Lösung 2.3:2c
Aus Online Mathematik Brückenkurs 1
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Version vom 14:07, 22. Okt. 2008
We start by completing the square of the left-hand side,
| \displaystyle \begin{align}
y^{2}+3y+4 &= \Bigl(y+\frac{3}{2}\Bigr)^{2} - \Bigl(\frac{3}{2}\Bigr)^{2}+4\\[5pt] &= \Bigl(y+\frac{3}{2}\Bigr)^{2} - \frac{9}{4} + \frac{16}{4}\\[5pt] &= \Bigl(y+\frac{3}{2}\Bigr)^{2} + \frac{7}{4}\,\textrm{.} \end{align} |
The equation is then
| \displaystyle \left( y+\frac{3}{2} \right)^{2}+\frac{7}{4}=0\,\textrm{.} |
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.
