Aus Online Mathematik Brückenkurs 1

Version vom 11:19, 12. Sep. 2008 (bearbeiten) Ian (Diskussion | Beiträge) ← Zum vorherigen Versionsunterschied		Version vom 11:20, 12. Sep. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) K Zum nächsten Versionsunterschied →
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	As always when completing the square, we focus on the quadratic and linear terms		As always when completing the square, we focus on the quadratic and linear terms
-	<math>2x-x^{2}</math>	+	<math>2x-x^{2}</math>, which we also can write as
-	, which we also can write as	+
	<math>-\left( x^{2}-2x \right)</math>		<math>-\left( x^{2}-2x \right)</math>
	. If we neglect the minus sign, we can complete square of the expression		. If we neglect the minus sign, we can complete square of the expression

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Version vom 11:20, 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}