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# 解の公式の求め方

## ■$3a{x}^{2}+bx+c=0\text{\hspace{5}\hspace{5}} \( a\neq 0 \)$  の解の公式の導出

ただし，$3\displaystyle{{b}^{2}- 4ac\geq 0}$ とする．

$3\displaystyle{\begin{array}{lll}a{x}^{2}+bx+c=0& \vspace{6}\\ a \{ {x}^{2}+\frac{\hspace{2}b\hspace{2}}{\hspace{2}a\hspace{2}}x+{ \( \frac{\hspace{2}b\hspace{2}}{\hspace{2} 2a \hspace{2}} \) }^{2} \} - a{ \( \frac{\hspace{2}b\hspace{2}}{\hspace{2} 2a \hspace{2}} \) }^{2}+c=0& \vspace{6}\\ a \{ {x}^{2}+\frac{\hspace{2}b\hspace{2}}{\hspace{2}a\hspace{2}}x+{ \( \frac{\hspace{2}b\hspace{2}}{\hspace{2} 2a \hspace{2}} \) }^{2} \} =a{ \( \frac{\hspace{2}b\hspace{2}}{\hspace{2} 2a \hspace{2}} \) }^{2}- c& \vspace{6}\\ \{ {x}^{2}+\frac{\hspace{2}b\hspace{2}}{\hspace{2}a\hspace{2}}x+{ \( \frac{\hspace{2}b\hspace{2}}{\hspace{2} 2a \hspace{2}} \) }^{2} \} ={ \( \frac{\hspace{2}b\hspace{2}}{\hspace{2} 2a \hspace{2}} \) }^{2}- \frac{\hspace{2}c\hspace{2}}{\hspace{2}a\hspace{2}}& \vspace{6}\\ { \( x+\frac{\hspace{2}b\hspace{2}}{\hspace{2} 2a \hspace{2}} \) }^{2}=\frac{\hspace{2} {b}^{2}- 4ac \hspace{2}}{\hspace{2} 4{a}^{2} \hspace{2}}& \vspace{6}\\ x+\frac{\hspace{2}b\hspace{2}}{\hspace{2} 2a \hspace{2}}=\pm \sqrt{ \frac{\hspace{2} {b}^{2}- 4ac \hspace{2}}{\hspace{2} 4{a}^{2} \hspace{2}} }& \vspace{6}\\ x=- \frac{\hspace{2}b\hspace{2}}{\hspace{2} 2a \hspace{2}}\pm \frac{\hspace{2} \sqrt{ {b}^{2}- 4ac } \hspace{2}}{\hspace{2} 2a \hspace{2}}& \vspace{6}\\ x=\frac{\hspace{2} - b\pm \sqrt{ {b}^{2}- 4ac } \hspace{2}}{\hspace{2} 2a \hspace{2}}& \vspace{6}\\ \end{array}}$

## ■$3a{x}^{2}+2bx+c=0\text{\hspace{5}\hspace{5}} \( a\neq 0 \)$  の解の公式の導出

ただし，$3\displaystyle{{b}^{2}- ac\geq 0}$ とする．

$3\displaystyle{\begin{array}{lll}a{x}^{2}+2bx+c=0& \vspace{6}\\ a \{ {x}^{2}+\frac{\hspace{2} 2b \hspace{2}}{\hspace{2}a\hspace{2}}x+{ \( \frac{\hspace{2}b\hspace{2}}{\hspace{2}a\hspace{2}} \) }^{2} \} - a{ \( \frac{\hspace{2}b\hspace{2}}{\hspace{2}a\hspace{2}} \) }^{2}+c=0& \vspace{6}\\ a \{ {x}^{2}+\frac{\hspace{2} 2b \hspace{2}}{\hspace{2}a\hspace{2}}x+{ \( \frac{\hspace{2}b\hspace{2}}{\hspace{2}a\hspace{2}} \) }^{2} \} =a{ \( \frac{\hspace{2}b\hspace{2}}{\hspace{2}a\hspace{2}} \) }^{2}- c& \vspace{6}\\ \{ {x}^{2}+\frac{\hspace{2} 2b \hspace{2}}{\hspace{2}a\hspace{2}}x+{ \( \frac{\hspace{2}b\hspace{2}}{\hspace{2}a\hspace{2}} \) }^{2} \} ={ \( \frac{\hspace{2}b\hspace{2}}{\hspace{2}a\hspace{2}} \) }^{2}- \frac{\hspace{2}c\hspace{2}}{\hspace{2}a\hspace{2}}& \vspace{6}\\ { \( x+\frac{\hspace{2}b\hspace{2}}{\hspace{2}a\hspace{2}} \) }^{2}=\frac{\hspace{2} {b}^{2}- ac \hspace{2}}{\hspace{2} {a}^{2} \hspace{2}}& \vspace{6}\\ x+\frac{\hspace{2}b\hspace{2}}{\hspace{2}a\hspace{2}}=\pm \sqrt{ \frac{\hspace{2} {b}^{2}- ac \hspace{2}}{\hspace{2} {a}^{2} \hspace{2}} }& \vspace{6}\\ x=- \frac{\hspace{2}b\hspace{2}}{\hspace{2}a\hspace{2}}\pm \frac{\hspace{2} \sqrt{ {b}^{2}- ac } \hspace{2}}{\hspace{2}a\hspace{2}}& \vspace{6}\\ x=\frac{\hspace{2} - b\pm \sqrt{ {b}^{2}- ac } \hspace{2}}{\hspace{2}a\hspace{2}}& \vspace{6}\\ \end{array}}$

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