Solution 3.1:4c
From Förberedande kurs i matematik 1
m  (Lösning 3.1:4c moved to Solution 3.1:4c: Robot: moved page)  | 
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| - | {{  | + | Each term in the expression can be simplified by breaking down the number under the root sign into its factors,  | 
| - | <  | + | |
| - | {{  | + | |
| + | <math>\begin{align}  | ||
| + | & 50=5\centerdot 10=5\centerdot 5\centerdot 2=2\centerdot 5^{2} \\   | ||
| + | & 20=2\centerdot 10=2\centerdot 2\centerdot 5=2^{2}\centerdot 5 \\   | ||
| + | & 18=2\centerdot 9=2\centerdot 3\centerdot 3=2\centerdot 3^{2} \\   | ||
| + | & 80=8\centerdot 10=\left( 2\centerdot 4 \right)\centerdot \left( 2\centerdot 5 \right)=\left( 2\centerdot 2\centerdot 2 \right)\centerdot \left( 2\centerdot 5 \right)=2^{4}\centerdot 5 \\   | ||
| + | \end{align}</math>  | ||
| + | |||
| + | and then taking the squares out from under the root sign.  | ||
| + | |||
| + | |||
| + | <math>\begin{align}  | ||
| + | & \sqrt{50}=\sqrt{2\centerdot 5^{2}}=5\sqrt{2} \\   | ||
| + | & \sqrt{20}=\sqrt{2^{2}\centerdot 5}=2\sqrt{5} \\   | ||
| + | & \sqrt{18}=\sqrt{2\centerdot 3^{2}}=3\sqrt{2} \\   | ||
| + | & \sqrt{80}=\sqrt{2^{4}\centerdot 5}=2^{2}\sqrt{5}=4\sqrt{5} \\   | ||
| + | \end{align}</math>  | ||
| + | |||
| + | |||
| + | All together, we get  | ||
| + | |||
| + | |||
| + | <math>\begin{align}  | ||
| + | & \sqrt{50}+4\sqrt{20}-3\sqrt{18}-2\sqrt{80} \\   | ||
| + | & =5\sqrt{2}+4\centerdot 2\sqrt{5}-3\centerdot 3\sqrt{2}-2\centerdot 4\sqrt{5} \\   | ||
| + | & =5\sqrt{2}+8\sqrt{5}-9\sqrt{2}-8\sqrt{5} \\   | ||
| + | & =\left( 5-9 \right)\sqrt{2}+\left( 8-8 \right)\sqrt{5}=-4\sqrt{2} \\   | ||
| + | \end{align}</math>  | ||
Revision as of 14:11, 22 September 2008
Each term in the expression can be simplified by breaking down the number under the root sign into its factors,
\displaystyle \begin{align}
& 50=5\centerdot 10=5\centerdot 5\centerdot 2=2\centerdot 5^{2} \\ 
& 20=2\centerdot 10=2\centerdot 2\centerdot 5=2^{2}\centerdot 5 \\ 
& 18=2\centerdot 9=2\centerdot 3\centerdot 3=2\centerdot 3^{2} \\ 
& 80=8\centerdot 10=\left( 2\centerdot 4 \right)\centerdot \left( 2\centerdot 5 \right)=\left( 2\centerdot 2\centerdot 2 \right)\centerdot \left( 2\centerdot 5 \right)=2^{4}\centerdot 5 \\ 
\end{align}
and then taking the squares out from under the root sign.
\displaystyle \begin{align}
& \sqrt{50}=\sqrt{2\centerdot 5^{2}}=5\sqrt{2} \\ 
& \sqrt{20}=\sqrt{2^{2}\centerdot 5}=2\sqrt{5} \\ 
& \sqrt{18}=\sqrt{2\centerdot 3^{2}}=3\sqrt{2} \\ 
& \sqrt{80}=\sqrt{2^{4}\centerdot 5}=2^{2}\sqrt{5}=4\sqrt{5} \\ 
\end{align}
All together, we get
\displaystyle \begin{align}
& \sqrt{50}+4\sqrt{20}-3\sqrt{18}-2\sqrt{80} \\ 
& =5\sqrt{2}+4\centerdot 2\sqrt{5}-3\centerdot 3\sqrt{2}-2\centerdot 4\sqrt{5} \\ 
& =5\sqrt{2}+8\sqrt{5}-9\sqrt{2}-8\sqrt{5} \\ 
& =\left( 5-9 \right)\sqrt{2}+\left( 8-8 \right)\sqrt{5}=-4\sqrt{2} \\ 
\end{align}
