四次方程系数满足一个条件则可化为二次方程

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Hall and Knight - Higher Algebra page 489:

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代入$r^2=p^2s$, 按$x^4,x^3,x$项系数配方\begin{aligned} x^{4}+p x^{3}+q x^{2}+r x+s & =x^{4}+p x^{3}+q x^{2}+r x+\frac{r^{2}}{p^{2}} \\ & =\left(x^{2}+\frac{p}{2} x+\frac{r}{p}\right)^{2}-\left(\frac{p^{2}}{4}+\frac{2 r}{p}-q\right) x^{2}\end{aligned}方程变为$$\left(x^{2}+\frac{p}{2} x+\frac{r}{p}\right)^{2}=\left(\frac{p^{2}}{4}+\frac{2 r}{p}-q\right) x^{2}$$设$a=\sqrt{\frac{p^{2}}{4}+\frac{2 r}{p}-q}$, 开方得二次方程$$x^{2}+\frac{p}{2} x+\frac{r}{p}=\pm a x$$

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逆命题是否成立即:
四次方程$x^{4}+p x^{3}+q x^{2}+r x+s=0$可化为二次方程, 是否有$r^2=p^2s$