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CIE 9231 2026 June Paper 14 Q2

A Level / CIE / FP1

CIE 9231 2026 June Paper 14 Paper · Question 2

题目

Problem

The equation 2x4+3x3+x+1=02x^4+3x^3+x+1=0 has roots α\alpha, β\beta, γ\gamma, δ\delta.

(a) Find a quartic equation whose roots are 1α3\dfrac{1}{\alpha^3}, 1β3\dfrac{1}{\beta^3}, 1γ3\dfrac{1}{\gamma^3}, 1δ3\dfrac{1}{\delta^3} and state the value of 1α3+1β3+1γ3+1δ3\dfrac{1}{\alpha^3}+\dfrac{1}{\beta^3}+\dfrac{1}{\gamma^3}+\dfrac{1}{\delta^3}.

(5)

(b) Find the value of 1α6+1β6+1γ6+1δ6\dfrac{1}{\alpha^6}+\dfrac{1}{\beta^6}+\dfrac{1}{\gamma^6}+\dfrac{1}{\delta^6}.

(2)

题目中文翻译

方程 2x4+3x3+x+1=02x^4+3x^3+x+1=0 的根为 α\alpha、β\beta、γ\gamma、δ\delta。

(a) 求一个以 1α3\dfrac{1}{\alpha^3}、1β3\dfrac{1}{\beta^3}、1γ3\dfrac{1}{\gamma^3}、1δ3\dfrac{1}{\delta^3} 为根的四次方程，并写出 1α3+1β3+1γ3+1δ3\dfrac{1}{\alpha^3}+\dfrac{1}{\beta^3}+\dfrac{1}{\gamma^3}+\dfrac{1}{\delta^3} 的值。

(5)

(b) 求 1α6+1β6+1γ6+1δ6\dfrac{1}{\alpha^6}+\dfrac{1}{\beta^6}+\dfrac{1}{\gamma^6}+\dfrac{1}{\delta^6} 的值。

(2)

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

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先取倒数根，再把其立方记为y；利用一次关系再立方来消去中间变量，避免复数立方根分支问题。

答题过程

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None of the original roots is zero. Put z=1/xz=1/x. Dividing the original equation by x4x^4 gives

z4+z3+3z+2=0.z^4+z^3+3z+2=0.

Set y=z3=1/x3y=z^3=1/x^3. Then z(y+3)=−(y+2)z(y+3)=-(y+2). Cubing this relation gives

y(y+3)3=−(y+2)3.y(y+3)^3=-(y+2)^3.

Expanding and collecting terms,

0= y(y3+9y2+27y+27) +y3+6y2+12y+8= y4+10y3+33y2+39y+8.\begin{align*} 0=&\,y(y^3+9y^2+27y+27)\\[4mm] &\,+y^3+6y^2+12y+8 \\[4mm] =&\,y^4+10y^3+33y^2+39y+8. \end{align*}

The required quartic is therefore

y4+10y3+33y2+39y+8=0.\boxed{y^4+10y^3+33y^2+39y+8=0}.

By Vieta’s formula, the requested sum is −10\boxed{-10}.

解法二

思路

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也可先把原方程分成仅含x立方的一边，以及x乘含x立方因子的一边；立方后所有幂次都能用新根表示。

答题过程

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Rearrange the original equation as

3x3+1=−x(2x3+1).3x^3+1=-x(2x^3+1).

Cubing both sides gives

27x9+27x6+9x3+1=−8x12−12x9−6x6−x3.27x^9+27x^6+9x^3+1 =-8x^{12}-12x^9-6x^6-x^3.

Thus

8x12+39x9+33x6+10x3+1=0.8x^{12}+39x^9+33x^6+10x^3+1=0.

Divide by x12x^{12} and put y=1/x3y=1/x^3 to obtain

y4+10y3+33y2+39y+8=0.\boxed{y^4+10y^3+33y^2+39y+8=0}.

Its root sum is −10\boxed{-10}, so

α−3+β−3+γ−3+δ−3=−10.\boxed{\alpha^{-3}+\beta^{-3}+\gamma^{-3}+\delta^{-3}=-10}.

(b)

解法一

思路

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所求是新四次方程根的平方和，由根和的平方扣掉两倍两两乘积和。

答题过程

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Let the four roots of the quartic in (a) be p,q,r,sp,q,r,s. Then

p+q+r+s= −10,pq+pr+ps+qr+qs+rs= 33.\begin{align*} p+q+r+s=&\,-10,\\[4mm] pq+pr+ps+qr+qs+rs=&\,33. \end{align*}

Therefore

α−6+β−6+γ−6+δ−6= p2+q2+r2+s2= (−10)2−2(33)=34.\begin{align*} \alpha^{-6}+\beta^{-6}+\gamma^{-6}+\delta^{-6} =&\,p^2+q^2+r^2+s^2 \\[4mm] =&\,(-10)^2-2(33)=\boxed{34}. \end{align*}