item Find the exact value of the following expression.% \begin{equation*} \dfrac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2% \sqrt{2}}=\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \end{equation*}% \begin{eqnarray*} &=&\dfrac{\sqrt{\left( \sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right) ^{2}}}{% \sqrt{\sqrt{5}+1}}-\sqrt{\left( \sqrt{2}-1\right) ^{2}} \\ &=&\dfrac{\sqrt{\left( \sqrt{5}+2\right) +\left( \sqrt{5}-2\right) +2\sqrt{% \sqrt{5}+2}\sqrt{\sqrt{5}-2}}}{\sqrt{\sqrt{5}+1}}-\left( \sqrt{2}-1\right) \\ &=&\dfrac{\sqrt{2\sqrt{5}+2\sqrt{\left( \sqrt{5}+2\right) \left( \sqrt{5}% -2\right) }}}{\sqrt{\sqrt{5}+1}}-\left( \sqrt{2}-1\right) \\ &=&\dfrac{\sqrt{2\sqrt{5}+2\sqrt{1}}}{\sqrt{\sqrt{5}+1}}-\left( \sqrt{2}% -1\right) =\dfrac{\sqrt{2\sqrt{5}+2}}{\sqrt{\sqrt{5}+1}}-\left( \sqrt{2}% -1\right) \\ &=&\dfrac{\sqrt{2}\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}}-\left( \sqrt{2}% -1\right) =\sqrt{2}-\left( \sqrt{2}-1\right) =1 \end{eqnarray*} \item We divided a line segment into two parts so that the ratio between the shorter and longer part is the same as the ratio between the longer part and the entire line segment. \ If $R$ represents this ratio, find the exact value of the following expression.% \begin{equation*} R^{\left( R^{\left( R^{2}+R^{-1}\right) }+R^{-1}\right) }+R^{-1} \end{equation*}% $2$ \item Find the integer part in $\left( \sqrt{3}+\sqrt{2}\right) ^{6}$. \ \ \ \ \ \ \ \ \ \ $969$\newline Solution: \ \begin{eqnarray*} \left( \sqrt{3}+\sqrt{2}\right) ^{6} &=&\left( \left( \sqrt{3}+\sqrt{2}% \right) ^{2}\right) ^{3}=\left( 5+2\sqrt{6}\right) ^{3} \\ &=&\left( 5+\sqrt{24}\right) ^{3}=125+3\left( 25\right) \sqrt{24}+3\left( 5\right) \left( 24\right) +24\sqrt{24} \\ &=&125+75\sqrt{24}+360+24\sqrt{24}=485+99\sqrt{24} \\ &=&485+\sqrt{235\,224}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\sqrt{% 235\,224}=484.\,\allowbreak 998\,96 \end{eqnarray*}% \begin{equation*} 485+484=969 \end{equation*} \item If \ $p,~q,$ \ and $r$ are solutions of the equation \ $% x^{3}-x^{2}+x-2=0,$ the find the exact value of $p^{3}+q^{3}+r^{3}$.% \begin{eqnarray*} x^{3}-x^{2}+x-2 &=&\left( x-p\right) \left( x-q\right) \left( x-r\right) \\ x^{3}-x^{2}+x-2 &=&x^{3}+x^{2}\left( -p-q-r\right) +x\left( pq+pr+qr\right) -pqr \\ -1 &=&-p-q-r \\ 1 &=&pq+pr+qr \\ -2 &=&-pqr \end{eqnarray*}% \begin{eqnarray*} 1 &=&p+q+r \\ 1 &=&pq+pr+qr \\ 2 &=&pqr \end{eqnarray*}% Let us first find the exact value of $p^{2}+q^{2}+r^{2}$. \ Since \ $\left( p+q+r\right) ^{2}=\allowbreak 2pq+2pr+2qr+p^{2}+q^{2}+r^{2},$ we have that \begin{equation*} p^{2}+q^{2}+r^{2}=\left( p+q+r\right) ^{2}-2\left( pq+pr+qr\right) =1^{2}-2\left( 1\right) =-1 \end{equation*}% which shows that not all three of $p$, $q$, and $r$ can be real. \ We start with $\left( p+q+r\right) ^{3}$.