Solution 3.1:3c

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We start by looking at the one part of the expression,
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We start by looking at one part of the expression <math>\sqrt{16}</math>. This subexpression can be simplified since <math>16 = 4\cdot 4 = 4^{2}</math> which gives that <math>\sqrt{16} = \sqrt{4^{2}} = 4</math> and the whole expression becomes
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<math>\sqrt{16}</math>. This root can be simplified since
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<math>16=4\centerdot 4=4^{2}</math>
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which gives that
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<math>\sqrt{16}=\sqrt{4^{2}}=4</math>
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and the whole expression becomes
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{{Displayed math||<math>\sqrt{16+\sqrt{16}} = \sqrt{16+4} = \sqrt{20}\,\textrm{.}</math>}}
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<math>\sqrt{16+\sqrt{16}}=\sqrt{16+4}=\sqrt{20}</math>
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Can <math>\sqrt{20}</math> be simplified? In order to answer this, we split 20 up into integer factors,
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{{Displayed math||<math>20 = 2\cdot 10 = 2\cdot 2\cdot 5 = 2^{2}\cdot 5</math>}}
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Can
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and see that 20 contains the square <math>2^2</math> as a factor and can therefore be taken outside the root sign,
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<math>\sqrt{20}</math>
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be simplified? In order to answer this, we split
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<math>\text{2}0</math>
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up into integer factors,
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{{Displayed math||<math>\sqrt{20} = \sqrt{2^{2}\centerdot 5} = 2\sqrt{5}\,\textrm{.}</math>}}
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<math>20=2\centerdot 10=2\centerdot 2\centerdot 5=2^{2}\centerdot 5</math>
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and see that
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<math>\text{2}0\text{ }</math>
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contains the square
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<math>\text{2}^{\text{2}}</math>
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as a factor and can therefore be taken outside the root sign,
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<math>\sqrt{20}=\sqrt{2^{2}\centerdot 5}^{2}=2\sqrt{5}</math>
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Current revision

We start by looking at one part of the expression \displaystyle \sqrt{16}. This subexpression can be simplified since \displaystyle 16 = 4\cdot 4 = 4^{2} which gives that \displaystyle \sqrt{16} = \sqrt{4^{2}} = 4 and the whole expression becomes

\displaystyle \sqrt{16+\sqrt{16}} = \sqrt{16+4} = \sqrt{20}\,\textrm{.}

Can \displaystyle \sqrt{20} be simplified? In order to answer this, we split 20 up into integer factors,

\displaystyle 20 = 2\cdot 10 = 2\cdot 2\cdot 5 = 2^{2}\cdot 5

and see that 20 contains the square \displaystyle 2^2 as a factor and can therefore be taken outside the root sign,

\displaystyle \sqrt{20} = \sqrt{2^{2}\centerdot 5} = 2\sqrt{5}\,\textrm{.}
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