Solution 3.1:5a
From Förberedande kurs i matematik 1
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- | {{ | + | If we multiply the top and bottom of the fraction by <math>\sqrt{12}</math>, the new denominator will be <math>\sqrt{12}\cdot\sqrt{12} = 12</math> and we will get rid of the root sign in the denominator |
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- | {{ | + | {{Displayed math||<math>\frac{2}{\sqrt{12}} = \frac{2}{\sqrt{12}}\cdot \frac{\sqrt{12}}{\sqrt{12}} = \frac{2\sqrt{12}}{12} = \frac{2\sqrt{12}}{2\cdot 6} = \frac{\sqrt{12}}{6}\,\textrm{.}</math>}} |
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+ | This expression can be simplified even further if we write <math>12 = 2\cdot 6 = 2\cdot 2\cdot 3 = 2^2\cdot 3</math> and take <math>2^2</math> out from under the root, we get | ||
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+ | {{Displayed math||<math>\frac{\sqrt{12}}{6} = \frac{2\sqrt{3}}{6} = \frac{2\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{3}\,\textrm{.}</math>}} |
Current revision
If we multiply the top and bottom of the fraction by \displaystyle \sqrt{12}, the new denominator will be \displaystyle \sqrt{12}\cdot\sqrt{12} = 12 and we will get rid of the root sign in the denominator
\displaystyle \frac{2}{\sqrt{12}} = \frac{2}{\sqrt{12}}\cdot \frac{\sqrt{12}}{\sqrt{12}} = \frac{2\sqrt{12}}{12} = \frac{2\sqrt{12}}{2\cdot 6} = \frac{\sqrt{12}}{6}\,\textrm{.} |
This expression can be simplified even further if we write \displaystyle 12 = 2\cdot 6 = 2\cdot 2\cdot 3 = 2^2\cdot 3 and take \displaystyle 2^2 out from under the root, we get
\displaystyle \frac{\sqrt{12}}{6} = \frac{2\sqrt{3}}{6} = \frac{2\sqrt{3}}{2\cdot 3} = \frac{\sqrt{3}}{3}\,\textrm{.} |