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}.