三次方程 $x^3+bx^2+d=0$ 的根为 $\alpha$、$\beta$、$\gamma$，其中 $\alpha=\beta$ 且 $d\ne0$。

(a) 证明 $4b^3+27d=0$。

(b) 已知 $2\alpha^2+\gamma^2=3b$，求 $b$ 和 $d$ 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。

两个重根代入韦达定理，二次对称和为零。利用d非零排除零根后，根之间成固定比例，代回三根积。

Since $\beta=\alpha$, Vieta’s formulae give

As $d\ne0$, both $\alpha$ and $\gamma$ are nonzero. Hence $\alpha(\alpha+2\gamma)=0$ implies $\alpha=-2\gamma$.

Substituting into the root sum gives $-3\gamma=-b$, so

Using the product,

Therefore $\boxed{4b^3+27d=0}$.

将上一问两根的表达式代入平方和条件，得到b的二次式；零解与d非零冲突。

Using the expressions in (a),

Thus $b^2=3b$, giving $b=0$ or $b=3$. The case $b=0$ would give $d=0$ by (a), which is excluded. Hence