Solution 3.1:5d
From Förberedande kurs i matematik 1
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| - | {{ | + | We can get rid of both square roots in the denominator if we multiply the top and bottom of the fraction by the conjugate expression <math>\sqrt{17}+\sqrt{13}</math>, and use the difference of two squares |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>(a-b)(a+b) = a^2-b^2</math>}} |
| + | |||
| + | with <math>a=\sqrt{17}</math> and <math>b=\sqrt{13}</math>. Both roots are squared away and we get | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \frac{1}{\sqrt{17}-\sqrt{13}} | ||
| + | &= \frac{1}{\sqrt{17}-\sqrt{13}}\cdot\frac{\sqrt{17}+\sqrt{13}}{\sqrt{17}+\sqrt{13}}\\[5pt] | ||
| + | &= \frac{\sqrt{17}+\sqrt{13}}{(\sqrt{17})^{2}-(\sqrt{13})^{2}}\\[5pt] | ||
| + | &= \frac{\sqrt{17}+\sqrt{13}}{17-13}\\[5pt] | ||
| + | &= \frac{\sqrt{17}+\sqrt{13}}{4}\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | This expression cannot be simplified any further because neither 17 nor 13 contain any squares as factors. | ||
Current revision
We can get rid of both square roots in the denominator if we multiply the top and bottom of the fraction by the conjugate expression \displaystyle \sqrt{17}+\sqrt{13}, and use the difference of two squares
| \displaystyle (a-b)(a+b) = a^2-b^2 |
with \displaystyle a=\sqrt{17} and \displaystyle b=\sqrt{13}. Both roots are squared away and we get
| \displaystyle \begin{align}
\frac{1}{\sqrt{17}-\sqrt{13}} &= \frac{1}{\sqrt{17}-\sqrt{13}}\cdot\frac{\sqrt{17}+\sqrt{13}}{\sqrt{17}+\sqrt{13}}\\[5pt] &= \frac{\sqrt{17}+\sqrt{13}}{(\sqrt{17})^{2}-(\sqrt{13})^{2}}\\[5pt] &= \frac{\sqrt{17}+\sqrt{13}}{17-13}\\[5pt] &= \frac{\sqrt{17}+\sqrt{13}}{4}\,\textrm{.} \end{align} |
This expression cannot be simplified any further because neither 17 nor 13 contain any squares as factors.
