Lösung 2.3:3f

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We can split up the first term on the left-hand side,
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<center> [[Image:2_3_3f.gif]] </center>
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<math>x\left( x^{2}-2x \right)</math>
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, into factors by taking
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<math>x</math>
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outside the bracket,
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<math>x\left( x^{2}-2x \right)=x\centerdot x\centerdot \left( x-2 \right)</math>
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and writing the other term as
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<math>x\centerdot \left( 2-x \right)=-x\left( x-2 \right)</math>. From this we see that both terms contain
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<math>x\left( x-2 \right)</math>
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as common factors and, if we take out those, the left-hand side becomes
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<math>\begin{align}
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& x\left( x^{2}-2x \right)+x\left( 2-x \right)=x^{2}\left( x-2 \right)-x\left( x-2 \right) \\
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& =x\left( x\left( x-2 \right)-\left( x-2 \right) \right)=x\left( x-2 \right)\left( x-1 \right). \\
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\end{align}</math>
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The whole equation can be written as
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<math>x\left( x-2 \right)\left( x-1 \right)=0</math>
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and this equation is satisfied only when one of the three factors
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<math>x</math>,
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<math>x-\text{2}</math>
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or
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<math>x-\text{1}</math>
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is zero, i.e. the solutions are
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<math>x=0</math>,
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<math>x=\text{2 }</math>
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and
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<math>x=\text{1}</math>.
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Because it is not completely obvious that x x=1 is a solution of the equation, we check that x=1 satisfies the equation, i.e. that we haven't calculated incorrectly:
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x=1: LHS
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<math>=1\centerdot \left( 1^{2}-2\centerdot 1 \right)+1\centerdot \left( 2-1 \right)=1\centerdot \left( -1 \right)+1\centerdot 1=0=</math>
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RHS

Version vom 15:28, 20. Sep. 2008

We can split up the first term on the left-hand side, \displaystyle x\left( x^{2}-2x \right) , into factors by taking \displaystyle x outside the bracket, \displaystyle x\left( x^{2}-2x \right)=x\centerdot x\centerdot \left( x-2 \right) and writing the other term as \displaystyle x\centerdot \left( 2-x \right)=-x\left( x-2 \right). From this we see that both terms contain \displaystyle x\left( x-2 \right) as common factors and, if we take out those, the left-hand side becomes


\displaystyle \begin{align} & x\left( x^{2}-2x \right)+x\left( 2-x \right)=x^{2}\left( x-2 \right)-x\left( x-2 \right) \\ & =x\left( x\left( x-2 \right)-\left( x-2 \right) \right)=x\left( x-2 \right)\left( x-1 \right). \\ \end{align}


The whole equation can be written as


\displaystyle x\left( x-2 \right)\left( x-1 \right)=0


and this equation is satisfied only when one of the three factors \displaystyle x, \displaystyle x-\text{2} or \displaystyle x-\text{1} is zero, i.e. the solutions are \displaystyle x=0, \displaystyle x=\text{2 } and \displaystyle x=\text{1}.

Because it is not completely obvious that x x=1 is a solution of the equation, we check that x=1 satisfies the equation, i.e. that we haven't calculated incorrectly:

x=1: LHS \displaystyle =1\centerdot \left( 1^{2}-2\centerdot 1 \right)+1\centerdot \left( 2-1 \right)=1\centerdot \left( -1 \right)+1\centerdot 1=0= RHS

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