Aus Online Mathematik Brückenkurs 1

Version vom 12:42, 1. Okt. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 08:40, 22. Okt. 2008 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
	Square both sides of the equation so that the root sign disappears,		Square both sides of the equation so that the root sign disappears,
-	{{Displayed math||<math>1-x = (2-x)^2\quad \Leftrightarrow \quad 1-x = 4-4x+x^2</math>}}	+	{{Abgesetzte Formel||<math>1-x = (2-x)^2\quad \Leftrightarrow \quad 1-x = 4-4x+x^2</math>}}
	and then solve the resulting second-order equation by completing the square,		and then solve the resulting second-order equation by completing the square,
-	{{Displayed math||<math>\begin{align}	+	{{Abgesetzte Formel||<math>\begin{align}
	x^{2}-3x+3 &= 0\,,\\[5pt]		x^{2}-3x+3 &= 0\,,\\[5pt]
	\Bigl(x-\frac{3}{2}\Bigr)^{2} - \Bigl(\frac{3}{2}\Bigr)^{2} + 3 &= 0\,,\\[5pt]		\Bigl(x-\frac{3}{2}\Bigr)^{2} - \Bigl(\frac{3}{2}\Bigr)^{2} + 3 &= 0\,,\\[5pt]

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Version vom 08:40, 22. Okt. 2008

Square both sides of the equation so that the root sign disappears,

\displaystyle 1-x = (2-x)^2\quad \Leftrightarrow \quad 1-x = 4-4x+x^2

and then solve the resulting second-order equation by completing the square,

\displaystyle \begin{align} x^{2}-3x+3 &= 0\,,\\[5pt] \Bigl(x-\frac{3}{2}\Bigr)^{2} - \Bigl(\frac{3}{2}\Bigr)^{2} + 3 &= 0\,,\\[5pt] \Bigl(x-\frac{3}{2}\Bigr)^{2} - \frac{9}{4} + \frac{12}{4} &= 0\,,\\[5pt] \Bigl(x-\frac{3}{2}\Bigr)^{2} + \frac{3}{4} &= 0\,\textrm{.} \end{align}

As can be seen, the second-order equation does not have any solutions (the left-hand side is always greater than or equal to 3/4, regardless of how x is chosen) so, the original root equation does not have any solutions.