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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オイラーの4平方恒等式
\[
\left(a_{0}^{\;2}+a_{1}^{\;2}+a_{2}^{\;2}+a_{3}^{\;2}\right)\left(b_{0}^{\;2}+b_{1}^{\;2}+b_{2}^{\;2}+b_{3}^{\;2}\right)=\left(a_{0}b_{0}-a_{1}b_{1}-a_{2}b_{2}-a_{3}b_{3}\right)^{2}+\left(a_{0}b_{1}+a_{1}b_{0}+a_{2}b_{3}-a_{3}b_{2}\right)^{2}+\left(a_{0}b_{2}-a_{1}b_{3}+a_{2}b_{0}+a_{3}b_{1}\right)^{2}+\left(a_{0}b_{3}+a_{1}b_{2}-a_{2}b_{1}+a_{3}b_{0}\right)^{2}
\]
因数分解による3次方程式の標準形の解
\[
x_{k}=\omega^{k}\sqrt[3]{-\frac{q}{2}+\sqrt{\left(\frac{q}{2}\right)^{2}+\left(\frac{p}{3}\right)^{3}}}-\omega^{3-k}\frac{p}{3}\frac{1}{\sqrt[3]{-\frac{q}{2}-\sqrt{\left(\frac{q}{2}\right)^{2}+\left(\frac{p}{3}\right)^{3}}}}\cnd{k\in\left\{ 0,1,2\right\} }
\]
ソフィー・ジェルマンの恒等式
\[
a^{4}+4b^{4}=\left(a^{2}+2ab+2b^{2}\right)\left(a^{2}-2ab+2b^{2}\right)
\]
4次方程式標準形の解き方
\[
y=\frac{\mp_{1}\sqrt{2u-p}\pm_{2}\sqrt{-p-2u-\frac{4q}{2\sqrt{2u-p}}}}{2}
\]