Aus Online Mathematik Brückenkurs 1

Version vom 08:56, 9. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 2.3:2b moved to Solution 2.3:2b: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 13:31, 20. Sep. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
-	{{NAVCONTENT_START}}	+	The first step when we solve the second-degree equation is to complete the square on the left-hand side:
-	<center> [[Image:2_3_2b.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
		+	<math>y^{2}+2y-15=\left( y+1 \right)^{2}-1^{2}-15=\left( y+1 \right)^{2}-16.</math>
		+
		+	The equation can now be written as
		+
		+
		+	<math>\left( y+1 \right)^{2}=16</math>
		+
		+
		+	and has, after taking the square root, the solutions
		+
		+
		+	<math>y+1=\sqrt{16}=4</math>
		+	which gives
		+	<math>y=-1+4=3</math>
		+
		+
		+	<math>y+1=-\sqrt{16}=-4</math>
		+	which gives
		+	<math>y=-1-4=-5</math>
		+
		+
		+	A quick check shows that
		+	<math>y=-\text{5 }</math>
		+	and
		+	<math>y=\text{3 }</math>
		+	satisfy the equation:
		+
		+
		+	<math>y=-\text{5 }</math>: LHS=
		+	<math>\left( -5 \right)^{2}+2\centerdot \left( -5 \right)-15=25-10-15=0</math>
		+	= RHS
		+
		+	<math>y=\text{3 }</math>: LHS=
		+	<math>3^{2}+2\centerdot 3-15=9+6-15=0</math>
		+	= RHS

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Version vom 13:31, 20. Sep. 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=\left( y+1 \right)^{2}-1^{2}-15=\left( y+1 \right)^{2}-16.

The equation can now be written as

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

and has, after taking the square root, the solutions

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

\displaystyle y+1=-\sqrt{16}=-4 which gives \displaystyle y=-1-4=-5

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

\displaystyle y=-\text{5 }: LHS= \displaystyle \left( -5 \right)^{2}+2\centerdot \left( -5 \right)-15=25-10-15=0 = RHS

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