34043中３・２次方程式・計算問題・総合3

(1) $\;\begin{eqnarray}\dfrac{\;x^{2}-4x\;}{12}=\dfrac{\;1\;}{\;2\;}(\,x+2\,)\;\end{eqnarray}$

$\begin{eqnarray}\dfrac{\;x^{2}-4x\;}{12}&=&\dfrac{\;1\;}{\;2\;}(\,x+2\,)\\[6pt]\Bigl(\,\dfrac{\;x^{2}-4x\;}{12}\Bigr)\color{red}\times 12&=&\dfrac{\;1\;}{\;2\;}(\,x+2\,)\color{red}\times 12\\[6pt]x^{2}-4x&=&6(\,x+2\,)\\[6pt]x^{2}-4x&=&6x+12\\[6pt]x^{2}-10x-12&=&0\\[6pt]x&=&\frac{\;10\pm\sqrt{\,10^{2}-4 \times 1 \times (\,-12\,)\,}\;}{2}\\[6pt]x&=&\frac{\;10\pm\sqrt{148\,}\;}{2}\\[6pt]x&=&\frac{\;10\pm2\sqrt{37\,}\;}{2}\\[6pt]x&=&5\pm\sqrt{37\,}\end{eqnarray}$

答$x=5\pm\sqrt{37\,}$

(2) $\;\begin{eqnarray}\dfrac{\;1\;}{\;3\;}(\,2x-3\,)^{2}=x^{2}+3\;\end{eqnarray}$

《 解法１》

$\begin{eqnarray}\dfrac{\;1\;}{\;3\;}(\,2x-3\,)^{2}&=&x^{2}+3\\[6pt]\dfrac{\;1\;}{\;3\;}(\,2x-3\,)^{2}\color{red} \times 3&=&(\,x^{2}+3\,)\color{red} \times 3\\[6pt](\,2x-3\,)^{2}&=&3x^{2}+9\\[6pt]4x^{2}-12x+9&=&3x^{2}+9\\[6pt]x^{2}-12x&=&0\\[6pt]x(\,x-12\,)&=&0\end{eqnarray}$

$x=0\;$ または $\;x-12=0$

よって，$x=0,\;x=12$

答$x=0,\;x=12$

《 解法２》

$\begin{eqnarray}\dfrac{\;1\;}{\;3\;}(\,2x-3\,)^{2}&=&x^{2}+3\\[6pt]\dfrac{\;1\;}{\;3\;}(\,2x-3\,)^{2}\color{red} \times 3&=&(\,x^{2}+3\,)\color{red} \times 3\\[6pt](\,2x-3\,)^{2}&=&3x^{2}+9\\[6pt]4x^{2}-12x+9&=&3x^{2}+9\\[6pt]x^{2}-12x&=&0\\[6pt]x&=&\frac{\;12\pm\sqrt{\,12^{2}-4 \times 1 \times 0\,}\;}{2}\\[6pt]x&=&\frac{\;12\pm\sqrt{144\,}\;}{2}\\[6pt]x&=&\frac{\;12\pm 12\;}{2}\\[6pt]x&=&6\pm 6\end{eqnarray}$

$x=0,\;x=12$

答$x=0,\;x=12$

(3) $\;\begin{eqnarray}\dfrac{\;(\,x-2\,)^{2}\;}{\;2\;}=\dfrac{\;(\,x+1\,)(\,x+3\,)\;}{\;3\;}-5\;\end{eqnarray}$

《 解法１》

$\begin{eqnarray}\dfrac{\;(\,x-2\,)^{2}\;}{\;2\;}&=&\dfrac{\;(\,x+1\,)(\,x+3\,)\;}{\;3\;}-5\\[6pt]\dfrac{\;(\,x-2\,)^{2}\;}{\;2\;}\color{red} \times 6&=&\Bigl\{\,\dfrac{\;(\,x+1\,)(\,x+3\,)\;}{\;3\;}-5\,\Bigr\} \color{red} \times 6 \\[6pt]3(\,x-2\,)^{2}&=&2(\,x+1\,)(\,x+3\,)-30 \\[6pt]3(\,x^{2}-4x+4\,)^{2}&=&2(\,x^{2}+4x+3\,)-30\\[6pt]3x^{2}-12x+12&=&2x^{2}+8x-24\\[6pt]x^{2}-20x+36&=&0\\[6pt](x-2)(x-18)&=&0\end{eqnarray}$

$x-2=0\;$ または $\;x-18=0$

よって，$x=2,\;x=18$

答$x=2,\;x=18$

《 解法２》

$\begin{eqnarray}\dfrac{\;(\,x-2\,)^{2}\;}{\;2\;}&=&\dfrac{\;(\,x+1\,)(\,x+3\,)\;}{\;3\;}-5\\[6pt]\dfrac{\;(\,x-2\,)^{2}\;}{\;2\;}\color{red} \times 6&=&\Bigl\{\,\dfrac{\;(\,x+1\,)(\,x+3\,)\;}{\;3\;}-5\,\Bigr\} \color{red} \times 6 \\[6pt]3(\,x-2\,)^{2}&=&2(\,x+1\,)(\,x+3\,)-30 \\[6pt]3(\,x^{2}-4x+4\,)^{2}&=&2(\,x^{2}+4x+3\,)-30\\[6pt]3x^{2}-12x+12&=&2x^{2}+8x-24\\[6pt]x^{2}-20x+36&=&0\\[6pt]x&=&\frac{\;20\pm\sqrt{\,20^{2}-4 \times 1 \times 36\,}\;}{2}\\[6pt]x&=&\frac{\;20\pm\sqrt{256\,}\;}{2}\\[6pt]x&=&\frac{\;20\pm 16\;}{2}\\[6pt]x&=&10\pm\ 8\end{eqnarray}$

$x=2,\;x=18$

答$x=2,\;x=18$