34011中３・２次方程式・計算問題・平方根の考え方で解く１

(1) $\;\begin{eqnarray}x^{2}=9\end{eqnarray}$

$\begin{eqnarray}x^{2}&=&9\\[3pt]x&=&\pm3\end{eqnarray}$

答$x=\pm3$

(2) $\;\begin{eqnarray}2x^{2}-14=0\end{eqnarray}$

$\begin{eqnarray}2x^{2}-14&=&0\\[3pt]2x^{2}&=&14\\[3pt]x^{2}&=&7\\[3pt]x&=&\pm\sqrt{7\,}\end{eqnarray}$

答$x=\pm\sqrt{7\,}$

(3) $\;\begin{eqnarray}3x^{2}-5=0\end{eqnarray}$

$\begin{eqnarray}3x^{2}-5&=&0\\[3pt]3x^{2}&=&5\\[3pt]x^{2}&=&\frac{5}{\;3\;}\\[3pt]x&=&\pm\sqrt{\frac{5}{\;3\;}\,}\\[3pt]&=&\pm\frac{\;\sqrt{15\,}\;}{3}\end{eqnarray}$

答$\displaystyle x=\pm\frac{\;\sqrt{15\,}\;}{3}$

(4) $\;\begin{eqnarray}12-x^{2}=0\end{eqnarray}$

$\begin{eqnarray}12-x^{2}&=&0\\[3pt]-x^{2}&=&-12\\[3pt]x^{2}&=&12\\[3pt]x&=&\pm\sqrt{12\,}\\[3pt]&=&\pm2\sqrt{3\,}\end{eqnarray}$

答$x=\pm2\sqrt{3\,}$

(5) $\;\begin{eqnarray}4x^{2}-49=0\end{eqnarray}$

$\begin{eqnarray}4x^{2}-49&=&0\\[3pt]4x^{2}&=&49\\[3pt]x^{2}&=&\frac{\;49\;}{4}\\[3pt]x&=&\pm\frac{7}{\;2\;}\end{eqnarray}$

答$x=\pm\dfrac{7}{\;2\;}$