対称式と2数を解とする2次方程式

\(x^2-2x+3=0\) の解と係数の関係より、

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\(~~~\left\{~\begin{array}{l}\alpha+\beta=-\displaystyle \frac{\,-2\,}{\,1\,}=2~~~\cdots{\small [\,1\,]}
\\[5pt]\alpha\beta=\displaystyle \frac{\,3\,}{\,1\,}=3~~~\hspace{28pt}\cdots{\small [\,2\,]}\end{array}\right.\)

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ここで、2数 \(\alpha+1~,~\beta+1\) の和と積を求めると、

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\(\begin{eqnarray}~~~\alpha+1+\beta+1&=&(\alpha+\beta)+2
\\[3pt]~~~&=&2+2\hspace{15pt}(\,∵~{\small [\,1\,]}\,)
\\[3pt]~~~&=&4\end{eqnarray}\)

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\(\begin{eqnarray}~~~(\alpha+1)(\beta+1)&=&\alpha\beta+\alpha+\beta+1
\\[3pt]~~~&=&3+2+1\hspace{15pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&6\end{eqnarray}\)

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よって、\(\alpha+1~,~\beta+1\) を解とする2次方程式の1つは、

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\(\begin{eqnarray}~~~x^2-4x+6=0\end{eqnarray}\)

 

2数 \(\alpha^2~,~\beta^2\) の和と積を求めると、

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\(\begin{eqnarray}~~~\alpha^2+\beta^2&=&(\alpha+\beta)^2-2\alpha\beta
\\[3pt]~~~&=&2^2-2\cdot 3\hspace{15pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&4-6
\\[3pt]~~~&=&-2\end{eqnarray}\)

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\(\begin{eqnarray}~~~\alpha^2\beta^2&=&(\alpha\beta)^2
\\[3pt]~~~&=&3^2\hspace{15pt}(\,∵~{\small [\,2\,]}\,)
\\[3pt]~~~&=&9\end{eqnarray}\)

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よって、\(\alpha^2~,~\beta^2\) を解とする2次方程式の1つは、

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\(\begin{eqnarray}~~~x^2-(-2)x+9&=&0
\\[3pt]~~~x^2+2x+9&=&0\end{eqnarray}\)