Solution 3.1:4d - Förberedande kurs i matematik 1

-                      Welcome to the course
                       
                        About the course 
                        Examinations
                      
                    
                      1. Numerical calculations
                       
                        1.1 Different types of numbers 
                        1.2 Fractions 
                        1.3 Powers
                      
                    
                      2. Algebra
                       
                        2.1 Algebraic expressions 
                        2.2 Linear expressions 
                        2.3 Quadratic expressions
                      
                    
                      3. Roots and logarithms
                       
                        3.1 Roots 
                        3.2 Equations with roots 
                        3.3 Logarithms 
                        3.4 Logarithmic equations
                      
                    
                      4. Trigonometry
                       
                        4.1 Angles and circles 
                        4.2 Functions 
                        4.3 Relations 
                        4.4 Equations
                      
                    
                      5. Written mathematics
                       
                        5.1 Writing formulas 
                        5.2 Mathematical text
                      
                    
                    
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+			Solution 3.1:4d
			
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 (Ny sida: {{NAVCONTENT_START}} <center> Bild:3_1_4d.gif </center> {{NAVCONTENT_STOP}})
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				Current revision (11:13, 30 September 2008) (edit) (undo)
Tek  (Talk | contribs) 
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- {{NAVCONTENT_START}} + We start by factorizing the numbers under the root sign,
- <center> [[Bild:3_1_4d.gif]] </center> +  
- {{NAVCONTENT_STOP}} + {{Displayed math||<math>\begin{align}
  + 48 &= 2\cdot 24 = 2\cdot 2\cdot 12 = 2\cdot 2\cdot 2\cdot 6 = 2\cdot 2\cdot 2\cdot 2\cdot 3 = 2^{4}\cdot 3\,,\\
  + 12 &= 2\cdot 6 = 2\cdot 2\cdot 3 = 2^{2}\cdot 3\,,\\ 
  + 3 &= 3\,,\\ 
  + 75 &= 3\cdot 25 = 3\cdot 5\cdot 5 = 3\cdot 5^{2}\,\textrm{.} 
  + \end{align}</math>}}
  +  
  + Now, we can take the squares out from under the root signs,
  +  
  + {{Displayed math||<math>\begin{align}
  + \sqrt{48} &= \sqrt{2^4\cdot 3} = 2^2\sqrt{3} = 4\sqrt{3}\,,\\[5pt] 
  + \sqrt{12} &= \sqrt{2^2\cdot 3} = 2\sqrt{3},\\[5pt] 
  + \sqrt{3} &= \sqrt{3}\,,\\[5pt] 
  + \sqrt{75} &= \sqrt{3\cdot 5^{2}} = 5\sqrt{3}\,, 
  + \end{align}</math>}}
  +  
  + and then simplify the whole expression
  +  
  + {{Displayed math||<math>\begin{align}
  + \sqrt{48} + \sqrt{12} + \sqrt{3} - \sqrt{75}
  + &= 4\sqrt{3} + 2\sqrt{3} + \sqrt{3} - 5\sqrt{3}\\[5pt] 
  + &= (4+2+1-5)\sqrt{3}\\[5pt]
  + &= 2\sqrt{3}\,\textrm{.} 
  + \end{align}</math>}}
Current revision
We start by factorizing the numbers under the root sign,





\displaystyle \begin{align}
48 &= 2\cdot 24 = 2\cdot 2\cdot 12 = 2\cdot 2\cdot 2\cdot 6 = 2\cdot 2\cdot 2\cdot 2\cdot 3 = 2^{4}\cdot 3\,,\\
12 &= 2\cdot 6 = 2\cdot 2\cdot 3 = 2^{2}\cdot 3\,,\\ 
3 &= 3\,,\\ 
75 &= 3\cdot 25 = 3\cdot 5\cdot 5 = 3\cdot 5^{2}\,\textrm{.} 
\end{align}



