Aus Online Mathematik Brückenkurs 1

Version vom 14:20, 26. Sep. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 08:31, 22. Okt. 2008 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
Zeile 3:		Zeile 3:
	By completing the square, the left-hand side becomes		By completing the square, the left-hand side becomes
-	{{Displayed math||<math>\underline{x^{2}-4x\vphantom{()}}+3 = \underline{(x-2)^{2}-2^{2}}+3 = (x-2)^{2}-1\,\textrm{,}</math>}}	+	{{Abgesetzte Formel||<math>\underline{x^{2}-4x\vphantom{()}}+3 = \underline{(x-2)^{2}-2^{2}}+3 = (x-2)^{2}-1\,\textrm{,}</math>}}
	where the underlined part on the right-hand side is the actual completed square. The equation can therefore be written as		where the underlined part on the right-hand side is the actual completed square. The equation can therefore be written as
-	{{Displayed math||<math>(x-2)^{2}-1 = 0</math>}}	+	{{Abgesetzte Formel||<math>(x-2)^{2}-1 = 0</math>}}
	which we solve by moving the "1" on the right-hand side and taking the square root. This gives the solutions:		which we solve by moving the "1" on the right-hand side and taking the square root. This gives the solutions:

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Version vom 08:31, 22. Okt. 2008

We solve the second order equation by combining together the x²- and x-terms by completing the square to obtain a quadratic term, and then solve the resulting equation by taking the root.

By completing the square, the left-hand side becomes

\displaystyle \underline{x^{2}-4x\vphantom{()}}+3 = \underline{(x-2)^{2}-2^{2}}+3 = (x-2)^{2}-1\,\textrm{,}

where the underlined part on the right-hand side is the actual completed square. The equation can therefore be written as

\displaystyle (x-2)^{2}-1 = 0

which we solve by moving the "1" on the right-hand side and taking the square root. This gives the solutions:

• \displaystyle x-2=\sqrt{1}=1\,,\ i.e. \displaystyle x=2+1=3\,,

• \displaystyle x-2=-\sqrt{1}=-1\,,\ i.e. \displaystyle x=2-1=1\,\textrm{.}

Because it is easy to make a mistake, we check the answer by substituting \displaystyle x=1 and \displaystyle x=3 into the original equation:

• x = 1: \displaystyle \ \text{LHS} = 1^{2}-4\cdot 1+3 = 1-4+3 = 0 = \text{RHS,}

• x = 3: \displaystyle \ \text{LHS} = 3^{2}-4\cdot 3+3 = 9-12+3 = 0 = \text{RHS.}