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CIE 9231 2025 June Paper 13 Q3

A Level / CIE / FP1

CIE 9231 2025 June Paper 13 Paper · Question 3

题目

Problem

The quartic equation x4+7x2+3x+22=0x^4+7x^2+3x+22=0 has roots α\alpha, β\beta, γ\gamma, δ\delta.

(a) Find the value of α2+β2+γ2+δ2\alpha^2+\beta^2+\gamma^2+\delta^2.

(2)

(b) Find the value of α4+β4+γ4+δ4\alpha^4+\beta^4+\gamma^4+\delta^4.

(2)

(c) Use standard results from the list of formulae (MF19) to find the value of

∑r=110((α2+r)2+(β2+r)2+(γ2+r)2+(δ2+r)2).\sum_{r=1}^{10}\big((\alpha^2+r)^2+(\beta^2+r)^2+(\gamma^2+r)^2+(\delta^2+r)^2\big).
(5)
题目中文翻译

四次方程 x4+7x2+3x+22=0x^4+7x^2+3x+22=0 的根为 α\alpha、β\beta、γ\gamma、δ\delta。

(a) 求 α2+β2+γ2+δ2\alpha^2+\beta^2+\gamma^2+\delta^2 的值。

(2)

(b) 求 α4+β4+γ4+δ4\alpha^4+\beta^4+\gamma^4+\delta^4 的值。

(2)

(c) 利用公式表（MF19）中的标准结果，求

∑r=110((α2+r)2+(β2+r)2+(γ2+r)2+(δ2+r)2)\sum_{r=1}^{10}\big((\alpha^2+r)^2+(\beta^2+r)^2+(\gamma^2+r)^2+(\delta^2+r)^2\big)

的值。

(5)
原卷通用作答要求

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。

解答

(a)

解法一

思路

展开

根和为零，两两乘积和为7，由此计算平方和。

答题过程

展开

By Vieta’s formulae, the sum of the four roots is 00 and their pairwise-product sum is 77. Hence

α2+β2+γ2+δ2=02−2(7)=−14.\alpha^2+\beta^2+\gamma^2+\delta^2 =0^2-2(7)=\boxed{-14}.

(b)

解法一

思路

展开

将各根满足的四次方程直接相加，常数22要累加四次。

答题过程

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Write Sj=αj+βj+γj+δjS_j=\alpha^j+\beta^j+\gamma^j+\delta^j. Summing the root equations gives

S4+7S2+3S1+88=0.S_4+7S_2+3S_1+88=0.

Using S2=−14S_2=-14 and S1=0S_1=0,

S4=−7(−14)−88=10.S_4=-7(-14)-88=\boxed{10}.

(c)

解法一

思路

展开

先对每一项展开平方，四个根的幂和用(a)(b)替换，再对r求和。

答题过程

展开

For each rr,

∑z∈{α,β,γ,δ}(z2+r)2= S4+2rS2+4r2= 10−28r+4r2.\begin{align*} \sum_{z\in\{\alpha,\beta,\gamma,\delta\}}(z^2+r)^2 =&\,S_4+2rS_2+4r^2 \\[4mm] =&\,10-28r+4r^2. \end{align*}

Therefore, using the standard sums,

∑r=110(10−28r+4r2)= 100−2810(11)2 +410(11)(21)6= 100−1540+1540=100.\begin{align*} \sum_{r=1}^{10}(10-28r+4r^2) =&\,100-28\frac{10(11)}2\\[4mm] &\,+4\frac{10(11)(21)}6 \\[4mm] =&\,100-1540+1540=\boxed{100}. \end{align*}