対称式と2数を解とする2次方程式

数研出版｜数学Ⅱ[709] p.54 練習18

\(2x^2-3x+5=0\) の解と係数の関係より、

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

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\({\small (1)}~\)2数 \(2\alpha-1~,~2\beta-1\) の和と積を求めると、

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

---

---

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

---

\(\begin{eqnarray}~~~x^2-x+8=0\end{eqnarray}\)

 

\({\small (2)}~\)2数 \(-\alpha~,~-\beta\) の和と積を求めると、

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

---

\(\begin{eqnarray}~~~(-\alpha)(-\beta)&=&\alpha\beta
\\[3pt]~~~&=&\displaystyle \frac{\,5\,}{\,2\,}\hspace{15pt}(\,∵~{\small [\,2\,]}\,)\end{eqnarray}\)

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

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\(\begin{eqnarray}~~~x^2-\left(-\displaystyle \frac{\,3\,}{\,2\,}\right)x+\displaystyle \frac{\,5\,}{\,2\,}&=&0
\\[5pt]~~~x^2+\displaystyle \frac{\,3\,}{\,2\,}x+\displaystyle \frac{\,5\,}{\,2\,}&=&0
\\[5pt]~~~2x^2+3x+5&=&0\end{eqnarray}\)

 

\({\small (3)}~\)2数 \(\alpha^2~,~\beta^2\) の和と積を求めると、

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\(\begin{eqnarray}~~~\alpha^2+\beta^2&=&(\alpha+\beta)^2-2\alpha\beta
\\[3pt]~~~&=&\left(\displaystyle \frac{\,3\,}{\,2\,}\right)^2-2\cdot\displaystyle \frac{\,5\,}{\,2\,}\hspace{15pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&\displaystyle \frac{\,9\,}{\,4\,}-5
\\[5pt]~~~&=&\displaystyle \frac{\,9\,}{\,4\,}-\displaystyle \frac{\,20\,}{\,4\,}
\\[5pt]~~~&=&-\displaystyle \frac{\,11\,}{\,4\,}\end{eqnarray}\)

---

\(\begin{eqnarray}~~~\alpha^2\beta^2&=&(\alpha\beta)^2
\\[3pt]~~~&=&\left(\displaystyle \frac{\,5\,}{\,2\,}\right)^2\hspace{15pt}(\,∵~{\small [\,2\,]}\,)
\\[5pt]~~~&=&\displaystyle \frac{\,25\,}{\,4\,}\end{eqnarray}\)

---

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

---

\(\begin{eqnarray}~~~x^2-\left(-\displaystyle \frac{\,11\,}{\,4\,}\right)x+\displaystyle \frac{\,25\,}{\,4\,}&=&0
\\[5pt]~~~x^2+\displaystyle \frac{\,11\,}{\,4\,}x+\displaystyle \frac{\,25\,}{\,4\,}&=&0
\\[5pt]~~~4x^2+11x+25&=&0\end{eqnarray}\)

数研出版｜数学Ⅱ[709] p.68 演習問題A 4
数研出版｜高等学校数学Ⅱ[710] p.65 章末問題B 5
数研出版｜新編数学Ⅱ[711] p.65 章末問題B 9

\(x^2+ax+b=0\) の2つの解を \(\alpha~,~\beta\) とすると、

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\(x^2+ax+b=0\) の解と係数の関係より、

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\(~~~\left\{~\begin{array}{l}\alpha+\beta=-a~~~\cdots{\small [\,1\,]}
\\[5pt]\alpha\beta=b~~~\hspace{18pt}\cdots{\small [\,2\,]}\end{array}\right.\)

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2つの解からそれぞれ1を引いた数は \(\alpha-1~,~\beta-1\) であり、

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これらを解にもつ2次方程式が \(x^2+bx+a=0\) であるから、

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\(x^2+bx+a=0\) の解と係数の関係より、

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\(~~~\left\{~\begin{array}{l}(\alpha-1)+(\beta-1)=-b~~~\cdots{\small [\,3\,]}
\\[5pt](\alpha-1)(\beta-1)=a~~~\hspace{18pt}\cdots{\small [\,4\,]}\end{array}\right.\)

---

\({\small [\,3\,]}\) より、

---

\(\begin{eqnarray}~~~(\alpha+\beta)-2&=&-b
\\[3pt]~~~-a-2&=&-b\hspace{15pt}(\,∵~{\small [\,1\,]}\,)
\\[3pt]~~~b&=&a+2~~~\cdots{\small [\,5\,]}\end{eqnarray}\)

---

\({\small [\,4\,]}\) より、

---

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

---

\({\small [\,5\,]}\)、\({\small [\,6\,]}\) より、

---

\(\begin{eqnarray}~~~-1&=&a+2
\\[3pt]~~~a&=&-3\end{eqnarray}\)

