Lösung 2.3:2c

Aus Online Mathematik Brückenkurs 1

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We start by completing the square of the left-hand side:
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<center> [[Image:2_3_2c.gif]] </center>
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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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The equation is then
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<math>\left( y+\frac{3}{2} \right)^{2}+\frac{7}{4}=0</math>
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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.

Version vom 13:37, 20. Sep. 2008

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


\displaystyle \begin{align} & y^{2}+3y+4=\left( y+\frac{3}{2} \right)^{2}-\left( \frac{3}{2} \right)^{2}+4 \\ & =\left( y+\frac{3}{2} \right)^{2}-\frac{9}{4}+\frac{16}{4}=\left( y+\frac{3}{2} \right)^{2}+\frac{7}{4} \\ \end{align}


The equation is then


\displaystyle \left( y+\frac{3}{2} \right)^{2}+\frac{7}{4}=0

The first term \displaystyle \left( y+\frac{3}{2} \right)^{2} is always greater than or equal to zero because it is a square and \displaystyle \frac{7}{4} is a positive number. This means that the left hand side cannot be zero, regardless of how \displaystyle 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“