Solution 3.1:6a

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We use the standard method and augment the fraction with the conjugate of the denominator <math>\sqrt{5}+2</math>. Then the formula for the difference of two squares gives
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<center> [[Image:3_1_6a.gif]] </center>
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{{Displayed math||<math>\begin{align}
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\frac{\sqrt{2}+3}{\sqrt{5}-2}
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&= \frac{\sqrt{2}+3}{\sqrt{5}-2}\cdot\frac{\sqrt{5}+2}{\sqrt{5}+2}\\[5pt]
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&= \frac{(\sqrt{2}+3)(\sqrt{5}+2)}{(\sqrt{5})^{2}-2^{2}}\\[5pt]
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&= \frac{\sqrt{2}\cdot\sqrt{5}+\sqrt{2}\cdot 2+3\cdot \sqrt{5}+3\cdot 2}{5-4}\\[5pt]
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&= \sqrt{2\cdot 5} + 2\sqrt{2} + 3\sqrt{5} + 6\\[5pt]
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&= 6+2\sqrt{2}+3\sqrt{5}+\sqrt{10}\,\textrm{.}
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\end{align}</math>}}

Current revision

We use the standard method and augment the fraction with the conjugate of the denominator \displaystyle \sqrt{5}+2. Then the formula for the difference of two squares gives

\displaystyle \begin{align}

\frac{\sqrt{2}+3}{\sqrt{5}-2} &= \frac{\sqrt{2}+3}{\sqrt{5}-2}\cdot\frac{\sqrt{5}+2}{\sqrt{5}+2}\\[5pt] &= \frac{(\sqrt{2}+3)(\sqrt{5}+2)}{(\sqrt{5})^{2}-2^{2}}\\[5pt] &= \frac{\sqrt{2}\cdot\sqrt{5}+\sqrt{2}\cdot 2+3\cdot \sqrt{5}+3\cdot 2}{5-4}\\[5pt] &= \sqrt{2\cdot 5} + 2\sqrt{2} + 3\sqrt{5} + 6\\[5pt] &= 6+2\sqrt{2}+3\sqrt{5}+\sqrt{10}\,\textrm{.} \end{align}

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