Lösung 3.1:3c

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We start by looking at the one part of the expression,
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<center> [[Image:3_1_3c.gif]] </center>
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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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<math>\sqrt{16+\sqrt{16}}=\sqrt{16+4}=\sqrt{20}</math>
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Can
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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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<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>

Version vom 12:59, 22. Sep. 2008

We start by looking at the one part of the expression, \displaystyle \sqrt{16}. This root can be simplified since \displaystyle 16=4\centerdot 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}


Can \displaystyle \sqrt{20} be simplified? In order to answer this, we split \displaystyle \text{2}0 up into integer factors,


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


and see that \displaystyle \text{2}0\text{ } contains the square \displaystyle \text{2}^{\text{2}} as a factor and can therefore be taken outside the root sign,


\displaystyle \sqrt{20}=\sqrt{2^{2}\centerdot 5}^{2}=2\sqrt{5}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:3c“