---

したがって、\(a=-3~,~b=-1\)

数研出版｜高等学校数学Ⅱ[710] p.52 練習19

\(x^2-3x-1=0\) の解と係数の関係より、

---

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

---

\({\small (1)}~\)2数 \(1-\alpha~,~1-\beta\) の和と積を求めると、

---

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

---

---

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

---

\(\begin{eqnarray}~~~x^2-(-1)x+(-3)&=&0
\\[3pt]~~~x^2+x-3&=&0\end{eqnarray}\)

 

\({\small (2)}~\)2数 \(\alpha^2~,~\beta^2\) の和と積を求めると、

---

\(\begin{eqnarray}~~~\alpha^2+\beta^2&=&(\alpha+\beta)^2-2\alpha\beta
\\[3pt]~~~&=&3^2-2\cdot(-1)\hspace{15pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&9+2
\\[3pt]~~~&=&11\end{eqnarray}\)

---

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

---

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

---

\(\begin{eqnarray}~~~x^2-11x+1=0\end{eqnarray}\)

数研出版｜高等学校数学Ⅱ[710] p.55 問題 5

\(x^2-7x-1=0\) の解と係数の関係より、

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\(~~~\left\{~\begin{array}{l}\alpha+\beta=-\displaystyle \frac{\,-7\,}{\,1\,}=7~~~\cdots{\small [\,1\,]}
\\[5pt]\alpha\beta=\displaystyle \frac{\,-1\,}{\,1\,}=-1~~~\cdots{\small [\,2\,]}\end{array}\right.\)

---

\({\small (1)}~\)2数 \(\alpha-2~,~\beta-2\) の和と積を求めると、

---

\(\begin{eqnarray}~~~(\alpha-2)+(\beta-2)&=&(\alpha+\beta)-4
\\[3pt]~~~&=&7-4\hspace{15pt}(\,∵~{\small [\,1\,]}\,)
\\[3pt]~~~&=&3\end{eqnarray}\)

---

---

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

---

\(\begin{eqnarray}~~~x^2-3x+(-11)&=&0
\\[3pt]~~~x^2-3x-11&=&0\end{eqnarray}\)

 

\({\small (2)}~\)2数 \(\displaystyle \frac{\,2\,}{\,\alpha\,}~,~\displaystyle \frac{\,2\,}{\,\beta\,}\) の和と積を求めると、

---

\(\begin{eqnarray}~~~\displaystyle \frac{\,2\,}{\,\alpha\,}+\displaystyle \frac{\,2\,}{\,\beta\,}&=&\displaystyle \frac{\,2\beta+2\alpha\,}{\,\alpha\beta\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2(\alpha+\beta)\,}{\,\alpha\beta\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2\cdot 7\,}{\,-1\,}\hspace{15pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&-14\end{eqnarray}\)

---

\(\begin{eqnarray}~~~\displaystyle \frac{\,2\,}{\,\alpha\,}\cdot\displaystyle \frac{\,2\,}{\,\beta\,}&=&\displaystyle \frac{\,4\,}{\,\alpha\beta\,}
\\[5pt]~~~&=&\displaystyle \frac{\,4\,}{\,-1\,}\hspace{15pt}(\,∵~{\small [\,2\,]}\,)
\\[5pt]~~~&=&-4\end{eqnarray}\)

---

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

---

\(\begin{eqnarray}~~~x^2-(-14)x+(-4)&=&0
\\[3pt]~~~x^2+14x-4&=&0\end{eqnarray}\)

 

\({\small (3)}~\)2数 \(\alpha+\beta~,~\alpha\beta\) の和と積を求めると、

---

\(\begin{eqnarray}~~~(\alpha+\beta)+\alpha\beta&=&7+(-1)\hspace{15pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&6\end{eqnarray}\)

---

\(\begin{eqnarray}~~~(\alpha+\beta)\cdot\alpha\beta&=&7\cdot(-1)\hspace{15pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&-7\end{eqnarray}\)

---

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

---

\(\begin{eqnarray}~~~x^2-6x+(-7)&=&0
\\[3pt]~~~x^2-6x-7&=&0\end{eqnarray}\)

数研出版｜新編数学Ⅱ[711] p.54 補充問題 3

\(x^2+2x+4=0\) の解と係数の関係より、

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\(~~~\left\{~\begin{array}{l}\alpha+\beta=-\displaystyle \frac{\,2\,}{\,1\,}=-2~~~\cdots{\small [\,1\,]}
\\[5pt]\alpha\beta=\displaystyle \frac{\,4\,}{\,1\,}=4~~~\hspace{20pt}\cdots{\small [\,2\,]}\end{array}\right.\)

