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FP1 2021 Jan Q4

A Level / Edexcel / FP1

IAL 2021 Jan Paper · Question 4

题目

Problem

4. The equation 2x2+5x+7=02x^2 + 5x + 7 = 0 has roots α\alpha and β\beta

Without solving the equation

(a) determine the exact value of α3+β3\alpha^3 + \beta^3

(3)

(b) form a quadratic equation, with integer coefficients, which has roots

α2βandβ2α\frac{\alpha^2}{\beta} \quad \text{and} \quad \frac{\beta^2}{\alpha}
(5)
题目中文翻译
  1. 方程 2x2+5x+7=02x^2 + 5x + 7 = 0 的两根为 α\alpha 和 β\beta。

不需求解该方程，

(a) 确定 α3+β3\alpha^3 + \beta^3 的精确值。

(3)

(b) 构造一个具有整数系数的二次方程，使其两根为

α2β和β2α\frac{\alpha^2}{\beta} \quad \text{和} \quad \frac{\beta^2}{\alpha}
(5)

解答

(a)

解法一

思路

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不直接求根，而是由韦达定理取得 α+β\alpha+\beta 与 αβ\alpha\beta。再利用恒等式

 α3+β3= (α+β)3 −3αβ(α+β)\begin{align*} &\,\alpha^3+\beta^3\\ =&\,(\alpha+\beta)^3\\ &\,\hspace{2pt}-3\alpha\beta(\alpha+\beta) \end{align*}

求值。

答题过程

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By Vieta’s formulae,

α+β= −52,αβ= 72.\begin{align*} \alpha+\beta=&\,-\frac52,\\ \alpha\beta=&\,\frac72. \end{align*}

Hence,

 α3+β3= (α+β)3 −3αβ(α+β)= (−52)3 −3(72) ×(−52)= −1258+2108= 858.\begin{align*} &\,\alpha^3+\beta^3\\ =&\,(\alpha+\beta)^3\\ &\,\hspace{2pt}-3\alpha\beta(\alpha+\beta)\\ =&\,\bigg(-\frac52\bigg)^3\\ &\,\hspace{2pt}-3\bigg(\frac72\bigg) \\ &\,\hspace{4pt}\times\bigg(-\frac52\bigg)\\ =&\,-\frac{125}{8}+\frac{210}{8}\\ =&\,\boxed{\frac{85}{8}}. \end{align*}

(b)

解法一

思路

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设新方程的两根为 u=α2/βu=\alpha^2/\beta 与 v=β2/αv=\beta^2/\alpha。分别用 (a) 的结果和韦达定理求 u+vu+v、uvuv，再代入根和与根积所确定的二次方程

 x2−(u+v)x +uv=0,\begin{align*} &\,x^2-(u+v)x\\ &\,\hspace{2pt}+uv=0, \end{align*}

最后清除分母。

答题过程

展开

Let

u= α2β,v= β2α.\begin{align*} u=&\,\frac{\alpha^2}{\beta},\\ v=&\,\frac{\beta^2}{\alpha}. \end{align*}

Using part (a), the sum of the new roots is

u+v= α2β +β2α= α3αβ +β3αβ= α3+β3αβ= 85872= 858 ×27= 17056= 8528.\begin{align*} u+v =&\,\frac{\alpha^2}{\beta}\\ &\,\hspace{2pt}+\frac{\beta^2}{\alpha}\\ =&\,\frac{\alpha^3}{\alpha\beta}\\ &\,\hspace{2pt}+\frac{\beta^3}{\alpha\beta}\\ =&\,\frac{\alpha^3+\beta^3}{\alpha\beta}\\ =&\,\frac{\frac{85}{8}}{\frac72}\\ =&\,\frac{85}{8}\\ &\,\hspace{2pt}\times\frac27\\ =&\,\frac{170}{56}\\ =&\,\frac{85}{28}. \end{align*}

Their product is

uv= α2β2αβ= αβ= 72.\begin{align*} uv =&\,\frac{\alpha^2\beta^2}{\alpha\beta}\\ =&\,\alpha\beta\\ =&\,\frac72. \end{align*}

Thus the required equation is initially

x2−8528x+72=0.x^2-\frac{85}{28}x+\frac72=0.

Multiplying by 2828 gives the equation with integer coefficients,

28x2−85x+98=0.\boxed{28x^2-85x+98=0}.