Lösung 2.3:3d

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Because both terms,
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<center> [[Image:2_3_3d.gif]] </center>
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<math>x\left( x+3 \right)</math>
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and
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<math>x\left( 2x-9 \right)</math>
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contain the factor
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<math>x</math>, we can take out
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<math>x</math> from the left-hand side and collect together the remaining expression:
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<math>\begin{align}
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& x\left( x+3 \right)-x\left( 2x-9 \right)=x\left( \left( x+3 \right)-\left( 2x-9 \right) \right) \\
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& =x\left( x+3-2x+9 \right)=x\left( -x+12 \right) \\
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\end{align}</math>
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The equation is thus
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<math>x\left( -x+12 \right)=0</math>
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and we obtain directly that the equation is satisfied if either
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<math>x</math>
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or
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<math>-x+\text{12}</math>
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is zero. The solutions to the equation are therefore
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<math>x=0\text{ }</math>
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and
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<math>x=\text{12}</math>.
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Here, it can be worth checking that
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<math>x=\text{12 }</math>
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is a solution (the case
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<math>x=0</math>
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is obvious):
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LHS
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<math>=12\centerdot \left( 12+3 \right)-12\centerdot \left( 2\centerdot 12-9 \right)=2\centerdot 15-12\centerdot 15=0=</math>
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RHS

Version vom 15:05, 20. Sep. 2008

Because both terms, \displaystyle x\left( x+3 \right) and \displaystyle x\left( 2x-9 \right) contain the factor \displaystyle x, we can take out \displaystyle x from the left-hand side and collect together the remaining expression:


\displaystyle \begin{align} & x\left( x+3 \right)-x\left( 2x-9 \right)=x\left( \left( x+3 \right)-\left( 2x-9 \right) \right) \\ & =x\left( x+3-2x+9 \right)=x\left( -x+12 \right) \\ \end{align}


The equation is thus


\displaystyle x\left( -x+12 \right)=0

and we obtain directly that the equation is satisfied if either \displaystyle x or \displaystyle -x+\text{12} is zero. The solutions to the equation are therefore \displaystyle x=0\text{ } and \displaystyle x=\text{12}.

Here, it can be worth checking that \displaystyle x=\text{12 } is a solution (the case \displaystyle x=0 is obvious):

LHS \displaystyle =12\centerdot \left( 12+3 \right)-12\centerdot \left( 2\centerdot 12-9 \right)=2\centerdot 15-12\centerdot 15=0= RHS

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