Lösung 2.3:5b

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K (Lösning 2.3:5b moved to Solution 2.3:5b: Robot: moved page)
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{{NAVCONTENT_START}}
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Instead of randomly trying different values of
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<center> [[Image:2_3_5b.gif]] </center>
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<math>x</math>
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{{NAVCONTENT_STOP}}
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, it is better investigate the second-degree expression by completing the square:
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<math>\begin{align}
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& 4x^{2}-28x+48=4\left( x^{2}-7x+12 \right)=4\left( \left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+12 \right) \\
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& =4\left( \left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{48}{4} \right)=4\left( \left( x-\frac{7}{2} \right)^{2}-\frac{1}{4} \right)=4\left( x-\frac{7}{2} \right)^{2}-1. \\
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\end{align}</math>
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In the expression in which the square has been completed, we see that if, e.g.
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<math>x={7}/{2}\;</math>, then the whole expression is negative and equal to
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<math>-\text{1}</math>.
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NOTE: All values of
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<math>x</math>
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between
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<math>\text{3}</math>
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and
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<math>\text{4}</math>
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give a negative value for the expression.

Version vom 09:57, 21. Sep. 2008

Instead of randomly trying different values of \displaystyle x , it is better investigate the second-degree expression by completing the square:


\displaystyle \begin{align} & 4x^{2}-28x+48=4\left( x^{2}-7x+12 \right)=4\left( \left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+12 \right) \\ & =4\left( \left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{48}{4} \right)=4\left( \left( x-\frac{7}{2} \right)^{2}-\frac{1}{4} \right)=4\left( x-\frac{7}{2} \right)^{2}-1. \\ \end{align}


In the expression in which the square has been completed, we see that if, e.g. \displaystyle x={7}/{2}\;, then the whole expression is negative and equal to \displaystyle -\text{1}.

NOTE: All values of \displaystyle x between \displaystyle \text{3} and \displaystyle \text{4} give a negative value for the expression.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:5b“