---

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

---

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

---

---

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

---

\(\begin{eqnarray}~~~x^2-(-4)x+7&=&0
\\[3pt]~~~x^2+4x+7&=&0\end{eqnarray}\)

数研出版｜新編数学Ⅱ[711] p.64 章末問題A 5

\(x^2-7x-1=0\) の解と係数の関係より、

---

\(~~~\left\{~\begin{array}{l}\alpha+\beta=-\displaystyle \frac{\,-7\,}{\,1\,}=7~~~\cdots{\small [\,1\,]}
\\[5pt]\alpha\beta=\displaystyle \frac{\,-1\,}{\,1\,}=-1~~~\cdots{\small [\,2\,]}\end{array}\right.\)

---

\({\small (1)}~\) 2数 \(\displaystyle \frac{\,2\,}{\,\alpha\,}~,~\displaystyle \frac{\,2\,}{\,\beta\,}\) の和と積を求めると、

---

\(\begin{eqnarray}~~~\displaystyle \frac{\,2\,}{\,\alpha\,}+\displaystyle \frac{\,2\,}{\,\beta\,}&=&\displaystyle \frac{\,2\beta+2\alpha\,}{\,\alpha\beta\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2(\alpha+\beta)\,}{\,\alpha\beta\,}
\\[5pt]~~~&=&\displaystyle \frac{\,2\cdot 7\,}{\,-1\,}\hspace{15pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&-14\end{eqnarray}\)

---

\(\begin{eqnarray}~~~\displaystyle \frac{\,2\,}{\,\alpha\,}\cdot\displaystyle \frac{\,2\,}{\,\beta\,}&=&\displaystyle \frac{\,4\,}{\,\alpha\beta\,}
\\[5pt]~~~&=&\displaystyle \frac{\,4\,}{\,-1\,}\hspace{15pt}(\,∵~{\small [\,2\,]}\,)
\\[5pt]~~~&=&-4\end{eqnarray}\)

---

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

---

\(\begin{eqnarray}~~~x^2-(-14)x+(-4)&=&0
\\[3pt]~~~x^2+14x-4&=&0\end{eqnarray}\)

 

\({\small (2)}~\) 2数 \(\alpha^2~,~\beta^2\) の和と積を求めると、

---

\(\begin{eqnarray}~~~\alpha^2+\beta^2&=&(\alpha+\beta)^2-2\alpha\beta
\\[3pt]~~~&=&7^2-2\cdot(-1)\hspace{15pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[3pt]~~~&=&49+2
\\[3pt]~~~&=&51\end{eqnarray}\)

---

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

---

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

---

\(\begin{eqnarray}~~~x^2-51x+1=0\end{eqnarray}\)

東京書籍｜Advanced数学Ⅱ[701] p.32 問22

\(x^2+2x+3=0\) の解と係数の関係より、

---

\(~~~\left\{~\begin{array}{l}\alpha+\beta=-\displaystyle \frac{\,2\,}{\,1\,}=-2~~~\cdots{\small [\,1\,]}
\\[5pt]\alpha\beta=\displaystyle \frac{\,3\,}{\,1\,}=3~~~\hspace{20pt}\cdots{\small [\,2\,]}\end{array}\right.\)

---

\({\small (1)}~\) 2数 \(2\alpha-1~,~2\beta-1\) の和と積を求めると、

---

---

---

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

---

\(\begin{eqnarray}~~~x^2-(-6)x+17&=&0
\\[3pt]~~~x^2+6x+17&=&0\end{eqnarray}\)

 

\({\small (2)}~\) 2数 \(\displaystyle \frac{\,1\,}{\,\alpha\,}~,~\displaystyle \frac{\,1\,}{\,\beta\,}\) の和と積を求めると、

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\(\begin{eqnarray}~~~\displaystyle \frac{\,1\,}{\,\alpha\,}+\displaystyle \frac{\,1\,}{\,\beta\,}&=&\displaystyle \frac{\,\beta+\alpha\,}{\,\alpha\beta\,}
\\[5pt]~~~&=&\displaystyle \frac{\,\alpha+\beta\,}{\,\alpha\beta\,}
\\[5pt]~~~&=&\displaystyle \frac{\,-2\,}{\,3\,}\hspace{15pt}(\,∵~{\small [\,1\,]}~,~{\small [\,2\,]}\,)
\\[5pt]~~~&=&-\displaystyle \frac{\,2\,}{\,3\,}\end{eqnarray}\)

