次の4次方程式を解け。
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【解答】
\newcommand\colNS[2]{\textcolor{#1}{#2}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}} \begin{align*} & \colFB{red}{2次式を探せ} \colMM{red}{x^2\ \Rightarrow\ x^4 指数が2倍!}\\ & \colMM{red}{x^2=A\ とおくと}\\ & \colMM{red}{A^2=(x^2)^2=x^4}\\ \\ & \begin{align*} x^4-x^2-2 &= 0\\ \colMM{orange}{(x^2)^2-x^2-2} & \colMM{orange}{=0}\\ \colMM{orange}{A^2-A-2} & \colMM{orange}{=0}\\ \colMM{orange}{(A-2)(A+1)} & \colMM{orange}{=0}\\ (x^2-2)(x^2+1) &= 0 \end{align*}\\ & よって\\ & x^2-2=0{\scriptsize または }x^2+1=0\\ \\ & \begin{align*} x^2-2 &= 0\\ x^2 &= 2\\ x &= \pm\sqrt{2} \end{align*}\\ \\ & \begin{align*} x^2+1 &= 0\\ x^2 &= -1\\ x &= \pm\sqrt{-1}\\ &= \pm\sqrt{1}\,i = \pm\,i \end{align*}\\ \\ & したがって\\ & x=\pm\sqrt{2},\ \pm\,i \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
【解答】
\newcommand\colNS[2]{\textcolor{#1}{#2}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}} \begin{align*} & \colFB{red}{2次式を探せ} \colMM{red}{x^2\ \Rightarrow\ x^4 指数が2倍!}\\ & \colMM{red}{x^2=A\ とおくと}\\ & \colMM{red}{A^2=(x^2)^2=x^4}\\ \\ & \begin{align*} x^4+x^2-12 &= 0\\ \colMM{orange}{(x^2)^2+x^2-12} & \colMM{orange}{=0}\\ \colMM{orange}{A^2+A-12} & \colMM{orange}{=0}\\ \colMM{orange}{(A-3)(A+4)} & \colMM{orange}{=0}\\ (x^2-3)(x^2+4) &= 0 \end{align*}\\ & よって\\ & x^2-3=0{\scriptsize または }x^2+4=0\\ \\ & \begin{align*} x^2-3 &= 0\\ x^2 &= 3\\ x &= \pm\sqrt{3} \end{align*}\\ \\ & \begin{align*} x^2+4 &= 0\\ x^2 &= -4\\ x &= \pm\sqrt{-4}\\ &= \pm\sqrt{4}\,i = \pm2\,i \end{align*}\\ \\ & したがって\\ & x=\pm\sqrt{3},\ \pm2\,i \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan
↓この問題へのリンクはこちら(右クリックで保存)
【解答】
\newcommand\colNS[2]{\textcolor{#1}{#2}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \newcommand\colFB[2]{\textcolor{#1}{\fbox{\scriptsize\bf\color{#1}#2}}} \begin{align*} & \colMM{magenta}{x^4-1 = x^4-0x^2-1\ と変形すれば・・・}\\ & \colFB{red}{2次式を探せ} \colMM{red}{x^2\ \Rightarrow\ x^4 指数が2倍!}\\ & \colMM{red}{x^2=A\ とおくと}\\ & \colMM{red}{A^2=(x^2)^2=x^4}\\ \\ & \begin{align*} x^4-1 &= 0\\ \colMM{orange}{(x^2)^2-1} & \colMM{orange}{=0}\\ \colMM{orange}{A^2-1} & \colMM{orange}{=0}\\ \colMM{orange}{(A+1)(A-1)} & \colMM{orange}{=0}\\ (x^2+1)(x^2-1) &= 0 \end{align*}\\ & よって\\ & x^2+1=0{\scriptsize または }x^2-1=0\\ \\ & \begin{align*} x^2+1 &= 0\\ x^2 &= -1\\ x &= \pm\sqrt{-1}\\ &= \pm\sqrt{1}\,i = \pm\,i \end{align*}\\ \\ & \begin{align*} x^2-1 &= 0\\ x^2 &= 1\\ x &= \pm\sqrt{1}\\ &= \pm1 \end{align*}\\ \\ & したがって\\ & x=\pm1,\ \pm\,i \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan