Solution 3.1:6a

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We use the standard method and augment the fraction with the conjugate of the denominator
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<center> [[Image:3_1_6a.gif]] </center>
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<math>\sqrt{5}+2</math>. Then the conjugate rule gives
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<math>\begin{align}
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& \frac{\sqrt{2}+3}{\sqrt{5}-2}=\frac{\sqrt{2}+3}{\sqrt{5}-2}\centerdot \frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\left( \sqrt{2}+3 \right)\left( \sqrt{5}+2 \right)}{\left( \sqrt{5} \right)^{2}-2^{2}} \\
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& =\frac{\sqrt{2}\centerdot \sqrt{5}+\sqrt{2}\centerdot 2+3\centerdot \sqrt{5}+3\centerdot 2}{5-4}=\sqrt{2\centerdot 5}+2\sqrt{2}+3\sqrt{5}+6 \\
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& =6+2\sqrt{2}+3\sqrt{5}+10 \\
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\end{align}</math>

Revision as of 08:50, 23 September 2008

We use the standard method and augment the fraction with the conjugate of the denominator \displaystyle \sqrt{5}+2. Then the conjugate rule gives


\displaystyle \begin{align} & \frac{\sqrt{2}+3}{\sqrt{5}-2}=\frac{\sqrt{2}+3}{\sqrt{5}-2}\centerdot \frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\left( \sqrt{2}+3 \right)\left( \sqrt{5}+2 \right)}{\left( \sqrt{5} \right)^{2}-2^{2}} \\ & =\frac{\sqrt{2}\centerdot \sqrt{5}+\sqrt{2}\centerdot 2+3\centerdot \sqrt{5}+3\centerdot 2}{5-4}=\sqrt{2\centerdot 5}+2\sqrt{2}+3\sqrt{5}+6 \\ & =6+2\sqrt{2}+3\sqrt{5}+10 \\ \end{align}

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