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第3題
\(\begin{align}
  & \alpha +\beta +\frac{\alpha +\beta }{2}=-a\quad ,\quad \alpha +\beta =-\frac{2}{3}a \\
& \frac{\alpha \beta \left( \alpha +\beta  \right)}{2}=-1\quad ,\quad \alpha \beta =\frac{-2}{\alpha +\beta }=\frac{3}{a} \\
& b=\alpha \beta +\left( \alpha +\beta  \right)\frac{\alpha +\beta }{2}=\frac{3}{a}+\frac{2{{a}^{2}}}{9}\in Z \\
&  \\
& a=-3,b=1 \\
& \alpha =1+\sqrt{2},\beta =1-\sqrt{2},\frac{\alpha +\beta }{2}=1 \\
\end{align}\)

\(\begin{align}
  & a=3,b=3 \\
& \alpha =-1,\beta =-1,\frac{\alpha +\beta }{2}=-1 \\
\end{align}\)(不合)

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