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Solution 2.3:1d

From Förberedande kurs i matematik 1

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m (Lösning 2.3:1d moved to Solution 2.3:1d: Robot: moved page)
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We apply the standard formula for completing the square,
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<center> [[Image:2_3_1d.gif]] </center>
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<math>x^{2}+ax=\left( x+\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}</math>
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on our expression and this gives
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<math>x^{2}+5x=\left( x+\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}</math>
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The whole expression becomes
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<math>\begin{align}
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& x^{2}+5x+3=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}+3=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{12}{4} \\
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& =\left( x+\frac{5}{2} \right)^{2}+\frac{12-25}{4}=\left( x+\frac{5}{2} \right)^{2}-\frac{13}{4} \\
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\end{align}</math>
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A quick check shows that we have calculated correctly.
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<math>\begin{align}
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& \left( x+\frac{5}{2} \right)^{2}-\frac{13}{4}=x^{2}+2\centerdot \frac{5}{2}\centerdot x+\left( \frac{5}{2} \right)^{2}-\frac{13}{4}=x^{2}+5x+\frac{25}{4}-\frac{13}{4} \\
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& =x^{2}+5x+\frac{12}{4}=x^{2}+5x+3 \\
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\end{align}</math>

Revision as of 13:02, 20 September 2008

We apply the standard formula for completing the square,


x2+ax=x+2a2−2a2 

on our expression and this gives


x2+5x=x+252−252=x+252−425 

The whole expression becomes


x2+5x+3=x+252−425+3=x+252−425+412=x+252+412−25=x+252−413


A quick check shows that we have calculated correctly.


x+252−413=x2+225x+252−413=x2+5x+425−413=x2+5x+412=x2+5x+3 

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