---

\(\begin{eqnarray}~~~\displaystyle \frac{\,1\,}{\,\alpha\,}\cdot\displaystyle \frac{\,1\,}{\,\beta\,}&=&\displaystyle \frac{\,1\,}{\,\alpha\beta\,}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,3\,}\hspace{15pt}(\,∵~{\small [\,2\,]}\,)\end{eqnarray}\)

---

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

---

\(\begin{eqnarray}~~~x^2-\left(-\displaystyle \frac{\,2\,}{\,3\,}\right)x+\displaystyle \frac{\,1\,}{\,3\,}&=&0
\\[5pt]~~~x^2+\displaystyle \frac{\,2\,}{\,3\,}x+\displaystyle \frac{\,1\,}{\,3\,}&=&0
\\[5pt]~~~3x^2+2x+1&=&0\end{eqnarray}\)

 

\({\small (3)}~\) 2数 \(\alpha^2~,~\beta^2\) の和と積を求めると、

---

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

---

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

---

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

---

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

東京書籍｜Advanced数学Ⅱ[701] p.35 問題 14(1)

\(x^2+mx+1=0\) の解と係数の関係より、

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\(~~~\left\{~\begin{array}{l}\alpha+\beta=-m~~~\cdots{\small [\,1\,]}
\\[5pt]\alpha\beta=1~~~\hspace{18pt}\cdots{\small [\,2\,]}\end{array}\right.\)

---

ここで、2数 \(\alpha-3~,~\beta-3\) の和と積を求めると、

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

---

---

よって、\(\alpha-3~,~\beta-3\) を解とする2次方程式の1つは、

---

\(\begin{eqnarray}~~~x^2-(-m-6)x+(3m+10)&=&0
\\[3pt]~~~x^2+(m+6)x+3m+10&=&0\end{eqnarray}\)

東京書籍｜Advanced数学Ⅱ[701] p.61 練習問題B 10

\(2x^2+3x+2=0\) の解と係数の関係より、

---

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

---

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

---

---

---

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

---

\(\begin{eqnarray}~~~x^2-(-3)x+4&=&0
\\[3pt]~~~x^2+3x+4&=&0\end{eqnarray}\)

東京書籍｜Standard数学Ⅱ[702] p.38 問22

\(2x^2-x-5=0\) の解と係数の関係より、

---

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

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

---

\(\begin{eqnarray}~~~(2\alpha+1)+(2\beta+1)&=&2(\alpha+\beta)+2
\\[3pt]~~~&=&2\cdot\displaystyle \frac{\,1\,}{\,2\,}+2\hspace{15pt}(\,∵~{\small [\,1\,]}\,)
\\[5pt]~~~&=&1+2
\\[3pt]~~~&=&3\end{eqnarray}\)

---

---

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

---

\(\begin{eqnarray}~~~x^2-3x+(-8)&=&0
\\[3pt]~~~x^2-3x-8&=&0\end{eqnarray}\)

 

\({\small (2)}~\) 2数 \(\alpha-1~,~\beta-1\) の和と積を求めると、

---

\(\begin{eqnarray}~~~(\alpha-1)+(\beta-1)&=&(\alpha+\beta)-2
\\[3pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}-2\hspace{15pt}(\,∵~{\small [\,1\,]}\,)
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}-\displaystyle \frac{\,4\,}{\,2\,}
\\[5pt]~~~&=&-\displaystyle \frac{\,3\,}{\,2\,}\end{eqnarray}\)

---

---

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

---

\(\begin{eqnarray}~~~x^2-\left(-\displaystyle \frac{\,3\,}{\,2\,}\right)x+(-2)&=&0
\\[5pt]~~~x^2+\displaystyle \frac{\,3\,}{\,2\,}x-2&=&0
\\[5pt]~~~2x^2+3x-4&=&0\end{eqnarray}\)

東京書籍｜Standard数学Ⅱ[702] p.39 Training 21

\(x^2-x+1=0\) の解と係数の関係より、

---

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

---

\({\small (1)}~\) 2数 \(2\alpha~,~2\beta\) の和と積を求めると、

---

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

---

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

---

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

---

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

 

\({\small (2)}~\) 2数 \(\alpha^2~,~\beta^2\) の和と積を求めると、

---

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

---

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

---

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

---

\(\begin{eqnarray}~~~x^2-(-1)x+1&=&0
\\[3pt]~~~x^2+x+1&=&0\end{eqnarray}\)

 

\({\small (3)}~\) 2数 \(\alpha^3~,~\beta^3\) の和と積を求めると、

---

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

---

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

---

よって、\(\alpha^3~,~\beta^3\) を解とする2次方程式の1つは、

---

\(\begin{eqnarray}~~~x^2-(-2)x+1&=&0
\\[3pt]~~~x^2+2x+1&=&0\end{eqnarray}\)