Solution 3.3:4b - Förberedande kurs i matematik 2

-                      Welcome to the course
                       
                        About the course 
                        Examinations
                      
                    
                      1. Derivatives
                       
                        1.1 Introduction 
                        1.2 Rules of differentiation 
                        1.3 Max and min problems
                      
                    
                      2. Integrals
                       
                        2.1 Introduction 
                        2.2 Substitution 
                        2.3 Integration by parts
                      
                    
                      3. Complex numbers
                       
                        3.1 Calculations 
                        3.2 Polar form 
                        3.3 Powers and roots 
                        3.4 Complex polynomials
                      
                    
                    
		       Search
+			Solution 3.3:4b
			
			  From Förberedande kurs i matematik 2
			  (Difference between revisions)
			  			                            Jump to: navigation, search
                          
		          
			
			
			
			
			
			
				Revision as of 13:41, 7 April 2008 (edit)
Oskar  (Talk | contribs)
 (Ny sida: {{NAVCONTENT_START}} <center> Bild:3_3_4b.gif </center> {{NAVCONTENT_STOP}})
← Previous diff
				Current revision (14:30, 30 October 2008) (edit) (undo)
Tek  (Talk | contribs) 
m 
 
			
		(3 intermediate revisions not shown.)
Line 1:
Line 1:
- {{NAVCONTENT_START}} + Typically, one solves a second-degree by completing the square, followed by taking the square root.
- <center> [[Bild:3_3_4b.gif]] </center> +  
- {{NAVCONTENT_STOP}} + If we complete the square of the left-hand side, we get
  +  
  + {{Displayed math||<math>\begin{align}
  + (z-2)^2-2^2+5&=0,\\[5pt] 
  + (z-2)^2+1&=0.
  + \end{align}</math>}}
  +  
  + Taking the square root then gives that the equation has roots <math>z-2=\pm i</math>, i.e. <math>z=2+i</math> and <math>z=2-i</math>.
  +  
  + If we want to be sure that we have found the correct solutions, we can substitute each solution into the equation and see whether the equation is satisfied.
  +  
  + <math>\begin{align}
  + z=2+i:\qquad z^2-4z+5
  + &= (2+i)^2-4(2+i)+5\\[5pt] 
  + &= 2^2+4i+i^2-8-4i+5\\[5pt]
  + &= 4+4i-1-8-4i+5\\[5pt]
  + &=0\,,\\[10pt] 
  + z={}\rlap{2-i:}\phantom{2+i:}{}\qquad z^2-4z+5
  + &= (2-i)^2-4(2-i)+5\\[5pt]
  + &= 2^2-4i+i^2-8+4i+5\\[5pt]
  + &= 4-4i-1-8+4i+5\\[5pt]
  + &= 0\,\textrm{.} 
  + \end{align}</math>
Current revision
Typically, one solves a second-degree by completing the square, followed by taking the square root.
If we complete the square of the left-hand side, we get





\displaystyle \begin{align}
(z-2)^2-2^2+5&=0,\\[5pt] 
(z-2)^2+1&=0.
\end{align}



Taking the square root then gives that the equation has roots \displaystyle z-2=\pm i, i.e. \displaystyle z=2+i and \displaystyle z=2-i.
If we want to be sure that we have found the correct solutions, we can substitute each solution into the equation and see whether the equation is satisfied.
\displaystyle \begin{align}
z=2+i:\qquad z^2-4z+5
&= (2+i)^2-4(2+i)+5\\[5pt] 
&= 2^2+4i+i^2-8-4i+5\\[5pt]
&= 4+4i-1-8-4i+5\\[5pt]
&=0\,,\\[10pt] 
z={}\rlap{2-i:}\phantom{2+i:}{}\qquad z^2-4z+5
&= (2-i)^2-4(2-i)+5\\[5pt]
&= 2^2-4i+i^2-8+4i+5\\[5pt]
&= 4-4i-1-8+4i+5\\[5pt]
&= 0\,\textrm{.} 
\end{align}

Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse2-ImperialCollege/index.php/Solution_3.3:4b"
@@ Line 1: / Line 1: @@
-{{NAVCONTENT_START}}
+Typically, one solves a second-degree by completing the square, followed by taking the square root.
-<center> [[Bild:3_3_4b.gif]] </center>
-{{NAVCONTENT_STOP}}
+If we complete the square of the left-hand side, we get
+{{Displayed math||<math>\begin{align}
+(z-2)^2-2^2+5&=0,\\[5pt]
+(z-2)^2+1&=0.
+\end{align}</math>}}
+Taking the square root then gives that the equation has roots <math>z-2=\pm i</math>, i.e. <math>z=2+i</math> and <math>z=2-i</math>.
+If we want to be sure that we have found the correct solutions, we can substitute each solution into the equation and see whether the equation is satisfied.
+<math>\begin{align}
+z=2+i:\qquad z^2-4z+5
+&= (2+i)^2-4(2+i)+5\\[5pt]
+&= 2^2+4i+i^2-8-4i+5\\[5pt]
+&= 4+4i-1-8-4i+5\\[5pt]
+&=0\,,\\[10pt]
+z={}\rlap{2-i:}\phantom{2+i:}{}\qquad z^2-4z+5
+&= (2-i)^2-4(2-i)+5\\[5pt]
+&= 2^2-4i+i^2-8+4i+5\\[5pt]
+&= 4-4i-1-8+4i+5\\[5pt]
+&= 0\,\textrm{.}
+\end{align}</math>
 \displaystyle \begin{align}
(z-2)^2-2^2+5&=0,\\[5pt] 
(z-2)^2+1&=0.
\end{align}