Lösung 2.3:1b

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When we complete the square, it is only the first two terms,
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<center> [[Image:2_3_1b.gif]] </center>
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<math>x^{2}+2x</math>
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, that are involved. The general
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formula for completing the square states that
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<math>x^{2}+ax</math>
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equals
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<math>\left( x+\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}</math>
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Note how the coefficient
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<math>a</math>
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in front of the
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<math>x</math>
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turns up halved in two places.
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If we use this formula, we obtain
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<math>x^{2}+2x=\left( x+\frac{2}{2} \right)^{2}-\left( \frac{2}{2} \right)^{2}=\left( x+1 \right)^{2}-1</math>
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and if we subtract the last "
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<math>1</math>
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" , we obtain
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<math>x^{2}+2x-1=\left( x+1 \right)^{2}-1-1=\left( x+1 \right)^{2}-2</math>
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To be completely certain that we have used the correct formula, we can expand the quadratic on the right-hand side,
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<math>\left( x+1 \right)^{2}-2=x^{2}+2x+1-2=x^{2}+2x-1</math>
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and see that the relation really holds.

Version vom 10:11, 12. Sep. 2008

When we complete the square, it is only the first two terms, \displaystyle x^{2}+2x , that are involved. The general formula for completing the square states that \displaystyle x^{2}+ax equals


\displaystyle \left( x+\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}


Note how the coefficient \displaystyle a in front of the \displaystyle x turns up halved in two places.

If we use this formula, we obtain


\displaystyle x^{2}+2x=\left( x+\frac{2}{2} \right)^{2}-\left( \frac{2}{2} \right)^{2}=\left( x+1 \right)^{2}-1


and if we subtract the last " \displaystyle 1 " , we obtain


\displaystyle x^{2}+2x-1=\left( x+1 \right)^{2}-1-1=\left( x+1 \right)^{2}-2


To be completely certain that we have used the correct formula, we can expand the quadratic on the right-hand side,


\displaystyle \left( x+1 \right)^{2}-2=x^{2}+2x+1-2=x^{2}+2x-1


and see that the relation really holds.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:1b“