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3次方程式の解と係数の関係 で確認できます。
問題アーカイブ01
\({\small (1)}~\) \( \alpha^2+\beta^2+\gamma^2 \)
\({\small (2)}~\) \( (\alpha+1)(\beta+1)(\gamma+1) \)
数研出版｜数学Ⅱ[709] p.66 発展 練習1
3次方程式 \( x^3+3x^2+4=0 \) の解と係数の関係より、
\(~~~\left\{~\begin{array}{l}\alpha+\beta+\gamma=-\displaystyle \frac{\,3\,}{\,1\,}=-3~~~\hspace{3pt}\cdots{\small [\,1\,]}\\[5pt]\alpha\beta+\beta\gamma+\gamma\alpha=\displaystyle \frac{\,0\,}{\,1\,}=0~~~\cdots{\small [\,2\,]}\\[5pt]\alpha\beta\gamma=-\displaystyle \frac{\,4\,}{\,1\,}=-4~~~\hspace{10pt}\cdots{\small [\,3\,]}\end{array}\right.\)
\({\small (1)}~\)
\(\begin{eqnarray}~~~&&\alpha^2+\beta^2+\gamma^2
\\[3pt]~~~&=&(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)
\\[3pt]~~~&=&(-3)^2-2\cdot 0\hspace{15pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&9-0
\\[3pt]~~~&=&9\end{eqnarray}\)
\({\small (2)}~\)
\\[3pt]~~~&=&(\alpha\beta+\alpha+\beta+1)(\gamma+1)
\\[3pt]~~~&=&\alpha\beta\gamma+\alpha\beta+\alpha\gamma+\alpha+\beta\gamma+\beta+\gamma+1
\\[3pt]~~~&=&\alpha\beta\gamma+(\alpha\beta+\beta\gamma+\gamma\alpha)+(\alpha+\beta+\gamma)+1
\\[3pt]~~~&=&-4+0+(-3)+1\hspace{15pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}~,~{\small [\,3\,]}\,)
\\[3pt]~~~&=&-6\end{eqnarray}\)
問題アーカイブ02
\({\small (1)}~\) \( \alpha^2+\beta^2+\gamma^2 \)
\({\small (2)}~\) \( (\alpha+1)(\beta+1)(\gamma+1) \)
数研出版｜高等学校数学Ⅱ[710] p.63 発展 練習1
数研出版｜新編数学Ⅱ[711] p.62 発展 練習1
3次方程式 \( x^3-3x^2+x+2=0 \) の解と係数の関係より、
\(~~~\left\{~\begin{array}{l}\alpha+\beta+\gamma=-\displaystyle \frac{\,-3\,}{\,1\,}=3~~~\hspace{3pt}\cdots{\small [\,1\,]}\\[5pt]\alpha\beta+\beta\gamma+\gamma\alpha=\displaystyle \frac{\,1\,}{\,1\,}=1~~~\cdots{\small [\,2\,]}\\[5pt]\alpha\beta\gamma=-\displaystyle \frac{\,2\,}{\,1\,}=-2~~~\hspace{10pt}\cdots{\small [\,3\,]}\end{array}\right.\)
\({\small (1)}~\)
\(\begin{eqnarray}~~~&&\alpha^2+\beta^2+\gamma^2
\\[3pt]~~~&=&(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)
\\[3pt]~~~&=&3^2-2\cdot 1\hspace{15pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&9-2
\\[3pt]~~~&=&7\end{eqnarray}\)
\({\small (2)}~\)
\\[3pt]~~~&=&(\alpha\beta+\alpha+\beta+1)(\gamma+1)
\\[3pt]~~~&=&\alpha\beta\gamma+\alpha\beta+\alpha\gamma+\alpha+\beta\gamma+\beta+\gamma+1
\\[3pt]~~~&=&\alpha\beta\gamma+(\alpha\beta+\beta\gamma+\gamma\alpha)+(\alpha+\beta+\gamma)+1
\\[3pt]~~~&=&-2+1+3+1\hspace{15pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}~,~{\small [\,3\,]}\,)
\\[3pt]~~~&=&3\end{eqnarray}\)
問題アーカイブ03
\({\small (1)}~\) \( (\alpha-1)(\beta-1)(\gamma-1) \)
\({\small (2)}~\) \( \displaystyle \frac{\,1\,}{\,\alpha\beta\,}+\frac{\,1\,}{\,\beta\gamma\,}+\frac{\,1\,}{\,\gamma\alpha\,} \)
東京書籍｜Advanced数学Ⅱ[701] p.59 発展 問1
3次方程式 \( 2x^3-4x^2+x-6=0 \) の解と係数の関係より、
\(~~~\left\{~\begin{array}{l}\alpha+\beta+\gamma=-\displaystyle \frac{\,-4\,}{\,2\,}=2~~~\hspace{3pt}\cdots{\small [\,1\,]}\\[5pt]\alpha\beta+\beta\gamma+\gamma\alpha=\displaystyle \frac{\,1\,}{\,2\,}~~~\cdots{\small [\,2\,]}\\[5pt]\alpha\beta\gamma=-\displaystyle \frac{\,-6\,}{\,2\,}=3~~~\hspace{10pt}\cdots{\small [\,3\,]}\end{array}\right.\)
\({\small (1)}~\)
\\[3pt]~~~&=&(\alpha\beta-\alpha-\beta+1)(\gamma-1)
\\[3pt]~~~&=&\alpha\beta\gamma-\alpha\beta-\alpha\gamma+\alpha-\beta\gamma+\beta+\gamma-1
\\[3pt]~~~&=&\alpha\beta\gamma-(\alpha\beta+\beta\gamma+\gamma\alpha)+(\alpha+\beta+\gamma)-1
\\[3pt]~~~&=&3-\displaystyle \frac{\,1\,}{\,2\,}+2-1\hspace{15pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}~,~{\small [\,3\,]}\,)
\\[5pt]~~~&=&\displaystyle \frac{\,7\,}{\,2\,}\end{eqnarray}\)
\({\small (2)}~\)
\(\begin{eqnarray}~~~&&\frac{\,1\,}{\,\alpha\beta\,}+\frac{\,1\,}{\,\beta\gamma\,}+\frac{\,1\,}{\,\gamma\alpha\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\gamma+\alpha+\beta\,}{\,\alpha\beta\gamma\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\alpha+\beta+\gamma\,}{\,\alpha\beta\gamma\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2\,}{\,3\,}\hspace{15pt}(\,∵~{\small [\,1\,]}~,~{\small [\,3\,]}\,)\end{eqnarray}\)
