Solution 3.1:5c
From Förberedande kurs i matematik 1
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} |