Lösung 2.3:2d

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As an extra check, we substitute ''x'' = 1/2 and ''x'' = 13/2 into the equation:
As an extra check, we substitute ''x'' = 1/2 and ''x'' = 13/2 into the equation:
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:*''x''&nbsp;=&nbsp;1/2: <math>\ \text{LHS} = 4\cdot\Bigl(\frac{1}{2}\Bigr)^{2} - 28\cdot\frac{1}{2}+13 = 4\cdot\frac{1}{4}-14+13 = 1-14+13 = \text{RHS,}</math>
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:*''x''&nbsp;=&nbsp;1/2: <math>\ \text{LHS} = 4\cdot\bigl(\tfrac{1}{2}\bigr)^{2} - 28\cdot\tfrac{1}{2}+13 = 4\cdot\tfrac{1}{4}-14+13 = \text{RHS,}</math>
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:*''x''&nbsp;=&nbsp;13/2: <math>\ \text{LHS} = 4\cdot\Bigl(\frac{13}{2}\Bigr)^{2} - 28\cdot\frac{13}{2}+13 = 4\cdot\frac{169}{4} - 14\cdot 13 + 13 = 169 - 182 + 13 = \text{RHS.}</math>
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:*''x''&nbsp;=&nbsp;13/2: <math>\ \text{LHS} = 4\cdot\bigl(\tfrac{13}{2}\bigr)^{2} - 28\cdot\tfrac{13}{2}+13 = 4\cdot\tfrac{169}{4} - 14\cdot 13 + 13 = \text{RHS.}</math>

Version vom 07:57, 29. Sep. 2008

The equation can be written in normalized form (i.e. the coefficient in front of x² is 1) by dividing both sides by 4,

Vorlage:Displayed math

Complete the square on the left-hand side,

Vorlage:Displayed math

The equation can therefore be written as

Vorlage:Displayed math

and taking the square root gives the solutions as

  • \displaystyle x-\frac{7}{2}=\sqrt{9}=3\,,\quad i.e. \displaystyle x=\frac{7}{2}+3=\frac{13}{2},
  • \displaystyle x-\frac{7}{2}=-\sqrt{9}=-3\,,\quad i.e. \displaystyle x=\frac{7}{2}-3=\frac{1}{2}.

As an extra check, we substitute x = 1/2 and x = 13/2 into the equation:

  • x = 1/2: \displaystyle \ \text{LHS} = 4\cdot\bigl(\tfrac{1}{2}\bigr)^{2} - 28\cdot\tfrac{1}{2}+13 = 4\cdot\tfrac{1}{4}-14+13 = \text{RHS,}
  • x = 13/2: \displaystyle \ \text{LHS} = 4\cdot\bigl(\tfrac{13}{2}\bigr)^{2} - 28\cdot\tfrac{13}{2}+13 = 4\cdot\tfrac{169}{4} - 14\cdot 13 + 13 = \text{RHS.}
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