Now, we can take the squares out from under the root signs,





\displaystyle \begin{align}
\sqrt{48} &= \sqrt{2^4\cdot 3} = 2^2\sqrt{3} = 4\sqrt{3}\,,\\[5pt] 
\sqrt{12} &= \sqrt{2^2\cdot 3} = 2\sqrt{3},\\[5pt] 
\sqrt{3} &= \sqrt{3}\,,\\[5pt] 
\sqrt{75} &= \sqrt{3\cdot 5^{2}} = 5\sqrt{3}\,, 
\end{align}



and then simplify the whole expression





\displaystyle \begin{align}
\sqrt{48} + \sqrt{12} + \sqrt{3} - \sqrt{75}
&= 4\sqrt{3} + 2\sqrt{3} + \sqrt{3} - 5\sqrt{3}\\[5pt] 
&= (4+2+1-5)\sqrt{3}\\[5pt]
&= 2\sqrt{3}\,\textrm{.} 
\end{align}




Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Solution_3.1:4d"
@@ Line 1: / Line 1: @@
-{{NAVCONTENT_START}}
+We start by factorizing the numbers under the root sign,
-<center> [[Bild:3_1_4d.gif]] </center>
-{{NAVCONTENT_STOP}}
+{{Displayed math||<math>\begin{align}
+&= 2\cdot 24 = 2\cdot 2\cdot 12 = 2\cdot 2\cdot 2\cdot 6 = 2\cdot 2\cdot 2\cdot 2\cdot 3 = 2^{4}\cdot 3\,,\\
+&= 2\cdot 6 = 2\cdot 2\cdot 3 = 2^{2}\cdot 3\,,\\
+&= 3\,,\\
+&= 3\cdot 25 = 3\cdot 5\cdot 5 = 3\cdot 5^{2}\,\textrm{.}
+\end{align}</math>}}
+Now, we can take the squares out from under the root signs,
+{{Displayed math||<math>\begin{align}
+\sqrt{48} &= \sqrt{2^4\cdot 3} = 2^2\sqrt{3} = 4\sqrt{3}\,,\\[5pt]
+\sqrt{12} &= \sqrt{2^2\cdot 3} = 2\sqrt{3},\\[5pt]
+\sqrt{3} &= \sqrt{3}\,,\\[5pt]
+\sqrt{75} &= \sqrt{3\cdot 5^{2}} = 5\sqrt{3}\,,
+\end{align}</math>}}
+and then simplify the whole expression
+{{Displayed math||<math>\begin{align}
+\sqrt{48} + \sqrt{12} + \sqrt{3} - \sqrt{75}
+&= 4\sqrt{3} + 2\sqrt{3} + \sqrt{3} - 5\sqrt{3}\\[5pt]
+&= (4+2+1-5)\sqrt{3}\\[5pt]
+&= 2\sqrt{3}\,\textrm{.}
+\end{align}</math>}}
 \displaystyle \begin{align}
48 &= 2\cdot 24 = 2\cdot 2\cdot 12 = 2\cdot 2\cdot 2\cdot 6 = 2\cdot 2\cdot 2\cdot 2\cdot 3 = 2^{4}\cdot 3\,,\\
12 &= 2\cdot 6 = 2\cdot 2\cdot 3 = 2^{2}\cdot 3\,,\\ 
3 &= 3\,,\\ 
75 &= 3\cdot 25 = 3\cdot 5\cdot 5 = 3\cdot 5^{2}\,\textrm{.} 
\end{align}
 \displaystyle \begin{align}
\sqrt{48} &= \sqrt{2^4\cdot 3} = 2^2\sqrt{3} = 4\sqrt{3}\,,\\[5pt] 
\sqrt{12} &= \sqrt{2^2\cdot 3} = 2\sqrt{3},\\[5pt] 
\sqrt{3} &= \sqrt{3}\,,\\[5pt] 
\sqrt{75} &= \sqrt{3\cdot 5^{2}} = 5\sqrt{3}\,, 
\end{align}
 \displaystyle \begin{align}
\sqrt{48} + \sqrt{12} + \sqrt{3} - \sqrt{75}
&= 4\sqrt{3} + 2\sqrt{3} + \sqrt{3} - 5\sqrt{3}\\[5pt] 
&= (4+2+1-5)\sqrt{3}\\[5pt]
&= 2\sqrt{3}\,\textrm{.} 
\end{align}