Solution 3.1:4d

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Current revision (11:13, 30 September 2008) (edit) (undo)
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We start by factorizing the numbers under the root sign,
We start by factorizing the numbers under the root sign,
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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48 &= 2\cdot 24 = 2\cdot 2\cdot 12 = 2\cdot 2\cdot 2\cdot 6 = 2\cdot 2\cdot 2\cdot 2\cdot 3 = 2^{4}\cdot 3\,,\\
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& 48=2\centerdot 24=2\centerdot 2\centerdot 12=2\centerdot 2\centerdot 2\centerdot 6=2\centerdot 2\centerdot 2\centerdot 2\centerdot 3=2^{4}\centerdot 3, \\
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12 &= 2\cdot 6 = 2\cdot 2\cdot 3 = 2^{2}\cdot 3\,,\\
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& 12=2\centerdot 6=2\centerdot 2\centerdot 3=2^{2}\centerdot 3, \\
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3 &= 3\,,\\
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& 3=3, \\
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75 &= 3\cdot 25 = 3\cdot 5\cdot 5 = 3\cdot 5^{2}\,\textrm{.}
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& 75=3\centerdot 25=3\centerdot 5\centerdot 5=3\centerdot 5^{2} \\
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\end{align}</math>}}
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\end{align}</math>
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Now, we can take the squares out from under the root signs,
Now, we can take the squares out from under the root signs,
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{{Displayed math||<math>\begin{align}
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\sqrt{48} &= \sqrt{2^4\cdot 3} = 2^2\sqrt{3} = 4\sqrt{3}\,,\\[5pt]
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\sqrt{12} &= \sqrt{2^2\cdot 3} = 2\sqrt{3},\\[5pt]
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\sqrt{3} &= \sqrt{3}\,,\\[5pt]
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\sqrt{75} &= \sqrt{3\cdot 5^{2}} = 5\sqrt{3}\,,
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\end{align}</math>}}
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<math>\begin{align}
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and then simplify the whole expression
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& \sqrt{48}=\sqrt{2^{4}\centerdot 3}=2^{2}\sqrt{3}=4\sqrt{3} \\
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& \sqrt{12}=\sqrt{2^{2}\centerdot 3}=2\sqrt{3} \\
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& \sqrt{3}=\sqrt{3} \\
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& \sqrt{75}=\sqrt{3\centerdot 5^{2}}=5\sqrt{3} \\
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\end{align}</math>
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and then simplify the whole expression:
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<math>\begin{align}
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{{Displayed math||<math>\begin{align}
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& \sqrt{48}+\sqrt{12}+\sqrt{3}-\sqrt{75}=4\sqrt{3}+2\sqrt{3}+\sqrt{3}-5\sqrt{3} \\
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\sqrt{48} + \sqrt{12} + \sqrt{3} - \sqrt{75}
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& =\left( 4+2+1-5 \right)\sqrt{3}=2\sqrt{3} \\
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&= 4\sqrt{3} + 2\sqrt{3} + \sqrt{3} - 5\sqrt{3}\\[5pt]
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\end{align}</math>
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&= (4+2+1-5)\sqrt{3}\\[5pt]
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&= 2\sqrt{3}\,\textrm{.}
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\end{align}</math>}}

Current revision

We start by factorizing the numbers under the root sign,

\displaystyle \begin{align}

48 &= 2\cdot 24 = 2\cdot 2\cdot 12 = 2\cdot 2\cdot 2\cdot 6 = 2\cdot 2\cdot 2\cdot 2\cdot 3 = 2^{4}\cdot 3\,,\\ 12 &= 2\cdot 6 = 2\cdot 2\cdot 3 = 2^{2}\cdot 3\,,\\ 3 &= 3\,,\\ 75 &= 3\cdot 25 = 3\cdot 5\cdot 5 = 3\cdot 5^{2}\,\textrm{.} \end{align}

Now, we can take the squares out from under the root signs,

\displaystyle \begin{align}

\sqrt{48} &= \sqrt{2^4\cdot 3} = 2^2\sqrt{3} = 4\sqrt{3}\,,\\[5pt] \sqrt{12} &= \sqrt{2^2\cdot 3} = 2\sqrt{3},\\[5pt] \sqrt{3} &= \sqrt{3}\,,\\[5pt] \sqrt{75} &= \sqrt{3\cdot 5^{2}} = 5\sqrt{3}\,, \end{align}

and then simplify the whole expression

\displaystyle \begin{align}

\sqrt{48} + \sqrt{12} + \sqrt{3} - \sqrt{75} &= 4\sqrt{3} + 2\sqrt{3} + \sqrt{3} - 5\sqrt{3}\\[5pt] &= (4+2+1-5)\sqrt{3}\\[5pt] &= 2\sqrt{3}\,\textrm{.} \end{align}

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