Solution 3.1:3b
From Förberedande kurs i matematik 1
When simplifying a radical expression, a common technique is to divide up the numbers under the root sign into their smallest possible integer factors and then take out the squares and see if common factors cancel each other out or can be combined together in a new way.
By successively dividing by 2 and 3, we see that
| \displaystyle \begin{align}
96 &= 2\cdot 48 = 2\cdot 2\cdot 24 = 2\cdot 2\cdot 2\cdot 12 = 2\cdot 2\cdot 2\cdot 2\cdot 6\\ &= 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3 = 2^{5}\cdot 3,\\[5pt] 18 &= 2\cdot 9 = 2\cdot 3\cdot 3 = 2\cdot 3^{2}. \end{align} |
Thus,
| \displaystyle \begin{align}
\sqrt{96} &= \sqrt{2^{5}\cdot 3} = \sqrt{2^{2}\cdot 2^{2}\cdot 2\cdot 3} = 2\cdot 2\cdot \sqrt{2}\cdot \sqrt{3}\,,\\[5pt] \sqrt{18} &= \sqrt{2\cdot 3^{2}} = 3\cdot\sqrt{2}\,, \end{align} |
and the whole quotient can be written as
| \displaystyle \frac{\sqrt{96}}{\sqrt{18}} = \frac{2\cdot 2\cdot \sqrt{2}\cdot \sqrt{3}}{3\cdot \sqrt{2}} = \frac{4\sqrt{3}}{3}\,\textrm{.} |
Note: If it is difficult to work with radicals, it is possible instead to write everything in power form
| \displaystyle \begin{align}
\frac{\sqrt{96}}{\sqrt{18}} &= \frac{96^{1/2}}{18^{1/2}} = \frac{(2^{5}\cdot 3)^{1/2}}{(2\cdot 3^{2})^{1/2}} = \frac{2^{5\cdot\frac{1}{2}}\cdot 3^{\frac{1}{2}}}{2^{\frac{1}{2}}\cdot 3^{2\cdot \frac{1}{2}}}\\[5pt] &= 2^{\frac{5}{2}-\frac{1}{2}}\cdot 3^{\frac{1}{2}-1} = 2^{2}\cdot 3^{-\frac{1}{2}} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}\,\textrm{.} \end{align} |
(In the last equality, we multiply top and bottom by \displaystyle \sqrt{3}.)
