Aus Online Mathematik Brückenkurs 1

Version vom 14:30, 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 →
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	The first step when we solve the second-degree equation is to complete the square on the left-hand side		The first step when we solve the second-degree equation is to complete the square on the left-hand side
-	{{Displayed math||<math>y^{2}+2y-15 = (y+1)^{2}-1^{2}-15 = (y+1)^{2}-16\,\textrm{.}</math>}}	+	{{Abgesetzte Formel||<math>y^{2}+2y-15 = (y+1)^{2}-1^{2}-15 = (y+1)^{2}-16\,\textrm{.}</math>}}
	The equation can now be written as		The equation can now be written as
-	{{Displayed math||<math>(y+1)^{2} = 16</math>}}	+	{{Abgesetzte Formel||<math>(y+1)^{2} = 16</math>}}
	and has, after taking the square root, the solutions:		and has, after taking the square root, the solutions:

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Version vom 08:31, 22. Okt. 2008

The first step when we solve the second-degree equation is to complete the square on the left-hand side

\displaystyle y^{2}+2y-15 = (y+1)^{2}-1^{2}-15 = (y+1)^{2}-16\,\textrm{.}

The equation can now be written as

\displaystyle (y+1)^{2} = 16

and has, after taking the square root, the solutions:

• \displaystyle y+1 = \sqrt{16} = 4\,\textrm{,}\ which gives \displaystyle y=-1+4=3\,\textrm{,}

• \displaystyle y+1 = -\sqrt{16} = -4\,\textrm{,}\ which gives \displaystyle y=-1-4=-5\,\textrm{.}

A quick check shows that \displaystyle y=-5 and \displaystyle y=3 satisfy the equation:

• y = -5: \displaystyle \ \text{LHS} = (-5)^{2} + 2\cdot (-5)-15 = 25-10-15 = 0 = \text{RHS,}

• y = 3: \displaystyle \ \text{LHS} = 3^{2} + 2\cdot 3 - 15 = 9+6-15 = 0 = \text{RHS.}