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求p,q的关系,使x3 + px + q的三个根成等腰直角三角形.
解:令α,β,γ为三个根,令γ-α=(β-α)i,从$\left\{ \begin{matrix}
\alpha + \beta + \gamma = 0 \\
\alpha\beta + \beta\gamma + \gamma\alpha = p \\
\alpha\beta\gamma = - q \\
\gamma - \alpha = (\beta - \alpha)i \\
\end{matrix} \right.\ $消去α,β,γ得27q2 = 50p3.具体地讲,消元的过程如下:
用α= − β − γ消去α,得$\left\{ \begin{matrix}
\beta\gamma - (\beta + \gamma)^{2} = p \\
(\beta + \gamma)\beta\gamma = q \\
\beta + 2\gamma = (2\beta + \gamma)i \\
\end{matrix} \right.\ $.(β+2γ)2 + (2β+γ)2 = 0⇔5β2 + 5γ2 + 8βγ = 0.
令u=β+γ,v=βγ,得uv=q,v − u2 = p, 5u2 − 2v = 0,则$q = uv = \frac{5u^{3}}{2},p = \frac{5u^{2}}{2} - u^{2} = \frac{3u^{2}}{2}$⇒$\left( \frac{2p}{3} \right)^{3} = u^{6} = \left( \frac{2q}{5} \right)^{2}$⇒$\frac{8p^{3}}{27} = \frac{4q^{2}}{25}$⇒50p3 = 27q2.
