Lösung 2.2:2c

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We can simplify the left-hand side in the equation by expanding the squares using the squaring rule:
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<center> [[Image:2_2_2c.gif]] </center>
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<math>\begin{align}
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& \left( x+3 \right)^{2}-\left( x-5 \right)^{2}=\left( x^{2}+2\centerdot 3x+3^{2} \right)-\left( x^{2}-2\centerdot 5x+5^{2} \right) \\
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& =x^{2}+6x+9-x^{2}+10x-25=16x-16 \\
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\end{align}</math>
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Thus, the equation is
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<math>16x-16=6x+4</math>
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Now, move all "
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<math>x</math>
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"s to the left-hand side (subtract
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<math>6x</math>
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from both sides) and the constants to the right-hand side (add
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<math>16</math>
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to both sides)
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<math>\begin{align}
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& 16x-6x=4+16 \\
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& 10x=20 \\
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\end{align}</math>
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Divide both sides by
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<math>10</math>
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to get the answer
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<math>x=\frac{20}{10}=2</math>
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Finally, we check that
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<math>x=2</math>
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satisfies the equation in the exercise:
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LHS =
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<math>\left( 2+3 \right)^{2}-\left( 2-5 \right)^{2}=5^{2}-\left( -3 \right)^{2}=25-9=16</math>
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RHS =
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<math>6\centerdot 2+4=12+4=16</math>

Version vom 12:58, 17. Sep. 2008

We can simplify the left-hand side in the equation by expanding the squares using the squaring rule:


\displaystyle \begin{align} & \left( x+3 \right)^{2}-\left( x-5 \right)^{2}=\left( x^{2}+2\centerdot 3x+3^{2} \right)-\left( x^{2}-2\centerdot 5x+5^{2} \right) \\ & =x^{2}+6x+9-x^{2}+10x-25=16x-16 \\ \end{align}


Thus, the equation is


\displaystyle 16x-16=6x+4


Now, move all " \displaystyle x "s to the left-hand side (subtract \displaystyle 6x from both sides) and the constants to the right-hand side (add \displaystyle 16 to both sides)


\displaystyle \begin{align} & 16x-6x=4+16 \\ & 10x=20 \\ \end{align}


Divide both sides by \displaystyle 10 to get the answer


\displaystyle x=\frac{20}{10}=2


Finally, we check that \displaystyle x=2 satisfies the equation in the exercise:

LHS = \displaystyle \left( 2+3 \right)^{2}-\left( 2-5 \right)^{2}=5^{2}-\left( -3 \right)^{2}=25-9=16


RHS = \displaystyle 6\centerdot 2+4=12+4=16

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