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
                      
                    
                    
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+			Solution 3.3:4b
			
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 (Ny sida: {{NAVCONTENT_START}} <center> Bild:3_3_4b.gif </center> {{NAVCONTENT_STOP}})
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- {{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/2009/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}