Solution 3.1:3b

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When simplifying a root 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 factor cancel each other out or can be combined together in a new way.
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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.
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By successively dividing by
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By successively dividing by 2 and 3, we see that
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<math>2</math>
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and
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<math>3</math>, we see that
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<math>\begin{align}
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& 96=2\centerdot 48=2\centerdot 2\centerdot 24=2\centerdot 2\centerdot 2\centerdot 12=2\centerdot 2\centerdot 2\centerdot 2\centerdot 6 \\
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& =2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 3=2^{5}\centerdot 3, \\
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& 18=2\centerdot 9=2\centerdot 3\centerdot 3=2\centerdot 3^{2}. \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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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\\
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&= 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 3 = 2^{5}\cdot 3,\\[5pt]
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18 &= 2\cdot 9 = 2\cdot 3\cdot 3 = 2\cdot 3^{2}.
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\end{align}</math>}}
Thus,
Thus,
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\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]
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& \sqrt{96}=\sqrt{2^{5}\centerdot 3}=\sqrt{2^{2}\centerdot 2^{2}\centerdot 2\centerdot 3}=2\centerdot 2\centerdot \sqrt{2}\centerdot \sqrt{3} \\
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\sqrt{18} &= \sqrt{2\cdot 3^{2}} = 3\cdot\sqrt{2}\,,
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& \sqrt{18}=\sqrt{2\centerdot 3^{2}}=3\centerdot \sqrt{2} \\
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\end{align}</math>}}
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\end{align}</math>
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and the whole quotient can be written as
and the whole quotient can be written as
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{{Displayed math||<math>\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{.}</math>}}
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<math>\frac{\sqrt{96}}{\sqrt{18}}=\frac{2\centerdot 2\centerdot \sqrt{2}\centerdot \sqrt{3}}{3\centerdot \sqrt{2}}=\frac{4\sqrt{3}}{3}</math>
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NOTE: If it is difficult to work with the root sign, it is possible instead to write everything in power form
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<math>\begin{align}
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Note: If it is difficult to work with radicals, it is possible instead to write everything in power form
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& \frac{\sqrt{96}}{\sqrt{18}}=\frac{\left( 96 \right)^{\frac{1}{2}}}{\left( 18 \right)^{{1}/{2}\;}}=\frac{\left( 2^{5}\centerdot 3 \right)^{\frac{1}{2}}}{\left( 2\centerdot 3^{2} \right)^{{1}/{2}\;}}=\frac{2^{5\centerdot \frac{1}{2}}\centerdot 3^{\frac{1}{2}}}{2^{\frac{1}{2}}\centerdot 3^{2\centerdot \frac{1}{2}}} \\
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& =2^{\frac{5}{2}-\frac{1}{2}}\centerdot 3^{\frac{1}{2}-1}=2^{2}\centerdot 3^{-\frac{1}{2}}=\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3} \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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\frac{\sqrt{96}}{\sqrt{18}}
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&= \frac{96^{1/2}}{18^{1/2}}
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= \frac{(2^{5}\cdot 3)^{1/2}}{(2\cdot 3^{2})^{1/2}}
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= \frac{2^{5\cdot\frac{1}{2}}\cdot 3^{\frac{1}{2}}}{2^{\frac{1}{2}}\cdot 3^{2\cdot \frac{1}{2}}}\\[5pt]
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&= 2^{\frac{5}{2}-\frac{1}{2}}\cdot 3^{\frac{1}{2}-1}
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= 2^{2}\cdot 3^{-\frac{1}{2}}
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= \frac{4}{\sqrt{3}}
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= \frac{4\sqrt{3}}{3}\,\textrm{.}
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\end{align}</math>}}
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(in the last line, we multiply top and bottom by
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(In the last equality, we multiply top and bottom by <math>\sqrt{3}</math>.)
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<math>\sqrt{3}</math>
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).
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Current revision

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}.)

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