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Lösung 3.3:4b

Aus Online Mathematik Brückenkurs 2

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K (Lösning 3.3:4b moved to Solution 3.3:4b: Robot: moved page)
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Typically, one solves a second-degree by completing the square, followed by taking the root.
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<center> [[Image:3_3_4b.gif]] </center>
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If we complete the square of the left-hand side, we get
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<math>\begin{align}
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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.
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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}
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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>

Version vom 09:48, 25. Okt. 2008

Typically, one solves a second-degree by completing the square, followed by taking the root.

If we complete the square of the left-hand side, we get


z−22−22+5=0z−22+1=0


Taking the root then gives that the equation has roots z−2=i i.e. z=2+i and 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.


z=2+i:z2−4z+5=2+i2−42+i+5=22+4i+i2−8−4i+5=4+4i−1−8−4i+5=0



z=2-i:z2−4z+5=2-i2−42-i+5=22−4i+i2−8+4i+5=4−4i−1−8+4i+5=0

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