Lösung 3.1:2e

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Looking first at <math>\sqrt{18}</math> this square root expression can be simplified by writing 18 as a product of its smallest possible integer factors
Looking first at <math>\sqrt{18}</math> this square root expression can be simplified by writing 18 as a product of its smallest possible integer factors
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{{Displayed math||<math>18 = 2\cdot 9 = 2\cdot 3\cdot 3 = 2\cdot 3^{2}</math>}}
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{{Abgesetzte Formel||<math>18 = 2\cdot 9 = 2\cdot 3\cdot 3 = 2\cdot 3^{2}</math>}}
and then we can take the quadratic out of the square root sign by using the rule
and then we can take the quadratic out of the square root sign by using the rule
<math>\sqrt{a^{2}b}=a\sqrt{b}</math> (valid for non-negative ''a'' and ''b''),
<math>\sqrt{a^{2}b}=a\sqrt{b}</math> (valid for non-negative ''a'' and ''b''),
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{{Displayed math||<math>\sqrt{18} = \sqrt{2\cdot 3^{2}} = 3\sqrt{2}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\sqrt{18} = \sqrt{2\cdot 3^{2}} = 3\sqrt{2}\,\textrm{.}</math>}}
In the same way, we write <math>8 = 2\cdot 4 = 2\cdot 2\cdot 2 = 2^{3}</math> and get
In the same way, we write <math>8 = 2\cdot 4 = 2\cdot 2\cdot 2 = 2^{3}</math> and get
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{{Displayed math||<math>\sqrt{8} = \sqrt{2\cdot 2^{2}} = 2\sqrt{2}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\sqrt{8} = \sqrt{2\cdot 2^{2}} = 2\sqrt{2}\,\textrm{.}</math>}}
All together, we get
All together, we get
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
\sqrt{18}\sqrt{8}
\sqrt{18}\sqrt{8}
&= 3\sqrt{2}\cdot 2\sqrt{2}\\[5pt]
&= 3\sqrt{2}\cdot 2\sqrt{2}\\[5pt]

Version vom 08:36, 22. Okt. 2008

Looking first at \displaystyle \sqrt{18} this square root expression can be simplified by writing 18 as a product of its smallest possible integer factors

\displaystyle 18 = 2\cdot 9 = 2\cdot 3\cdot 3 = 2\cdot 3^{2}

and then we can take the quadratic out of the square root sign by using the rule \displaystyle \sqrt{a^{2}b}=a\sqrt{b} (valid for non-negative a and b),

\displaystyle \sqrt{18} = \sqrt{2\cdot 3^{2}} = 3\sqrt{2}\,\textrm{.}

In the same way, we write \displaystyle 8 = 2\cdot 4 = 2\cdot 2\cdot 2 = 2^{3} and get

\displaystyle \sqrt{8} = \sqrt{2\cdot 2^{2}} = 2\sqrt{2}\,\textrm{.}

All together, we get

\displaystyle \begin{align}

\sqrt{18}\sqrt{8} &= 3\sqrt{2}\cdot 2\sqrt{2}\\[5pt] &= 3\cdot 2\cdot \bigl(\sqrt{2}\bigr)^{2}\\[5pt] &= 3\cdot 2\cdot 2\\[5pt] &= 12\,\textrm{.} \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:2e“