% \begin{equation*} \left( p+q+r\right) ^{3}=p^{3}+q^{3}+r^{3}+3pq^{2}+3p^{2}q+3pr^{2}+3\allowbreak p^{2}r+3qr^{2}+3q^{2}r+6pqr \end{equation*}% Let us introduce a new variable $A=pq^{2}+p^{2}q+pr^{2}+p^{2}r+qr^{2}+q^{2}r$% . \ With this notation,% \begin{eqnarray*} p^{3}+q^{3}+r^{3} &=&\left( p+q+r\right) ^{3}-3A-6pqr \\ &=&1^{3}-3A-6\left( 2\right) \\ &=&-11-3A \end{eqnarray*}% Let us now similarly compute $\left( p+q+r\right) \left( p^{2}+q^{2}+r^{2}\right) $.% \begin{equation*} \left( p+q+r\right) \left( p^{2}+q^{2}+r^{2}\right) =p^{3}+q^{3}+r^{3}+pq^{2}+p^{2}q+pr^{2}+p^{2}r+qr^{2}+q^{2}\allowbreak r \end{equation*}% and so \begin{eqnarray*} p^{3}+q^{3}+r^{3} &=&\left( p+q+r\right) \left( p^{2}+q^{2}+r^{2}\right) -\left( pq^{2}+p^{2}q+pr^{2}+p^{2}r+qr^{2}+q^{2}\allowbreak r\right) \\ &=&1\cdot \left( -1\right) -A=-1-A \end{eqnarray*}% If we denote $p^{3}+q^{3}+r^{3}$ by $X,$ we have a simple system of linear equations we can solve% \begin{eqnarray*} X &=&-11-3A \\ X &=&-1-A \end{eqnarray*}% and obtain that $A=-5$ and $X=4$. \item Is the number \ $\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}$ \ a solution of the equation \ $x^{3}-3x-14=0$?\newline Solution: Let us denote $\sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}$ by $x$ and\ compute $x^{3}$ first.% \begin{equation*} \left( \sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right) ^{3}= \end{equation*}% \begin{eqnarray*} &=&\left( \sqrt[3]{7+4\sqrt{3}}\right) ^{3}+3\left( \sqrt[3]{7+4\sqrt{3}}% \right) ^{2}\left( \sqrt[3]{7-4\sqrt{3}}\right) + \\ &&\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }+3\left( \sqrt[3]{7+4\sqrt{3}}\right) \left( \sqrt[3]{7-4\sqrt{3}}% \right) ^{2}+\left( \sqrt[3]{7-4\sqrt{3}}\right) ^{3} \\ &=&7+4\sqrt{3}+3\sqrt[3]{\left( 7+4\sqrt{3}\right) ^{2}\left( 7-4\sqrt{3}% \right) }+3\sqrt[3]{\left( 7+4\sqrt{3}\right) \left( 7-4\sqrt{3}\right) ^{2}}% +7-4\sqrt{3} \\ &=&14+3\left( \sqrt[3]{\left( 7+4\sqrt{3}\right) ^{2}\left( 7-4\sqrt{3}% \right) }+\sqrt[3]{\left( 7+4\sqrt{3}\right) \left( 7-4\sqrt{3}\right) ^{2}}% \right) \\ &=&14+3\sqrt[3]{\left( 7+4\sqrt{3}\right) \left( \sqrt[3]{7-4\sqrt{3}}% \right) }\left( \sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}\right) \\ &=&14+3\sqrt[3]{49-48}\left( \sqrt[3]{7+4\sqrt{3}}+\sqrt[3]{7-4\sqrt{3}}% \right) \\ &=&14+3\cdot 1\cdot x=14+3x \end{eqnarray*} Since we have that $x^{3}=3x+14,$ it is indeed the solution of the equation \ $x^{3}-3x-14=0$. \end{enumerate} \vspace{0.5in}\vspace{0.4in}\vspace{2in}\vspace{0.2in} %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}