Lösung 2.3:2b

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The first step when we solve the second-degree equation is to complete the square on the left-hand side:
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The first step when we solve the second-degree equation is to complete the square on the left-hand side
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{{Displayed math||<math>y^{2}+2y-15 = (y+1)^{2}-1^{2}-15 = (y+1)^{2}-16\,\textrm{.}</math>}}
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<math>y^{2}+2y-15=\left( y+1 \right)^{2}-1^{2}-15=\left( y+1 \right)^{2}-16.</math>
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The equation can now be written as
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The equation can now be written as
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{{Displayed math||<math>(y+1)^{2} = 16</math>}}
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and has, after taking the square root, the solutions:
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<math>\left( y+1 \right)^{2}=16</math>
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:*<math>y+1 = \sqrt{16} = 4\,\textrm{,}\ </math> which gives <math>y=-1+4=3\,\textrm{,}</math>
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:*<math>y+1 = -\sqrt{16} = -4\,\textrm{,}\ </math> which gives <math>y=-1-4=-5\,\textrm{.}</math>
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and has, after taking the square root, the solutions
 
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A quick check shows that <math>y=-5</math> and <math>y=3</math> satisfy the equation:
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<math>y+1=\sqrt{16}=4</math>
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:*''y''&nbsp;=&nbsp;-5: <math>\ \text{LHS} = (-5)^{2} + 2\cdot (-5)-15 = 25-10-15 = 0 = \text{RHS,}</math>
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which gives
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<math>y=-1+4=3</math>
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:*''y''&nbsp;=&nbsp;3: <math>\ \text{LHS} = 3^{2} + 2\cdot 3 - 15 = 9+6-15 = 0 = \text{RHS.}</math>
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<math>y+1=-\sqrt{16}=-4</math>
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which gives
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<math>y=-1-4=-5</math>
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A quick check shows that
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<math>y=-\text{5 }</math>
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and
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<math>y=\text{3 }</math>
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satisfy the equation:
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<math>y=-\text{5 }</math>: LHS=
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<math>\left( -5 \right)^{2}+2\centerdot \left( -5 \right)-15=25-10-15=0</math>
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= RHS
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<math>y=\text{3 }</math>: LHS=
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<math>3^{2}+2\centerdot 3-15=9+6-15=0</math>
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= RHS
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Version vom 14:30, 26. Sep. 2008

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

Vorlage:Displayed math

The equation can now be written as

Vorlage:Displayed math

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.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:2b“