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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複二次式の定義と因数分解
\[
a_{4}x^{4}+a_{2}x^{2}+a_{0}=\frac{1}{4a_{4}}\left(2a_{4}x^{2}+a_{2}+\sqrt{a_{2}^{\;2}-4a_{4}a_{0}}\right)\left(2a_{4}x^{2}+a_{2}-\sqrt{a_{2}^{\;2}-4a_{4}a_{0}}\right)
\]
交代式の因数分解
\[
\text{交代式}=\text{差積}\times\text{対称式}
\]
オイラーの4平方恒等式
\[
\left(a_{0}^{\;2}+a_{1}^{\;2}+a_{2}^{\;2}+a_{3}^{\;2}\right)\left(b_{0}^{\;2}+b_{1}^{\;2}+b_{2}^{\;2}+b_{3}^{\;2}\right)=\left(a_{0}b_{0}-a_{1}b_{1}-a_{2}b_{2}-a_{3}b_{3}\right)^{2}+\left(a_{0}b_{1}+a_{1}b_{0}+a_{2}b_{3}-a_{3}b_{2}\right)^{2}+\left(a_{0}b_{2}-a_{1}b_{3}+a_{2}b_{0}+a_{3}b_{1}\right)^{2}+\left(a_{0}b_{3}+a_{1}b_{2}-a_{2}b_{1}+a_{3}b_{0}\right)^{2}
\]
ブラーマグプタ2平方恒等式
\[
\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=\left(ac\pm bd\right)^{2}+\left(ad\mp bc\right)^{2}
\]