Solution 3.1:5c

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The trick is to use the conjugate rule
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The trick is to use the formula for the difference of two squares
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<math>\left( a-b \right)(a+b)=a^{\text{2}}-b^{\text{2}}</math>
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<math>(a-b)(a+b) = a^{2}-b^{2}</math> and multiply the top and bottom of the fraction by <math>3-\sqrt{7}</math> (note the minus sign), since then the new denominator will be <math>(3+\sqrt{7})(3-\sqrt{7}) = 3^{2} - (\sqrt{7})^{2} = 9-7 = 2</math> (the formula with <math>a=3</math> and <math>b=\sqrt{7}\,</math>), i.e. the root sign is squared away.
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and multiply the top and bottom of the fraction by
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<math>3-\sqrt{7}</math>
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(note the minus sign), since then the new denominator will be
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<math>\left( 3+\sqrt{7} \right)\left( 3-\sqrt{7} \right)=3^{2}-\left( \sqrt{7} \right)^{2}=9-7=2</math>
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(conjugate rule with
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<math>a=\text{3 }</math>
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and
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<math>b=\sqrt{\text{7}}</math>
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), i.e. the root sign is squared away.
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The whole calculation is
The whole calculation is
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\frac{2}{3+\sqrt{7}}
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& \frac{2}{3+\sqrt{7}}=\frac{2}{3+\sqrt{7}}\centerdot \frac{3-\sqrt{7}}{3-\sqrt{7}}=\frac{2\left( 3-\sqrt{7} \right)}{3^{2}-\left( \sqrt{7} \right)^{2}} \\
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&= \frac{2}{3+\sqrt{7}}\cdot\frac{3-\sqrt{7}}{3-\sqrt{7}}
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& =\frac{2\centerdot 3-2\sqrt{7}}{2}=3-\sqrt{7} \\
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= \frac{2(3-\sqrt{7}\,)}{3^{2}-(\sqrt{7}\,)^{2}}\\[5pt]
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\end{align}</math>
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&= \frac{2\cdot 3-2\sqrt{7}}{2} = 3-\sqrt{7}\,\textrm{.}
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\end{align}</math>}}

Current revision

The trick is to use the formula for the difference of two squares \displaystyle (a-b)(a+b) = a^{2}-b^{2} and multiply the top and bottom of the fraction by \displaystyle 3-\sqrt{7} (note the minus sign), since then the new denominator will be \displaystyle (3+\sqrt{7})(3-\sqrt{7}) = 3^{2} - (\sqrt{7})^{2} = 9-7 = 2 (the formula with \displaystyle a=3 and \displaystyle b=\sqrt{7}\,), i.e. the root sign is squared away.

The whole calculation is

\displaystyle \begin{align}

\frac{2}{3+\sqrt{7}} &= \frac{2}{3+\sqrt{7}}\cdot\frac{3-\sqrt{7}}{3-\sqrt{7}} = \frac{2(3-\sqrt{7}\,)}{3^{2}-(\sqrt{7}\,)^{2}}\\[5pt] &= \frac{2\cdot 3-2\sqrt{7}}{2} = 3-\sqrt{7}\,\textrm{.} \end{align}

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