Lösung 2.3:1d

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K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel))
Zeile 1: Zeile 1:
We apply the standard formula for completing the square,
We apply the standard formula for completing the square,
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{{Displayed math||<math>x^{2}+ax = \Bigl(x+\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{2}\,\textrm{,}</math>}}
+
{{Abgesetzte Formel||<math>x^{2}+ax = \Bigl(x+\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{2}\,\textrm{,}</math>}}
on our expression and this gives
on our expression and this gives
-
{{Displayed math||<math>x^{2}+5x = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>x^{2}+5x = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}\,\textrm{.}</math>}}
The whole expression becomes
The whole expression becomes
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
x^{2}+5x+3
x^{2}+5x+3
&= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}+3\\[5pt]
&= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}+3\\[5pt]
Zeile 19: Zeile 19:
A quick check shows that we have calculated correctly
A quick check shows that we have calculated correctly
-
{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{13}{4}
\Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{13}{4}
&= x^{2} + 2\cdot\frac{5}{2}\cdot x + \Bigl(\frac{5}{2}\Bigr)^{2} - \frac{13}{4}\\[5pt]
&= x^{2} + 2\cdot\frac{5}{2}\cdot x + \Bigl(\frac{5}{2}\Bigr)^{2} - \frac{13}{4}\\[5pt]

Version vom 08:31, 22. Okt. 2008

We apply the standard formula for completing the square,

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

on our expression and this gives

\displaystyle x^{2}+5x = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}\,\textrm{.}

The whole expression becomes

\displaystyle \begin{align}

x^{2}+5x+3 &= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}+3\\[5pt] &= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{12}{4}\\[5pt] &= \Bigl(x+\frac{5}{2}\Bigr)^{2} + \frac{12-25}{4}\\[5pt] &= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{13}{4}\,\textrm{.} \end{align}

A quick check shows that we have calculated correctly

\displaystyle \begin{align}

\Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{13}{4} &= x^{2} + 2\cdot\frac{5}{2}\cdot x + \Bigl(\frac{5}{2}\Bigr)^{2} - \frac{13}{4}\\[5pt] &= x^{2} + 5x + \frac{25}{4} - \frac{13}{4}\\[5pt] &= x^{2} + 5x + \frac{12}{4}\\[5pt] &= x^{2}+5x+3\,\textrm{.} \end{align}

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