Lösung 2.3:2a

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We solve the second order equation by combining together the x2- and
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We solve the second order equation by combining together the ''x''²- and ''x''-terms by completing the square to obtain a quadratic term, and then solve the resulting equation by taking the root.
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<math>x</math>
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-terms by completing the square to obtain a quadratic term, and then solve the resulting equation by taking the root.
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By completing the square, the left-hand side becomes
By completing the square, the left-hand side becomes
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{{Displayed math||<math>\underline{x^{2}-4x\vphantom{()}}+3 = \underline{(x-2)^{2}-2^{2}}+3 = (x-2)^{2}-1\,\textrm{,}</math>}}
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<math>\underline{x^{2}-4x}+3=\underline{\left( x-2 \right)^{2}-2^{2}}+3=\left( x-2 \right)^{2}-1</math>
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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
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{{Displayed math||<math>(x-2)^{2}-1 = 0</math>}}
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<math>\left( x-2 \right)^{2}-1=0</math>
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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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which we solve by moving the "
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<math>1</math>
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" on the right-hand side and taking the square root. This gives the solutions
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<math>x-2=\sqrt{1}=1</math>
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:*<math>x-2=\sqrt{1}=1\,,\ </math> i.e. <math>x=2+1=3\,,</math>
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i.e.
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<math>x=2+1=3</math>
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<math>x-2=-\sqrt{1}=-1</math>
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i.e.
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<math>x=2-1=1</math>
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:*<math>x-2=-\sqrt{1}=-1\,,\ </math> i.e. <math>x=2-1=1\,\textrm{.}</math>
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Because it is easy to make a mistake, we check the answer by substituting
 
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<math>x=1</math>
 
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and
 
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<math>x=3</math>
 
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into the original equation.:
 
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Because it is easy to make a mistake, we check the answer by substituting <math>x=1</math> and <math>x=3</math> into the original equation:
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<math>x=\text{1}</math>: LHS=
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:*''x''&nbsp;=&nbsp;1: <math>\ \text{LHS} = 1^{2}-4\cdot 1+3 = 1-4+3 = 0 = \text{RHS,}</math>
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<math>1^{2}-4\centerdot 1+3=1-4+3=0</math>
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= RHS
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<math>x=3</math>: LHS=
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:*''x''&nbsp;=&nbsp;3: <math>\ \text{LHS} = 3^{2}-4\cdot 3+3 = 9-12+3 = 0 = \text{RHS.}</math>
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<math>3^{2}-4\centerdot 3+3=9-12+3=0</math>
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= RHS
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Version vom 14:20, 26. Sep. 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

Vorlage:Displayed math

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

Vorlage:Displayed math

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