Lösung 2.3:2c

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We start by completing the square of the left-hand side:
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We start by completing the square of the left-hand side,
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<math>\begin{align}
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& y^{2}+3y+4=\left( y+\frac{3}{2} \right)^{2}-\left( \frac{3}{2} \right)^{2}+4 \\
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& =\left( y+\frac{3}{2} \right)^{2}-\frac{9}{4}+\frac{16}{4}=\left( y+\frac{3}{2} \right)^{2}+\frac{7}{4} \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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y^{2}+3y+4 &= \Bigl(y+\frac{3}{2}\Bigr)^{2} - \Bigl(\frac{3}{2}\Bigr)^{2}+4\\[5pt]
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&= \Bigl(y+\frac{3}{2}\Bigr)^{2} - \frac{9}{4} + \frac{16}{4}\\[5pt]
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&= \Bigl(y+\frac{3}{2}\Bigr)^{2} + \frac{7}{4}\,\textrm{.}
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\end{align}</math>}}
The equation is then
The equation is then
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{{Displayed math||<math>\left( y+\frac{3}{2} \right)^{2}+\frac{7}{4}=0\,\textrm{.}</math>}}
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<math>\left( y+\frac{3}{2} \right)^{2}+\frac{7}{4}=0</math>
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The first term <math>\bigl(y+\tfrac{3}{2}\bigr)^{2}</math> is always greater than or equal to zero because it is a square and <math>\tfrac{7}{4}</math> is a positive number. This means that the left hand side cannot be zero, regardless of how ''y'' is chosen. The equation has no solution.
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The first term
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<math>\left( y+\frac{3}{2} \right)^{2}</math>
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is always greater than or equal to zero because it is a square and
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<math>\frac{7}{4}</math>
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is a positive number. This means that the left hand side cannot be zero, regardless of how
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<math>y</math>
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is chosen. The equation has no solution.
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Version vom 07:38, 29. Sep. 2008

We start by completing the square of the left-hand side,

Vorlage:Displayed math

The equation is then

Vorlage:Displayed math

The first term \displaystyle \bigl(y+\tfrac{3}{2}\bigr)^{2} is always greater than or equal to zero because it is a square and \displaystyle \tfrac{7}{4} is a positive number. This means that the left hand side cannot be zero, regardless of how y is chosen. The equation has no solution.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.3:2c“