Solution 3.1:3a

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First expand the expression
First expand the expression
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{{Displayed math||<math>\begin{align}
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\bigl(\sqrt{5}-\sqrt{2}\bigr)\bigl(\sqrt{5}+\sqrt{2}\bigr)
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&= \sqrt{5}\cdot\sqrt{5} + \sqrt{5}\cdot\sqrt{2} - \sqrt{2}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2}\\[5pt]
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&= \sqrt{5}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2}\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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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
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& \left( \sqrt{5}-\sqrt{2} \right)\left( \sqrt{5}3\sqrt{2} \right)=\sqrt{5}\centerdot \sqrt{5}+\sqrt{5}\centerdot \sqrt{2}-\sqrt{2}\centerdot \sqrt{5}-\sqrt{2}\centerdot \sqrt{2} \\
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& =\sqrt{5}\centerdot \sqrt{5}-\sqrt{2}\centerdot \sqrt{2} \\
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\end{align}</math>
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{{Displayed math||<math>\sqrt{5}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2} = 5-2 = 3\,\textrm{.}</math>}}
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Because
 
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<math>\sqrt{5}</math>
 
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and
 
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<math>\sqrt{2}</math>
 
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are defined as those numbers which, when multiplied with themselves give
 
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<math>\text{5}</math>
 
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and
 
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<math>2</math> respectively,
 
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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>.
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<math>\sqrt{5}\centerdot \sqrt{5}-\sqrt{2}\centerdot \sqrt{2}=5-2=3</math>
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NOTE: The expansion of
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<math>\left( \sqrt{5}-\sqrt{2} \right)\left( \sqrt{5}3\sqrt{2} \right)</math>
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can also be done directly with the conjugate rule
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<math>\left( a-b \right)(a+b)=a^{\text{2}}-b^{\text{2}}</math>
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using
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<math>a=\sqrt{5}</math>
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and
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<math>b=\sqrt{2}</math>.
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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}.

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