方程 $x^4-x^2+2x+5=0$ 的根为 $\alpha$、$\beta$、$\gamma$、$\delta$。

(a) 求一个以 $\alpha^2$、$\beta^2$、$\gamma^2$、$\delta^2$ 为根的四次方程，并写出 $\alpha^2+\beta^2+\gamma^2+\delta^2$ 的值。

(b) 求 $\dfrac1{\alpha^2}+\dfrac1{\beta^2}+\dfrac1{\gamma^2}+\dfrac1{\delta^2}$ 的值。

(c) 求 $\alpha^4+\beta^4+\gamma^4+\delta^4$ 的值。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

令y=x平方，把原式中唯一的奇次项隔离后平方，消去x。这样不需要为复数根选择平方根符号。

Put $y=x^2$ for a root $x$ of the original quartic. Then

Squaring gives

Hence a required quartic is

Its roots are $\alpha^2,\beta^2,\gamma^2,\delta^2$, whose sum is

四个新根的倒数和，等于新根的三次对称和除以四根积。

From the quartic in (a), the sum of the four triple products of its roots is $14$ and their product is $25\ne0$. Therefore

四次和等于新根的平方和，由新根一次和及二次对称和得到。

The roots of the new quartic have sum $2$ and pairwise product sum $11$. Hence

也可直接把四个原根满足的方程相加，利用根和为零以及(a)的平方和。

The original quartic has root sum $\alpha+\beta+\gamma+\delta=0$. Summing $x^4=x^2-2x-5$ over its four roots gives