4次方程式の標準形
4次方程式の標準形
4次方程式\(a_{4}\ne0\)
\[ a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}=0 \] は以下のように変形できる。
\[ X^{4}+pX^{2}+qX+r=0 \] ただし、
\[ \begin{cases} X=x+\frac{a_{3}}{4a_{4}}\\ p=-\frac{3a_{3}^{\;2}}{8a_{4}^{\;2}}+\frac{a_{2}}{a_{4}}\\ q=\frac{a_{3}^{\;3}}{8a_{4}^{\;3}}-\frac{a_{3}a_{2}}{2a_{4}^{\;2}}+\frac{a_{1}}{a_{4}}\\ r=-\frac{3a_{3}^{\;4}}{256a_{4}^{\;4}}+\frac{a_{3}^{\;2}a_{2}}{16a_{4}^{\;3}}-\frac{a_{3}a_{1}}{4a_{4}^{\;2}}+\frac{a_{0}}{a_{4}} \end{cases} \]
4次方程式\(a_{4}\ne0\)
\[ a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}=0 \] は以下のように変形できる。
\[ X^{4}+pX^{2}+qX+r=0 \] ただし、
\[ \begin{cases} X=x+\frac{a_{3}}{4a_{4}}\\ p=-\frac{3a_{3}^{\;2}}{8a_{4}^{\;2}}+\frac{a_{2}}{a_{4}}\\ q=\frac{a_{3}^{\;3}}{8a_{4}^{\;3}}-\frac{a_{3}a_{2}}{2a_{4}^{\;2}}+\frac{a_{1}}{a_{4}}\\ r=-\frac{3a_{3}^{\;4}}{256a_{4}^{\;4}}+\frac{a_{3}^{\;2}a_{2}}{16a_{4}^{\;3}}-\frac{a_{3}a_{1}}{4a_{4}^{\;2}}+\frac{a_{0}}{a_{4}} \end{cases} \]
\[
X=x+\frac{a_{3}}{4a_{4}}
\]
とおくと、
\begin{align*} a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0} & =a_{4}\left\{ \left(X-\frac{a_{3}}{4a_{4}}\right)^{4}+\frac{a_{3}}{a_{4}}\left(X-\frac{a_{3}}{4a_{4}}\right)^{3}+\frac{a_{2}}{a_{4}}\left(X-\frac{a_{3}}{4a_{4}}\right)^{2}+\frac{a_{1}}{a_{4}}\left(X-\frac{a_{3}}{4a_{4}}\right)+\frac{a_{0}}{a_{4}}\right\} \\ & =a_{4}\left\{ X^{4}+\left(-\frac{3a_{3}^{\;2}}{8a_{4}^{\;2}}+\frac{a_{2}}{a_{4}}\right)X^{2}+\left(\frac{a_{3}^{\;3}}{8a_{4}^{\;3}}-\frac{a_{3}a_{2}}{2a_{4}^{\;2}}+\frac{a_{1}}{a_{4}}\right)X-\frac{3a_{3}^{\;4}}{256a_{4}^{\;4}}+\frac{a_{3}^{\;2}a_{2}}{16a_{4}^{\;3}}-\frac{a_{3}a_{1}}{4a_{4}^{\;2}}+\frac{a_{0}}{a_{4}}\right\} \end{align*} となるので、
\[ \begin{cases} p=-\frac{3a_{3}^{\;2}}{8a_{4}^{\;2}}+\frac{a_{2}}{a_{4}}\\ q=\frac{a_{3}^{\;3}}{8a_{4}^{\;3}}-\frac{a_{3}a_{2}}{2a_{4}^{\;2}}+\frac{a_{1}}{a_{4}}\\ r=-\frac{3a_{3}^{\;4}}{256a_{4}^{\;4}}+\frac{a_{3}^{\;2}a_{2}}{16a_{4}^{\;3}}-\frac{a_{3}a_{1}}{4a_{4}^{\;2}}+\frac{a_{0}}{a_{4}} \end{cases} \] とおくと、
\begin{align*} 0 & =a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}\\ & =a_{4}\left\{ X^{4}+pX^{2}+qX+r\right\} \end{align*} 両辺を\(a_{4}\ne0\)で割って、
\[ X^{4}+pX^{2}+qX+r=0 \] となる。
\begin{align*} a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0} & =a_{4}\left\{ \left(X-\frac{a_{3}}{4a_{4}}\right)^{4}+\frac{a_{3}}{a_{4}}\left(X-\frac{a_{3}}{4a_{4}}\right)^{3}+\frac{a_{2}}{a_{4}}\left(X-\frac{a_{3}}{4a_{4}}\right)^{2}+\frac{a_{1}}{a_{4}}\left(X-\frac{a_{3}}{4a_{4}}\right)+\frac{a_{0}}{a_{4}}\right\} \\ & =a_{4}\left\{ X^{4}+\left(-\frac{3a_{3}^{\;2}}{8a_{4}^{\;2}}+\frac{a_{2}}{a_{4}}\right)X^{2}+\left(\frac{a_{3}^{\;3}}{8a_{4}^{\;3}}-\frac{a_{3}a_{2}}{2a_{4}^{\;2}}+\frac{a_{1}}{a_{4}}\right)X-\frac{3a_{3}^{\;4}}{256a_{4}^{\;4}}+\frac{a_{3}^{\;2}a_{2}}{16a_{4}^{\;3}}-\frac{a_{3}a_{1}}{4a_{4}^{\;2}}+\frac{a_{0}}{a_{4}}\right\} \end{align*} となるので、
\[ \begin{cases} p=-\frac{3a_{3}^{\;2}}{8a_{4}^{\;2}}+\frac{a_{2}}{a_{4}}\\ q=\frac{a_{3}^{\;3}}{8a_{4}^{\;3}}-\frac{a_{3}a_{2}}{2a_{4}^{\;2}}+\frac{a_{1}}{a_{4}}\\ r=-\frac{3a_{3}^{\;4}}{256a_{4}^{\;4}}+\frac{a_{3}^{\;2}a_{2}}{16a_{4}^{\;3}}-\frac{a_{3}a_{1}}{4a_{4}^{\;2}}+\frac{a_{0}}{a_{4}} \end{cases} \] とおくと、
\begin{align*} 0 & =a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}\\ & =a_{4}\left\{ X^{4}+pX^{2}+qX+r\right\} \end{align*} 両辺を\(a_{4}\ne0\)で割って、
\[ X^{4}+pX^{2}+qX+r=0 \] となる。
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3次式の実数の範囲で因数分解
\[
a^{3}\pm b^{3}=\left(a\pm b\right)\left(a^{2}\mp ab+b^{2}\right)
\]
因数分解による3次方程式の標準形の解
\[
x_{k}=\omega^{k}\sqrt[3]{-\frac{q}{2}+\sqrt{\left(\frac{q}{2}\right)^{2}+\left(\frac{p}{3}\right)^{3}}}-\omega^{3-k}\frac{p}{3}\frac{1}{\sqrt[3]{-\frac{q}{2}-\sqrt{\left(\frac{q}{2}\right)^{2}+\left(\frac{p}{3}\right)^{3}}}}\cnd{k\in\left\{ 0,1,2\right\} }
\]
相反方程式の定義と解法
\[
\sum_{k=0}^{n}a_{k}x^{k}=0
\]
3次方程式の標準形
\[
X^{3}+pX+q=0
\]