Solution 3.1:3a
From Förberedande kurs i matematik 1
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| - | {{  | + | First expand the expression  | 
| - | <  | + | |
| - | {{  | + | {{Displayed math||<math>\begin{align}  | 
| + | \bigl(\sqrt{5}-\sqrt{2}\bigr)\bigl(\sqrt{5}+\sqrt{2}\bigr)  | ||
| + | &= \sqrt{5}\cdot\sqrt{5} + \sqrt{5}\cdot\sqrt{2} - \sqrt{2}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2}\\[5pt]   | ||
| + | &= \sqrt{5}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2}\,\textrm{.}   | ||
| + | \end{align}</math>}}  | ||
| + | |||
| + | Because <math>\sqrt{5}</math> and <math>\sqrt{2}</math> are defined as those numbers which, when multiplied with themselves give 5 and 2 respectively, we have that  | ||
| + | |||
| + | {{Displayed math||<math>\sqrt{5}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2} = 5-2 = 3\,\textrm{.}</math>}}  | ||
| + | |||
| + | |||
| + | Note: The expansion of <math>\bigl(\sqrt{5}-\sqrt{2}\bigr)\bigl(\sqrt{5}+\sqrt{2}\bigr)</math> can also be done directly with the formula for difference of two squares <math>(a-b)(a+b) = a^{2} - b^{2}</math> using <math>a=\sqrt{5}</math> and <math>b=\sqrt{2}</math>.  | ||
Current revision
First expand the expression
| \displaystyle \begin{align}
 \bigl(\sqrt{5}-\sqrt{2}\bigr)\bigl(\sqrt{5}+\sqrt{2}\bigr) &= \sqrt{5}\cdot\sqrt{5} + \sqrt{5}\cdot\sqrt{2} - \sqrt{2}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2}\\[5pt] &= \sqrt{5}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2}\,\textrm{.} \end{align}  | 
Because \displaystyle \sqrt{5} and \displaystyle \sqrt{2} are defined as those numbers which, when multiplied with themselves give 5 and 2 respectively, we have that
| \displaystyle \sqrt{5}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2} = 5-2 = 3\,\textrm{.} | 
Note: The expansion of \displaystyle \bigl(\sqrt{5}-\sqrt{2}\bigr)\bigl(\sqrt{5}+\sqrt{2}\bigr) can also be done directly with the formula for difference of two squares \displaystyle (a-b)(a+b) = a^{2} - b^{2} using \displaystyle a=\sqrt{5} and \displaystyle b=\sqrt{2}.
