Lösung 2.3:1c

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<math>2x-x^{2}</math>, which we also can write as <math>-(x^{2}-2x)</math>. If we neglect the minus sign, we can complete square of the expression <math>2x-x^{2}</math> by using the formula
<math>2x-x^{2}</math>, which we also can write as <math>-(x^{2}-2x)</math>. If we neglect the minus sign, we can complete square of the expression <math>2x-x^{2}</math> by using the formula
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{{Displayed math||<math>x^{2}-ax = \Bigl(x-\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{2}</math>}}
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{{Abgesetzte Formel||<math>x^{2}-ax = \Bigl(x-\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{2}</math>}}
and we obtain
and we obtain
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{{Displayed math||<math>x^{2}-2x = \Bigl(x-\frac{2}{2}\Bigr)^{2} - \Bigl(\frac{2}{2}\Bigr)^{2} = (x-1)^{2}-1\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>x^{2}-2x = \Bigl(x-\frac{2}{2}\Bigr)^{2} - \Bigl(\frac{2}{2}\Bigr)^{2} = (x-1)^{2}-1\,\textrm{.}</math>}}
This means that
This means that
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
5+2x-x^{2} &= 5-(x^{2}-2x) = 5-\bigl((x-1)^{2}-1\bigr)\\[5pt]
5+2x-x^{2} &= 5-(x^{2}-2x) = 5-\bigl((x-1)^{2}-1\bigr)\\[5pt]
&= 5-(x-1)^{2}+1 = 6-(x-1)^{2}\textrm{.}
&= 5-(x-1)^{2}+1 = 6-(x-1)^{2}\textrm{.}
Zeile 17: Zeile 17:
A quick check shows that we have completed the square correctly
A quick check shows that we have completed the square correctly
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{{Displayed math||<math>\begin{align}
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{{Abgesetzte Formel||<math>\begin{align}
6-(x-1)^{2}
6-(x-1)^{2}
&= 6-(x^{2}-2x+1)\\[5pt]
&= 6-(x^{2}-2x+1)\\[5pt]

Version vom 08:31, 22. Okt. 2008

As always when completing the square, we focus on the quadratic and linear terms \displaystyle 2x-x^{2}, which we also can write as \displaystyle -(x^{2}-2x). If we neglect the minus sign, we can complete square of the expression \displaystyle 2x-x^{2} by using the formula

\displaystyle x^{2}-ax = \Bigl(x-\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{2}

and we obtain

\displaystyle x^{2}-2x = \Bigl(x-\frac{2}{2}\Bigr)^{2} - \Bigl(\frac{2}{2}\Bigr)^{2} = (x-1)^{2}-1\,\textrm{.}

This means that

\displaystyle \begin{align}

5+2x-x^{2} &= 5-(x^{2}-2x) = 5-\bigl((x-1)^{2}-1\bigr)\\[5pt] &= 5-(x-1)^{2}+1 = 6-(x-1)^{2}\textrm{.} \end{align}

A quick check shows that we have completed the square correctly

\displaystyle \begin{align}

6-(x-1)^{2} &= 6-(x^{2}-2x+1)\\[5pt] &= 6-x^{2}+2x-1\\[5pt] & =5+2x-x^{2}\textrm{.} \end{align}

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