Lösung 2.3:2f

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We divide both sides by
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<center> [[Image:2_3_2f.gif]] </center>
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<math>3</math>
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and complete the square on the left-hand side:
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<math>\begin{align}
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& x^{2}-\frac{10}{3}x+\frac{8}{3}=\left( x-\frac{5}{3} \right)^{2}-\left( -\frac{5}{3} \right)^{2}+\frac{8}{3}=\left( x-\frac{5}{3} \right)^{2}-\frac{25}{9}+\frac{24}{9} \\
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& =\left( x-\frac{5}{3} \right)^{2}-\frac{1}{9} \\
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\end{align}</math>
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The equation then becomes
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<math>\left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}</math>
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and taking the root gives the solutions as
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<math>x-\frac{5}{3}=\sqrt{\frac{1}{9}}=\frac{1}{3}</math>
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i.e.
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<math>x=\frac{5}{3}+\frac{1}{3}=\frac{6}{3}=2.</math>
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<math>x-\frac{5}{3}=-\sqrt{\frac{1}{9}}=-\frac{1}{3}</math>
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i.e.
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<math>x=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}.</math>
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Check:
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<math>x=\text{4}/\text{3}</math>: LHS
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<math>=3\centerdot \left( \frac{4}{3} \right)^{2}-10\centerdot \frac{4}{3}+8=3\centerdot \frac{16}{9}-\frac{40}{3}+\frac{8\centerdot 3}{3}=\frac{16-40+24}{3}=0=</math>
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RHS
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<math>x=\text{2}</math>: LHS
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<math>=3\centerdot 2^{2}-10\centerdot 2+8=12-20+8=0=</math>
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RHS

Version vom 14:31, 20. Sep. 2008

We divide both sides by \displaystyle 3 and complete the square on the left-hand side:


\displaystyle \begin{align} & x^{2}-\frac{10}{3}x+\frac{8}{3}=\left( x-\frac{5}{3} \right)^{2}-\left( -\frac{5}{3} \right)^{2}+\frac{8}{3}=\left( x-\frac{5}{3} \right)^{2}-\frac{25}{9}+\frac{24}{9} \\ & =\left( x-\frac{5}{3} \right)^{2}-\frac{1}{9} \\ \end{align}


The equation then becomes


\displaystyle \left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}

and taking the root gives the solutions as


\displaystyle x-\frac{5}{3}=\sqrt{\frac{1}{9}}=\frac{1}{3} i.e. \displaystyle x=\frac{5}{3}+\frac{1}{3}=\frac{6}{3}=2.


\displaystyle x-\frac{5}{3}=-\sqrt{\frac{1}{9}}=-\frac{1}{3} i.e. \displaystyle x=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}.


Check:


\displaystyle x=\text{4}/\text{3}: LHS \displaystyle =3\centerdot \left( \frac{4}{3} \right)^{2}-10\centerdot \frac{4}{3}+8=3\centerdot \frac{16}{9}-\frac{40}{3}+\frac{8\centerdot 3}{3}=\frac{16-40+24}{3}=0= RHS

\displaystyle x=\text{2}: LHS \displaystyle =3\centerdot 2^{2}-10\centerdot 2+8=12-20+8=0= RHS

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