Solution 3.1:5c
From Förberedande kurs i matematik 1
(Difference between revisions)
m (Lösning 3.1:5c moved to Solution 3.1:5c: Robot: moved page) |
m |
||
(One intermediate revision not shown.) | |||
Line 1: | Line 1: | ||
- | {{ | + | The trick is to use the formula for the difference of two squares |
- | < | + | <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. |
- | {{ | + | |
+ | The whole calculation is | ||
+ | |||
+ | {{Displayed math||<math>\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}</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} |