34012中３・２次方程式・計算問題・平方根の考え方で解く2
banno計算問題 》平方根の考え方で解く②
次の 2 次方程式を解きなさい。
(1) $(\,x-3\, )^{2}=16$
(2) $(\,x+2\,)^{2}=7$
(3) $(\,x-8\,)^{2}-25=0$
(4) $3(\,x+5\,)^{2}-24=0$
(5) $2\Bigl(\,x+\dfrac{2}{3}\,\Bigr)^{2}-\dfrac{2}{\;9\;}=0$
解答・解説
(1) $\;\begin{eqnarray}(\,x-3\,)^{2}=16\end{eqnarray}$
$\begin{eqnarray}(\,x-3\,)^{2}&=&16\\[3pt]x-3&=&\pm4\\[3pt]x&=&3\pm4\end{eqnarray}$
よって，$\;x=-1,\;x=7$
答$x=-1,\;x=7$
(2) $\;\begin{eqnarray}(\,x+2\,)^{2}=7\end{eqnarray}$
$\begin{eqnarray}(\,x+2\,)^{2}&=&7\\[3pt]x+2&=&\pm\sqrt{7\,}\\[3pt]x&=&-2\pm\sqrt{7\,}\end{eqnarray}$
答$x=-2\pm\sqrt{7\,}$
(3) $\;\begin{eqnarray}(\,x-8\,)^{2}-25=0\end{eqnarray}$
$\begin{eqnarray}(\,x-8\,)^{2}-25&=&0\\[3pt](\,x-8\,)^{2}&=&25\\[3pt]x-8&=&\pm5\\[3pt]x&=&8\pm5\end{eqnarray}$
よって，$\;x=3,\;x=13$
答$x=3,\;x=13$
(4) $\;\begin{eqnarray}3(\,x+5\,)^{2}-24=0\end{eqnarray}$
$\begin{eqnarray}3(\,x+5\,)^{2}-24&=&0\\[3pt]3(\,x+5\,)^{2}&=&24\\[3pt](\,x+5\,)^{2}&=&8\\[3pt]x+5&=&\pm\sqrt{8\,}\\[3pt]x&=&-5\pm2\sqrt{2\,}\end{eqnarray}$
答$x=-5\pm2\sqrt{2\,}$
(5) $\;\begin{eqnarray}2\Bigl(\,x+\frac{2}{\;3\;}\,\Bigr)^{2}-\frac{2}{\;9\;}=0\end{eqnarray}$
$\begin{eqnarray}2\Bigl(\,x+\frac{2}{\;3\;}\,\Bigr)^{2}-\frac{2}{\;9\;}&=&0\\[3pt]2\Bigl(\,x+\frac{2}{\;3\;}\,\Bigr)^{2}&=&\frac{2}{\;9\;}\\[3pt]\Bigl(\,x+\frac{2}{\;3\;}\,\Bigr)^{2}&=&\frac{1}{\;9\;}\\[3pt]x+\frac{2}{\;3\;}&=&\pm\frac{1}{\;3\;}\\[3pt]x&=&-\frac{2}{\;3\;}\pm\frac{1}{3}\end{eqnarray}$
よって，$\;x=-1,\;x=-\dfrac{1}{\;3\;}$
答$x=-1,\;x=-\dfrac{1}{\;3\;}$
