Lösung 2.3:1c
Aus Online Mathematik Brückenkurs 1
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| - | {{ | + | As always when completing the square, we focus on the quadratic and linear terms  | 
| - | < | + | <math>2x-x^{2}</math> | 
| - | {{ | + | , which we also can write as  | 
| + | <math>-\left( x^{2}-2x \right)</math> | ||
| + | . If we neglect the minus sign, we can complete square of the expression  | ||
| + | <math>2x-x^{2}</math> | ||
| + | by using the formula | ||
| + | |||
| + | |||
| + | <math>x^{2}-ax=\left( x-\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}</math> | ||
| + | |||
| + | |||
| + | and we obtain | ||
| + | |||
| + | |||
| + | <math>x^{2}-2x=\left( x-\frac{2}{2} \right)^{2}-\left( \frac{2}{2} \right)^{2}=\left( x-1 \right)^{2}-1</math> | ||
| + | |||
| + | |||
| + | This means that | ||
| + | |||
| + | |||
| + | <math>\begin{align} | ||
| + | & 5+2x-x^{2}=5-\left( x^{2}-2x \right)=5-\left( \left( x-1 \right)^{2}-1 \right) \\  | ||
| + | &  \\  | ||
| + | & =5-\left( x-1 \right)^{2}+1=6-\left( x-1 \right)^{2} \\  | ||
| + | &  \\  | ||
| + | \end{align}</math> | ||
| + | |||
| + | |||
| + | A quick check shows that we have completed the square correctly.: | ||
| + | |||
| + | |||
| + | <math>\begin{align} | ||
| + | & 6-\left( x-1 \right)^{2}=6-\left( x^{2}-2x+1 \right)=6-x^{2}+2x-1 \\  | ||
| + | &  \\  | ||
| + | & =5+2x-x^{2} \\  | ||
| + | &  \\  | ||
| + | \end{align}</math> | ||
Version vom 11:19, 12. Sep. 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 -\left( x^{2}-2x \right) . 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=\left( x-\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}
and we obtain
\displaystyle x^{2}-2x=\left( x-\frac{2}{2} \right)^{2}-\left( \frac{2}{2} \right)^{2}=\left( x-1 \right)^{2}-1
This means that
\displaystyle \begin{align}
& 5+2x-x^{2}=5-\left( x^{2}-2x \right)=5-\left( \left( x-1 \right)^{2}-1 \right) \\ 
&  \\ 
& =5-\left( x-1 \right)^{2}+1=6-\left( x-1 \right)^{2} \\ 
&  \\ 
\end{align}
A quick check shows that we have completed the square correctly.:
\displaystyle \begin{align}
& 6-\left( x-1 \right)^{2}=6-\left( x^{2}-2x+1 \right)=6-x^{2}+2x-1 \\ 
&  \\ 
& =5+2x-x^{2} \\ 
&  \\ 
\end{align}
 
		  