4次方程式標準形の解き方

\begin{align*} y^{4}+py^{2}+qy+r & =\left(y^{2}+u\right)^{2}-2uy^{2}-u^{2}+py^{2}+qy+r\\ & =\left(y^{2}+u\right)^{2}-\left(2u-p\right)y^{2}+qy+r-u^{2}\\ & =\left(y^{2}+u\right)^{2}-\left(2u-p\right)\left(y+\frac{q}{2\left(2u-p\right)}\right)^{2}-\frac{q^{2}}{4\left(2u-p\right)}+r-u^{2} \end{align*} ここで右辺第3項以降が0になるように\(u\)を選ぶ、
\begin{align*} 0 & =-\frac{q^{2}}{4\left(2u-p\right)}+r-u^{2}\\ & =\frac{1}{4\left(2u-p\right)}\left(q^{2}+4\left(2u-p\right)\left(r-u^{2}\right)\right)\\ & =\frac{1}{4\left(2u-p\right)}\left(q^{2}-8u^{3}+4pu^{2}+8ru-4pr\right)\\ & =\frac{-1}{4\left(2u-p\right)}\left(8u^{3}-4pu^{2}-8ru+4pr-q^{2}\right) \end{align*} \(u\ne\frac{p}{2}\)の場合を考える。\(u=\frac{p}{2}\)のときは\(q=0\)なので複2次式として解ける。
これより、\(u\)についての3次方程式を解き、\(u\)の解を1つ選べば、
\begin{align*} y^{4}+py^{2}+qy+r & =\left(y^{2}+u\right)^{2}-\left(2u-p\right)\left(y+\frac{q}{2\left(2u-p\right)}\right)^{2}\\ & =\left\{ \left(y^{2}+u\right)+\sqrt{2u-p}\left(y+\frac{q}{2\left(2u-p\right)}\right)\right\} \left\{ \left(y^{2}+u\right)-\sqrt{2u-p}\left(y+\frac{q}{2\left(2u-p\right)}\right)\right\} \\ & =\left\{ y^{2}+\sqrt{2u-p}y+u+\frac{q}{2\sqrt{2u-p}}\right\} \left\{ y^{2}-\sqrt{2u-p}y+u+\frac{q}{2\sqrt{2u-p}}\right\} \\ & =\left\{ y^{2}+\sqrt{2u-p}y+u+\frac{q}{2\sqrt{2u-p}}\right\} \left\{ y^{2}-\sqrt{2u-p}y+u+\frac{q}{2\sqrt{2u-p}}\right\} \end{align*} これより、解は
\[ y^{2}\pm_{1}\sqrt{2u-p}y+u+\frac{q}{2\sqrt{2u-p}}=0 \] の解になる。すなわち、
\begin{align*} y & =\frac{\mp_{1}\sqrt{2u-p}\pm_{2}\sqrt{2u-p-4\left(u+\frac{q}{2\sqrt{2u-p}}\right)}}{2}\\ & =\frac{\mp_{1}\sqrt{2u-p}\pm_{2}\sqrt{-p-2u-\frac{4q}{2\sqrt{2u-p}}}}{2} \end{align*} となる。