Solution 3.1:4c
From Förberedande kurs i matematik 1
m   | 
			|||
| Line 1: | Line 1: | ||
Each term in the expression can be simplified by breaking down the number under the root sign into its factors,  | Each term in the expression can be simplified by breaking down the number under the root sign into its factors,  | ||
| + | {{Displayed math||<math>\begin{align}  | ||
| + | 50 &= 5\cdot 10 = 5\cdot 5\cdot 2 = 2\cdot 5^{2}\,,\\[5pt]  | ||
| + | 20 &= 2\cdot 10 = 2\cdot 2\cdot 5 = 2^{2}\cdot 5\,,\\[5pt]  | ||
| + | 18 &= 2\cdot 9 = 2\cdot 3\cdot 3 = 2\cdot 3^{2}\,,\\[5pt]  | ||
| + | 80 &= 8\cdot 10 = (2\cdot 4)\cdot (2\cdot 5) = (2\cdot 2\cdot 2)\cdot (2\cdot 5) = 2^{4}\cdot 5\,,   | ||
| + | \end{align}</math>}}  | ||
| - | + | and then taking the squares out from under the root sign,  | |
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | and then taking the squares out from under the root sign  | + | |
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| + | {{Displayed math||<math>\begin{align}  | ||
| + | \sqrt{50} &= \sqrt{2\cdot 5^2} = 5\sqrt{2}\,,\\   | ||
| + | \sqrt{20} &= \sqrt{2^2\cdot 5} = 2\sqrt{5}\,,\\   | ||
| + | \sqrt{18} &= \sqrt{2\cdot 3^2} = 3\sqrt{2}\,,\\   | ||
| + | \sqrt{80} &= \sqrt{2^4\cdot 5} = 2^{2}\sqrt{5} = 4\sqrt{5}\,\textrm{.}  | ||
| + | \end{align}</math>}}  | ||
All together, we get  | All together, we get  | ||
| - | + | {{Displayed math||<math>\begin{align}  | |
| - | <math>\begin{align}  | + | \sqrt{50} + 4\sqrt{20} - 3\sqrt{18} - 2\sqrt{80}   | 
| - | + | &= 5\sqrt{2} + 4\cdot 2\sqrt{5} - 3\cdot 3\sqrt{2} - 2\cdot 4\sqrt{5}\\[5pt]   | |
| - | & =5\sqrt{2}+4\  | + | &= 5\sqrt{2} + 8\sqrt{5} - 9\sqrt{2} - 8\sqrt{5}\\[5pt]   | 
| - | & =5\sqrt{2}+8\sqrt{5}-9\sqrt{2}-8\sqrt{5} \\   | + | &= (5-9)\sqrt{2} + (8-8)\sqrt{5} = -4\sqrt{2}\,\textrm{.}   | 
| - | & =  | + | \end{align}</math>}}  | 
| - | \end{align}</math>  | + | |
Current revision
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\cdot 10 = 5\cdot 5\cdot 2 = 2\cdot 5^{2}\,,\\[5pt] 20 &= 2\cdot 10 = 2\cdot 2\cdot 5 = 2^{2}\cdot 5\,,\\[5pt] 18 &= 2\cdot 9 = 2\cdot 3\cdot 3 = 2\cdot 3^{2}\,,\\[5pt] 80 &= 8\cdot 10 = (2\cdot 4)\cdot (2\cdot 5) = (2\cdot 2\cdot 2)\cdot (2\cdot 5) = 2^{4}\cdot 5\,, \end{align}  | 
and then taking the squares out from under the root sign,
| \displaystyle \begin{align}
 \sqrt{50} &= \sqrt{2\cdot 5^2} = 5\sqrt{2}\,,\\ \sqrt{20} &= \sqrt{2^2\cdot 5} = 2\sqrt{5}\,,\\ \sqrt{18} &= \sqrt{2\cdot 3^2} = 3\sqrt{2}\,,\\ \sqrt{80} &= \sqrt{2^4\cdot 5} = 2^{2}\sqrt{5} = 4\sqrt{5}\,\textrm{.} \end{align}  | 
All together, we get
| \displaystyle \begin{align}
 \sqrt{50} + 4\sqrt{20} - 3\sqrt{18} - 2\sqrt{80} &= 5\sqrt{2} + 4\cdot 2\sqrt{5} - 3\cdot 3\sqrt{2} - 2\cdot 4\sqrt{5}\\[5pt] &= 5\sqrt{2} + 8\sqrt{5} - 9\sqrt{2} - 8\sqrt{5}\\[5pt] &= (5-9)\sqrt{2} + (8-8)\sqrt{5} = -4\sqrt{2}\,\textrm{.} \end{align}  | 
