Lösung 2.3:2d

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The equation can be written in normalized form (i.e. the coefficient in front of
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<center> [[Image:2_3_2d.gif]] </center>
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<math>x^{\text{2}}</math>
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is
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<math>1</math>
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) by dividing both sides by
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<math>4</math>,
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<math>x^{2}-7x+\frac{13}{4}=0</math>
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Completing the square on the left-hand side,
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<math>\begin{align}
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& x^{2}-7x+\frac{13}{4}=\left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+\frac{13}{4}=\left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{13}{4} \\
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& =\left( x-\frac{7}{2} \right)^{2}-\frac{36}{4}=\left( x-\frac{7}{2} \right)^{2}-9 \\
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\end{align}</math>
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The equation can therefore be written as
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<math>\left( x-\frac{7}{2} \right)^{2}-9=0</math>
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and taking the square root gives the solutions as
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<math>x-\frac{7}{2}=\sqrt{9}=3</math>
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i.e.
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<math>x=\frac{7}{2}+3=\frac{13}{2},</math>
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<math>x-\frac{7}{2}=-\sqrt{9}=-3</math>
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i.e.
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<math>x=\frac{7}{2}-3=\frac{1}{2}.</math>
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As an extra check, we substitute x=1/2 and x=13/2 into the equation:
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<math>x=\text{1}/\text{2}</math>: LHS
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<math>=4\centerdot \left( \frac{1}{2} \right)^{2}-28\centerdot \frac{1}{2}+13=4\centerdot \frac{1}{4}-14+13=1-14+13=</math>
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RHS
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<math>x=\text{13}/\text{2}</math>: LHS
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<math>=4\centerdot \left( \frac{13}{2} \right)^{2}-28\centerdot \frac{13}{2}+13=4\centerdot \frac{169}{4}-14\centerdot 13+13=169-182+13=</math>
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RHS

Version vom 13:53, 20. Sep. 2008

The equation can be written in normalized form (i.e. the coefficient in front of \displaystyle x^{\text{2}} is \displaystyle 1 ) by dividing both sides by \displaystyle 4,


\displaystyle x^{2}-7x+\frac{13}{4}=0


Completing the square on the left-hand side,


\displaystyle \begin{align} & x^{2}-7x+\frac{13}{4}=\left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+\frac{13}{4}=\left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{13}{4} \\ & =\left( x-\frac{7}{2} \right)^{2}-\frac{36}{4}=\left( x-\frac{7}{2} \right)^{2}-9 \\ \end{align}

The equation can therefore be written as


\displaystyle \left( x-\frac{7}{2} \right)^{2}-9=0

and taking the square root gives the solutions as


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


\displaystyle x-\frac{7}{2}=-\sqrt{9}=-3 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:


\displaystyle x=\text{1}/\text{2}: LHS \displaystyle =4\centerdot \left( \frac{1}{2} \right)^{2}-28\centerdot \frac{1}{2}+13=4\centerdot \frac{1}{4}-14+13=1-14+13= RHS

\displaystyle x=\text{13}/\text{2}: LHS \displaystyle =4\centerdot \left( \frac{13}{2} \right)^{2}-28\centerdot \frac{13}{2}+13=4\centerdot \frac{169}{4}-14\centerdot 13+13=169-182+13= RHS

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