Solution 3.1:6b
From Förberedande kurs i matematik 1
The root sign in the denominator lies in a quadratic term and we therefore expand first the quadratic
| \displaystyle \begin{align}
(\sqrt{3}-2)^2 &= (\sqrt{3}\,)^{2} - 2\cdot\sqrt{3}\cdot 2 + 2^{2}\\[5pt] &= 3-4\sqrt{3}+4\\[5pt] &= 7-4\sqrt{3}\,\textrm{.} \end{align} |
Thus,
| \displaystyle \frac{1}{(\sqrt{3}-2)^{2}-2} = \frac{1}{7-4\sqrt{3}-2} = \frac{1}{5-4\sqrt{3}} |
and in the expression we can get rid of the root sign from the denominator by multiplying the top and bottom of the equation by the conjugate \displaystyle 5+4\sqrt{3},
| \displaystyle \begin{align}
\frac{1}{5-4\sqrt{3}} &= \frac{1}{5-4\sqrt{3}}\cdot \frac{5+4\sqrt{3}}{5+4\sqrt{3}}\\[5pt] &= \frac{5+4\sqrt{3}}{5^{2}-(4\sqrt{3})^{2}}\\[5pt] &= \frac{5+4\sqrt{3}}{5^{2}-4^{2}(\sqrt{3})^{2}}\\[5pt] &= \frac{5+4\sqrt{3}}{25-16\cdot 3}\\[5pt] &= \frac{5+4\sqrt{3}}{-23}\\[5pt] &= -\frac{5+4\sqrt{3}}{23}\,\textrm{.} \end{align} |
