Lösung 3.3:4b

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Typically, one solves a second-degree by completing the square, followed by taking the root.
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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
If we complete the square of the left-hand side, we get
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{{Displayed math||<math>\begin{align}
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(z-2)^2-2^2+5&=0,\\[5pt]
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(z-2)^2+1&=0.
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\end{align}</math>}}
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<math>\begin{align}
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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>.
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& \left( z-2 \right)^{2}-2^{2}+5=0, \\
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& \left( z-2 \right)^{2}+1=0. \\
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\end{align}</math>
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Taking the root then gives that the equation has roots
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<math>z-2=\pm i,</math>
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i.e.
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<math>z=\text{2}+i</math>
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and
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<math>z=\text{2}-i</math>.
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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.
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.
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<math>\begin{align}
 
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& z=\text{2}+i:\quad z^{2}-4z+5=\left( \text{2}+i \right)^{2}-4\left( \text{2}+i \right)+5 \\
 
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& =2^{2}+4i+i^{2}-8-4i+5 \\
 
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& =4+4i-1-8-4i+5=0 \\
 
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\end{align}</math>
 
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<math>\begin{align}
<math>\begin{align}
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& z=\text{2-}i:\quad z^{2}-4z+5=\left( \text{2-}i \right)^{2}-4\left( \text{2-}i \right)+5 \\
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z=2+i:\qquad z^2-4z+5
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& =2^{2}-4i+i^{2}-8+4i+5 \\
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&= (2+i)^2-4(2+i)+5\\[5pt]
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& =4-4i-1-8+4i+5=0 \\
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&= 2^2+4i+i^2-8-4i+5\\[5pt]
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&= 4+4i-1-8-4i+5\\[5pt]
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&=0\,,\\[10pt]
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z={}\rlap{2-i:}\phantom{2+i:}{}\qquad z^2-4z+5
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&= (2-i)^2-4(2-i)+5\\[5pt]
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&= 2^2-4i+i^2-8+4i+5\\[5pt]
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&= 4-4i-1-8+4i+5\\[5pt]
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&= 0\,\textrm{.}
\end{align}</math>
\end{align}</math>

Version vom 14:30, 30. Okt. 2008

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}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse2-TU-Berlin/index.php/L%C3%B6sung_3.3:4b“