This is an extension of the things you learned in \emph{Topic 1: Algebra}. We will be using 3 main results and then replacing the $a$'s and $b$'s with \emph{surds}. \begin{align} & (a+b)^{2}=a^{2}+2 a b+b^{2} \\ & (a-b)^{2}=a^{2}-2 a b+b^{2} \\ & (a-b)(a+b)=a^{2}-b^{2} \end{align} \begin{examplebox} \subsection{Examples:} \begin{questions} \Question[5] \begin{doublespace} \begin{parts} \part \[(\sqrt{5}+\sqrt{3})^{2}=\] \begin{solutionordottedlines}[2cm] $(\sqrt{5}+\sqrt{3})^{2} = (\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{3}) + (\sqrt{3})^2 = 5 + 2\sqrt{15} + 3 = 8 + 2\sqrt{15}$ \end{solutionordottedlines} \part \[(3 \sqrt{2}-2 \sqrt{3})^{2}=\] \begin{solutionordottedlines}[2cm] $(3 \sqrt{2}-2 \sqrt{3})^{2} = (3\sqrt{2})^2 - 2(3\sqrt{2})(2\sqrt{3}) + (2\sqrt{3})^2 = 9\cdot2 - 2\cdot6\sqrt{6} + 4\cdot3 = 18 - 12\sqrt{6} + 12$ \end{solutionordottedlines} \part \[(2 \sqrt{3}+4 \sqrt{6})^{2}=\] \begin{solutionordottedlines}[2cm] $(2 \sqrt{3}+4 \sqrt{6})^{2} = (2\sqrt{3})^2 + 2(2\sqrt{3})(4\sqrt{6}) + (4\sqrt{6})^2 = 4\cdot3 + 2\cdot8\sqrt{18} + 16\cdot6 = 12 + 16\sqrt{18} + 96$ \end{solutionordottedlines} \part \[(\sqrt{7}-5)(\sqrt{7}+5)=\] \begin{solutionordottedlines}[2cm] $(\sqrt{7}-5)(\sqrt{7}+5) = (\sqrt{7})^2 - (5)^2 = 7 - 25 = -18$ \end{solutionordottedlines} \part \[(5 \sqrt{6}-2 \sqrt{5})(5 \sqrt{6}+2 \sqrt{5})=\] \begin{solutionordottedlines}[2cm] $(5 \sqrt{6}-2 \sqrt{5})(5 \sqrt{6}+2 \sqrt{5}) = (5\sqrt{6})^2 - (2\sqrt{5})^2 = 25\cdot6 - 4\cdot5 = 150 - 20 = 130$ \end{solutionordottedlines} \end{parts} \end{doublespace} \end{questions} \end{examplebox} \begin{exercisebox} \subsection{Exercises:} \begin{questions} \Question[12] \begin{parts}\begin{multicols}{2} \part \((5+\sqrt{3})^{2}\) \begin{solutionordottedlines}[2cm] $(5+\sqrt{3})^{2} = 5^2 + 2(5)(\sqrt{3}) + (\sqrt{3})^2 = 25 + 10\sqrt{3} + 3 = 28 + 10\sqrt{3}$ \end{solutionordottedlines} \part \((\sqrt{2}+6)^{2}\) \begin{solutionordottedlines}[2cm] $(\sqrt{2}+6)^{2} = (\sqrt{2})^2 + 2(\sqrt{2})(6) + 6^2 = 2 + 12\sqrt{2} + 36 = 38 + 12\sqrt{2}$ \end{solutionordottedlines} \part \((5 \sqrt{6}+2 \sqrt{3})^{2}\) \begin{solutionordottedlines}[2cm] $(5 \sqrt{6}+2 \sqrt{3})^{2} = (5\sqrt{6})^2 + 2(5\sqrt{6})(2\sqrt{3}) + (2\sqrt{3})^2 = 25\cdot6 + 2\cdot10\sqrt{18} + 4\cdot3 = 150 + 20\sqrt{18} + 12$ \end{solutionordottedlines} \part \((\sqrt{21}+\sqrt{3})^{2}\) \begin{solutionordottedlines}[2cm] $(\sqrt{21}+\sqrt{3})^{2} = (\sqrt{21})^2 + 2(\sqrt{21})(\sqrt{3}) + (\sqrt{3})^2 = 21 + 2\sqrt{63} + 3 = 24 + 2\sqrt{63}$ \end{solutionordottedlines} \part \((4+2 \sqrt{5})^{2}\) \begin{solutionordottedlines}[2cm] $(4+2 \sqrt{5})^{2} = 4^2 + 2(4)(2\sqrt{5}) + (2\sqrt{5})^2 = 16 + 16\sqrt{5} + 4\cdot5 = 16 + 16\sqrt{5} + 20 = 36 16\sqrt{5}$ \end{solutionordottedlines} \part \((\sqrt{7}-2)^{2}\) \begin{solutionordottedlines}[2cm] $(\sqrt{7}-2)^{2} = (\sqrt{7})^2 - 2(\sqrt{7})(2) + 2^2 = 7 - 4\sqrt{7} + 4 = 11 - 4\sqrt{7}$ \end{solutionordottedlines} \part \((4-\sqrt{3})^{2}\) \begin{solutionordottedlines}[2cm] $(4-\sqrt{3})^{2} = 4^2 - 2(4)(\sqrt{3}) + (\sqrt{3})^2 = 16 - 8\sqrt{3} + 3 = 19 - 8\sqrt{3}$ \end{solutionordottedlines} \part \((2 \sqrt{35}+\sqrt{5})^{2}\) \begin{solutionordottedlines}[2cm] $(2 \sqrt{35}+\sqrt{5})^{2} = (2\sqrt{35})^2 + 2(2\sqrt{35})(\sqrt{5}) + (\sqrt{5})^2 = 4\cdot35 + 2\cdot2\sqrt{175} 5 = 140 + 4\sqrt{175} + 5$ \end{solutionordottedlines} \part \((\sqrt{70}-3 \sqrt{10})^{2}\) \begin{solutionordottedlines}[2cm] $(\sqrt{70}-3 \sqrt{10})^{2} = (\sqrt{70})^2 - 2(\sqrt{70})(3\sqrt{10}) + (3\sqrt{10})^2 = 70 - 2\cdot3\sqrt{700} + 9\cdot10 = 70 - 6\sqrt{700} + 90$ \end{solutionordottedlines} \part \((3-\sqrt{5})(3+\sqrt{5})\) \begin{solutionordottedlines}[2cm] $(3-\sqrt{5})(3+\sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4$ \end{solutionordottedlines} \part \((2 \sqrt{5}-1)^{2}\) \begin{solutionordottedlines}[2cm] $(2 \sqrt{5}-1)^{2} = (2\sqrt{5})^2 - 2(2\sqrt{5})(1) + 1^2 = 4\cdot5 - 2\cdot2\sqrt{5} + 1 = 20 - 4\sqrt{5} + 1 = 2 - 4\sqrt{5}$ \end{solutionordottedlines} \part \((\sqrt{6}-1)(\sqrt{6}+1)\) \begin{solutionordottedlines}[2cm] $(\sqrt{6}-1)(\sqrt{6}+1) = (\sqrt{6})^2 - 1^2 = 6 - 1 = 5$ \end{solutionordottedlines} %\item \((2 \sqrt{3}-5 \sqrt{6})(2 \sqrt{3}+5 \sqrt{6})\) % \mdots{2} %\item \((\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})\) % \mdots{2} \end{multicols}\end{parts} \end{questions} \end{exercisebox}