Lösung 2.3:1d

Aus Online Mathematik Brückenkurs 1

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K (Lösning 2.3:1d moved to Solution 2.3:1d: Robot: moved page)
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We apply the standard formula for completing the square,
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<center> [[Image:2_3_1d.gif]] </center>
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<math>x^{2}+ax=\left( x+\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}</math>
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on our expression and this gives
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<math>x^{2}+5x=\left( x+\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}</math>
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The whole expression becomes
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<math>\begin{align}
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& x^{2}+5x+3=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}+3=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{12}{4} \\
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& =\left( x+\frac{5}{2} \right)^{2}+\frac{12-25}{4}=\left( x+\frac{5}{2} \right)^{2}-\frac{13}{4} \\
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\end{align}</math>
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A quick check shows that we have calculated correctly.
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<math>\begin{align}
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& \left( x+\frac{5}{2} \right)^{2}-\frac{13}{4}=x^{2}+2\centerdot \frac{5}{2}\centerdot x+\left( \frac{5}{2} \right)^{2}-\frac{13}{4}=x^{2}+5x+\frac{25}{4}-\frac{13}{4} \\
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& =x^{2}+5x+\frac{12}{4}=x^{2}+5x+3 \\
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\end{align}</math>

Version vom 13:02, 20. Sep. 2008

We apply the standard formula for completing the square,


\displaystyle x^{2}+ax=\left( x+\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}

on our expression and this gives


\displaystyle x^{2}+5x=\left( x+\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}

The whole expression becomes


\displaystyle \begin{align} & x^{2}+5x+3=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}+3=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{12}{4} \\ & =\left( x+\frac{5}{2} \right)^{2}+\frac{12-25}{4}=\left( x+\frac{5}{2} \right)^{2}-\frac{13}{4} \\ \end{align}


A quick check shows that we have calculated correctly.


\displaystyle \begin{align} & \left( x+\frac{5}{2} \right)^{2}-\frac{13}{4}=x^{2}+2\centerdot \frac{5}{2}\centerdot x+\left( \frac{5}{2} \right)^{2}-\frac{13}{4}=x^{2}+5x+\frac{25}{4}-\frac{13}{4} \\ & =x^{2}+5x+\frac{12}{4}=x^{2}+5x+3 \\ \end{align}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:1d“