Lösung 2.3:6c

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If we complete the square of the expression, we have that
If we complete the square of the expression, we have that
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{{Displayed math||<math>\begin{align}
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x^{2} - 5x + 7 &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} + 7\\[5pt]
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&= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{28}{4}\\[5pt]
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&= \Bigl(x-\frac{5}{2}\Bigr)^{2} + \frac{3}{4}
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\end{align}</math>}}
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<math>\begin{align}
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and because <math>\bigl(x-\tfrac{5}{2}\bigr)^{2}</math> is a quadratic, this term assumes a minimal value zero when <math>x=5/2\,</math>. This shows that the polynomial's smallest value is <math>\tfrac{3}{4}</math>.
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& x^{2}-5x+7=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+7 \\
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& =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{28}{4}=\left( x-\frac{5}{2} \right)^{2}+\frac{3}{4} \\
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\end{align}</math>
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and because
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<math>\left( x-\frac{5}{2} \right)^{2}</math>
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is a quadratic, this term is at least equal to zero when
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<math>x={5}/{2}\;</math>. This shows that the polynomial's smallest value is
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<math>\frac{3}{4}</math>.
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Version vom 11:45, 29. Sep. 2008

If we complete the square of the expression, we have that

Vorlage:Displayed math

and because \displaystyle \bigl(x-\tfrac{5}{2}\bigr)^{2} is a quadratic, this term assumes a minimal value zero when \displaystyle x=5/2\,. This shows that the polynomial's smallest value is \displaystyle \tfrac{3}{4}.

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