Lösung 3.3:4c

Aus Online Mathematik Brückenkurs 2

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 3.3:4c moved to Solution 3.3:4c: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
We complete the square on the left-hand side:
-
<center> [[Image:3_3_4c.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
 +
<math>\begin{align}
 +
& \left( z+1 \right)^{\text{2}}-1^{2}+3=0 \\
 +
& \left( z+1 \right)^{\text{2}}+2=0 \\
 +
\end{align}</math>
 +
 
 +
 
 +
Taking the root now gives
 +
<math>z+1=\pm i\sqrt{2}</math>
 +
i.e.
 +
<math>z=-1+i\sqrt{2}</math>
 +
and
 +
<math>z=-1-i\sqrt{2}</math>.
 +
 
 +
We test the solutions in the equation to ascertain that we have calculated correctly.
 +
 
 +
 
 +
<math>\begin{align}
 +
& z=-1+i\sqrt{2}:\quad z^{2}+2z+3=\left( -1+i\sqrt{2} \right)^{2}+2\left( -1+i\sqrt{2} \right)+3 \\
 +
& =\left( -1 \right)^{2}-2\centerdot i\sqrt{2}+i^{2}\left( \sqrt{2} \right)^{2}-2+2i\sqrt{2}+3 \\
 +
& =1-2\centerdot i\sqrt{2}-2-2+2i\sqrt{2}+3=0, \\
 +
\end{align}</math>
 +
 
 +
 
 +
 
 +
<math>\begin{align}
 +
& z=-1-i\sqrt{2}:\quad z^{2}+2z+3=\left( -1-i\sqrt{2} \right)^{2}+2\left( -1-i\sqrt{2} \right)+3 \\
 +
& =\left( -1 \right)^{2}+2\centerdot i\sqrt{2}+i^{2}\left( \sqrt{2} \right)^{2}-2-2i\sqrt{2}+3 \\
 +
& =1+2\centerdot i\sqrt{2}-2-2-2\sqrt{2}i+3=0, \\
 +
\end{align}</math>

Version vom 10:32, 25. Okt. 2008

We complete the square on the left-hand side:


\displaystyle \begin{align} & \left( z+1 \right)^{\text{2}}-1^{2}+3=0 \\ & \left( z+1 \right)^{\text{2}}+2=0 \\ \end{align}


Taking the root now gives \displaystyle z+1=\pm i\sqrt{2} i.e. \displaystyle z=-1+i\sqrt{2} and \displaystyle z=-1-i\sqrt{2}.

We test the solutions in the equation to ascertain that we have calculated correctly.


\displaystyle \begin{align} & z=-1+i\sqrt{2}:\quad z^{2}+2z+3=\left( -1+i\sqrt{2} \right)^{2}+2\left( -1+i\sqrt{2} \right)+3 \\ & =\left( -1 \right)^{2}-2\centerdot i\sqrt{2}+i^{2}\left( \sqrt{2} \right)^{2}-2+2i\sqrt{2}+3 \\ & =1-2\centerdot i\sqrt{2}-2-2+2i\sqrt{2}+3=0, \\ \end{align}


\displaystyle \begin{align} & z=-1-i\sqrt{2}:\quad z^{2}+2z+3=\left( -1-i\sqrt{2} \right)^{2}+2\left( -1-i\sqrt{2} \right)+3 \\ & =\left( -1 \right)^{2}+2\centerdot i\sqrt{2}+i^{2}\left( \sqrt{2} \right)^{2}-2-2i\sqrt{2}+3 \\ & =1+2\centerdot i\sqrt{2}-2-2-2\sqrt{2}i+3=0, \\ \end{align}