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If $$z = \frac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }$$ find $\lfloor z \rfloor$.

 Aug 4, 2021
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If
\(z = \dfrac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} }\)
find \(\lfloor z \rfloor\).

 

\(\begin{array}{|rcll|} \hline z &=& \dfrac{ \left\{ \sqrt{3} \right\}^2 - 2 \left\{ \sqrt{2} \right\}^2 }{ \left\{ \sqrt{3} \right\} - 2 \left\{ \sqrt{2} \right\} } \\\\ z &=& \dfrac{ (\sqrt{3}-1)^2 - 2 (\sqrt{2}-1)^2 } { \sqrt{3}-1 - 2 (\sqrt{2}-1) } \\\\ z &=& \dfrac{ 3-2\sqrt{3}+1 -2(2-2\sqrt{2}+1) } { \sqrt{3}-2\sqrt{2}+1 } \\\\ z &=& \dfrac{ -2\sqrt{3}+4\sqrt{2}-2 } { \sqrt{3}-2\sqrt{2}+1 } \\\\ z &=& \dfrac{ -2(\sqrt{3}-2\sqrt{2}+1) } { \sqrt{3}-2\sqrt{2}+1 } \\\\ z &=& -2 \\\\ \lfloor z \rfloor &=& -2 \\ \hline \end{array}\)

 

laugh

 Aug 4, 2021

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